A 10-kg block on a rough horizontal surface is attached to a light spring (force constant = 1400 N/m). The block is pulled 10.0 cm to the right from its equilibrium position and released from rest. The frictional force between the block and surface has a magnitude of 30 N. What is the kinetic energy of the block as it passes through its equilibrium position?

Answers

Answer 1

To develop this problem it is necessary to apply the concepts related to the conservation of Energy. In this case the definition concerning kinetic energy from the simple harmonic movement.

From the conservation of energy we know that the kinetic energy would be conserved through the work done by the frictional force and the simple harmonic potential energy, in other words:

[tex]KE = PE_s +W_f[/tex]

[tex]KE = \frac{1}{2}kA^2 + W_f[/tex]

Where,

[tex]KE =[/tex] Kinetic Energy

[tex]PE_s =[/tex]Potential Harmonic Simple Energy

[tex]W_f =[/tex] Work made by friction.

Our values are given as,

[tex]m = 10Kg  \rightarrow[/tex] mass

[tex]k = 1400N/m \rightarrow[/tex] Spring constant

[tex]A = 0.1m \rightarrow[/tex]Amplitude

[tex]f_f = 30N \rightarrow[/tex] Frictional Force

Replacing we have,

[tex]KE = \frac{1}{2}kA^2 + W_f[/tex]

[tex]KE =\frac{1}{2} 1400 * 0.1^2 + ( - 30 * 0.1)[/tex]

[tex]KE = 4 J[/tex]

Therefore the Kinetic Energy of the block as it passes through its equlibrium position is 4J.

Answer 2
Final answer:

The kinetic energy of the 10-kg block as it passes through its equilibrium position is 4 Joules. This is calculated by converting the potential energy stored in the spring to kinetic energy and then subtracted the energy lost due to friction.

Explanation:

The first step in solving this problem is to understand the two forces acting on the block in this question: the spring force and the frictional force. The spring potential energy when the block is pulled 10 cm to the right is given by the formula U = 1/2kx^2, where k is the force constant and x is the displacement. Substituting the given values, we have U = 1/2(1400 N/m)(0.1 m)^2 = 7 Joules. This is the initial potential energy stored in the spring. As the block passes through its equilibrium position, this potential energy is fully converted to kinetic energy.

We also need to take into account the work done against frictional force which is equal to the frictional force times the displacement, i.e., W_friction = Friction * displacement = 30N * 0.1m = 3 Joules. This is the energy lost due to friction.

Finally, we subtract the work done by the frictional force from the potential energy to achieve the kinetic energy. Therefore, the kinetic energy of the block as it passes through its equilibrium position is K = U - W_friction = 7J - 3J = 4 Joules.

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Related Questions

If two electrons in the same atom have the same four quantum numbers, then they must have the same energy.a. trueb. falsec. They cannot both have the same four quantum numbers.

Answers

Final answer:

The statement that two electrons in the same atom with the same four quantum numbers must have the same energy is false.

Explanation:

The statement that two electrons in the same atom with the same four quantum numbers must have the same energy is false. The Pauli exclusion principle states that no two electrons in an atom can have the same set of four quantum numbers. The four quantum numbers are the principal quantum number (n), the angular momentum quantum number (l), the magnetic quantum number (m), and the spin quantum number (s).

Since the quantum numbers determine the energy and other properties of the electrons, if two electrons in the same atom have the same set of quantum numbers, they must have different spins in order to obey the Pauli exclusion principle. Therefore, they would have different energies.

Lasers can be constructed that produce an extremely high intensity electromagnetic wave for a brief time—called pulsed lasers. They are used to ignite nuclear fusion, for example. Such a laser may produce an electromagnetic wave with a maximum electric field strength of 1.52\times 10^{11}~\text{V/m}1.52×10 ​11 ​​ V/m for a time of 1.00 ns. What energy does it deliver on a 1.00~\mathrm{mm^2}1.00 mm ​2 ​​ area?

Answers

Answer:

30643 J

Explanation:

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi \times 10^{-7}\ H/m[/tex]

t = Time taken = 1 ns

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

[tex]E_0[/tex] = Maximum electric field strength = [tex]1.52\times 10^{11}\ V/m[/tex]

A = Area = [tex]1\ mm^2[/tex]

Magnitude of magnetic field is given by

[tex]B_0=\dfrac{E_0}{c}\\\Rightarrow B_0=\dfrac{1.52\times 10^{11}}{3\times 10^8}\\\Rightarrow B_0=506.67\ T[/tex]

Intensity is given by

[tex]I=\dfrac{cB_0^2}{2\mu_0}\\\Rightarrow I=\dfrac{3\times 10^8\times 506.67^2}{2\times 4\pi \times 10^{-7}}\\\Rightarrow I=3.0643\times 10^{19}\ W/m^2[/tex]

Power, intensity and time have the relation

[tex]E=IAt\\\Rightarrow E=3.0643\times 10^{19}\times 1\times 10^{-6}\times 1\times 10^{-9}\\\Rightarrow E=30643\ J[/tex]

The energy it delivers is 30643 J

A flight attendant pulls her 70.0 N flight bag a distance of 253 m along a level airport floor at a constant velocity. The force she exerts is 40.0 N at an angle of 52 degrees above the horizontal. Find the following: a) the work she does on the flight bag b) the work done on the flight bag c) the coefficient of kinetic friction between the flight bag and the floor.

Answers

Answer:

6230.49413 J

0 J

0.63998

Explanation:

F = Force = 40 N

[tex]\theta[/tex] = Angle = 52°

Work done is given by

[tex]W=Fscos50\\\Rightarrow W=40\times 253\times cos52\\\Rightarrow W=6230.49413\ J[/tex]

The work she does on the flight bag is 6230.49413 J

The work done on the flight bag will be the opposite of the work done by the flight attendant

[tex]W=-6230.49413\ J[/tex]

So net work will be

[tex]W_n=6230.49413-6230.49413\\\Rightarrow W_n=0[/tex]

The net work done on the flight bag is 0 J

Coefficient of friction is given by

[tex]\mu=\dfrac{F_f}{F_N}\\\Rightarrow \mu=\dfrac{F_f}{F_g-F_{app}}\\\Rightarrow \mu=\dfrac{40cos52}{70-40sin52}\\\Rightarrow \mu=0.63998[/tex]

The coefficient of friction is 0.63998

(a)  The work done by the attendant on the flight bag is 6230.49 J.

(b)  The work on the flight bag is 0 J .

(c)  The coefficient of kinetic friction between the flight bag and the floor is 0.35.

Given data:

The weight of flight bag is, W = 70.0 N.

The distance covered by the bag is, d = 253 m.

The magnitude of force exerted on bag is, F = 40.0 N.

The angle of inclination with horizontal is, [tex]\theta =52^{\circ}[/tex].

Work done defined as the product of force and distance covered due to applied force.

(a)

The work done on the flight bag is given as,

[tex]W'=F \times dcos\theta[/tex]

Solving as,

[tex]W'=40 \times 253cos52\\W'=6230.49 \;\rm J[/tex]

Thus, the work done by the attendant on the flight bag is 6230.49 J.

(b)

The work done on the flight bag will be the opposite of the work done by the flight attendant. So,

W'' = - W

W'' = - 6230.49 J

Then net work done on the flight bag is,

[tex]W_{net}=W'+W''\\W_{net}=6230.49 - 6230.49\\W_{net}=0[/tex]

Thus, the net work on the flight bag is 0 J .

(c)

The expression for the frictional force is given as,

[tex]f = \mu \times W[/tex]

[tex]\mu[/tex] is the coefficient of kinetic friction. And the frictional force is due to the horizontal component of applied force. Then,

[tex]f=Fcos\theta[/tex]

So,

[tex]Fcos\theta= \mu \times W\\\\40 \times cos52=\mu \times 70\\\\\mu=\dfrac{40 \times cos52}{70} \\\\\mu = 0.35[/tex]

Thus, the coefficient of kinetic friction between the flight bag and the floor is 0.35.

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I place a 500-g ice cube (initially at 0°C) in a Styrofoam box with wall thickness 1.0 cm and total surface area 600 cm2 .

If the air surrounding the box is at 20°C and after 4 hours the ice is completely melted, what is the conductivity of the Styrofoam material? (Lf = 80 cal/g)
a. 9.6 × 10−5cal/s⋅cm⋅°C
b. 2.8 × 10−6cal/s⋅cm⋅°C
c. 1.15 × 10−2cal/s⋅cm⋅°C
d. 2.3 × 10−4cal/s⋅cm⋅°C

Answers

Final answer:

The conductivity of the Styrofoam material is found using the heat required to melt the ice and the conduction formula, resulting in 2.8 × 10^⁻⁶cal/s·cm·°C.

Explanation:

To find the conductivity of the Styrofoam material, we need to calculate the amount of heat transferred to the ice cube to melt it completely and then relate this to the conductivity formula. First, we calculate the total heat required to melt the 500-g ice cube using the latent heat of fusion (Lf). The total heat (Q) required is the mass (m) of the ice times the latent heat of fusion (Lf).

Q = m × Lf = 500 g × 80 cal/g = 40000 cal

Next, we use the formula for heat conduction, which is Q = (k × A × ΔT × t) / d. Here, Q is the heat transferred, k is the thermal conductivity, A is the area through which heat is transferred, ΔT is the temperature difference between the inside and outside of the box, t is the time, and d is the thickness of the walls.

Since all values except k are known, we rearrange the equation to solve for k:

k = (Q × d) / (A × ΔT × t)

k = (40000 cal × 1.0 cm) / (600 cm² × 20°C × (4 × 3600 s))

Upon calculation, we find the conductivity k equals 2.8 × 10-6cal/s·cm·°C, which corresponds to option (b).

Ray baked a cake. The total mass of the cake was equal to the total mass of the ingredients. This is an example of what?

Answers

Answer:

The Law of Conservation of Mass-Energy

Explanation:

It states that nothing can be created or destroyed, which is why the cakes mass is the same as the ingredients.

Answer : This is an example of law of conservation of mass.

Explanation :

Law of conservation of mass : It states that mass can neither be created nor destroyed but it can only be transformed from one form to another form.

The balanced chemical reaction always follow the law of conservation of mass.

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

For example :

[tex]2H_2+O_2\rightarrow 2H_2O[/tex]

This reaction is a balanced chemical reaction in which number of atoms of hydrogen and oxygen are equal on the both side of the reaction. So, this reaction obey the law of conservation of mass.

As per question, the total mass of the cake was equal to the total mass of the ingredients. That means, they obey the law of conservation of mass.

Hence, this is an example of law of conservation of mass.

It is weigh-in time for the local under-85-kg rugby team. The bathroom scale used to assess eligibility can be described by Hooke’s law and is depressed 0.75 cm by its maximum load of 120 kg. (a) What is the spring’s effective force constant? (b) A player stands on the scales and depresses it by 0.48 cm. Is he eligible to play on this under-85-kg team?

Answers

Answer:

(a)spring effective force constant =1568N/cm

(b) Yes

Explanation:

Hooke's law is represented mathematically as, F=ke

where

F is force applied to elastic material,k= spring constante= extension

(a).After the maximum load is exceeded, Hooke's law doesn't apply anymore. from the problem, its stated that a maximum load of 120kg will cause an extension of 0.75cm, we will use this to determine the spring constant.

F=mg, g=9.8[tex]m/s^{2}[/tex]

F= 120*9.8 =1176N

From Hooke's law, k=[tex]\frac{F}{e}[/tex]

k=[tex]\frac{1176}{0.75}[/tex]

k=1568N/cm

(b). The players who stands on the scale causes a 0.48cm extension.

    e= 0.48cm, k= 1568N/cm

F=Ke

F= 1568*0.48

F= 752.64N

To calculate the mass of the player we divide this force by g=9.8[tex]m/s^{2}[/tex]

F=mg

m=F/g

[tex]m= \frac{ 752.64}{9.8} \\\\m=76.8kg[/tex]

since the rugby team is an under 85kg team, this player with 76.8kg mass is eligible

The energy E of the electron in a hydrogen atom can be calculated from the Bohr formula: =E−Ryn2 In this equation Ry stands for the Rydberg energy, and n stands for the principal quantum number of the orbital that holds the electron. (You can find the value of the Rydberg energy using the Data button on the ALEKS toolbar.) Calculate the wavelength of the line in the emission line spectrum of hydrogen caused by the transition of the electron from an orbital with =n10 to an orbital with =n8. Round your answer to 3 significant digits.

Answers

Final answer:

The energy change in an electron's transition in a hydrogen atom can be calculated using the Bohr formula. This energy corresponds to the energy of the emitted photon in the transition, and can be used to calculate the wavelength of the light emitted during this transition.

Explanation:

The energy change associated with the transition of an electron between two energy levels in a hydrogen atom can be calculated using the equation E = 13.6 eV/n². In this equation, 'E' represents the energy of the electron, and 'n' is the quantum number of the orbit that the electron occupies. If we consider a transition of an electron from an orbital with n = 10 to an orbital with n = 8, we can calculate the energy change (∆E) using the difference in energies of these two states. This energy change corresponds to the energy of the emitted photon when the electron transition occurs. The energy of the photon can be connected to its wavelength through the equation E = hf, where 'h' is Planck's constant, and 'f' is the frequency, which is related to the wavelength (λ) by the speed of light (c) as f = c/λ. Therefore, you can first calculate ∆E using the given energy equation, then use the result to find 'f' using E = hf, and finally substitute 'f' in f = c/λ to find λ, the wavelength of the emitted light.

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For many years Colonel John P. Stapp, USAF, held the world's land speed record. He participated in studying whether a jet pilot could survive emergency ejection. On March 19, 1954, he rode a rocket-propelled sled that moved down a track at a speed of 632 mi/h. He and the sled were safely brought to rest in 1.40s

(a) Determine the negative acceleration he experienced.
__ m/s2
(b) Determine the distance he traveled during this negative acceleration.
__ m.

Answers

Answer:

(a) -202 m/s²

(b) 198 m

Explanation:

Given data

Initial speed (v₀): 283 m/s

[tex]\frac{632mi}{h} .\frac{1609.34m}{1mi} .\frac{1h}{3600s} =283m/s[/tex]

Final speed (vf): 0 (rest)Time (t): 1.40 s

(a) The acceleration (a) is the change in the speed over the time elapsed.

a = (vf - v₀)/t = (0 - 283 m/s)/ 1.40s = -202 m/s²

(b) We can find the distance traveled (d) using the following kinematic expression.

y = v₀ × t + 1/2 × a × t²

y = 283 m/s × 1.40 s + 1/2 × (-202 m/s²) × (1.40 s)²

y = 198 m

The Reynolds number, rho VD/mu, is a very important parameter in fluid mechanics. Verify that the Reynolds number is dimensionless, using both the FLT system and the MLT system for basic dimensions, and determine its value for methane flowing at a velocity of 4 m/s through a 2-in-diameter pipe.

Answers

Answer:

Re = 1 10⁴

Explanation:

Reynolds number is

         Re = ρ v D /μ

The units of each term are

       ρ = [kg / m³]

       v = [m / s]

      D = [m]

      μ = [Pa s]

The pressure

      Pa = [N / m²] = [Kg m / s²] 1 / [m²] = [kg / m s²]

      μ = [Pa s] = [kg / m s²] [s] = [kg / m s]

We substitute the units in the equation

      Re = [kg / m³] [m / s] [m] / [kg / m s]

      Re = [kg / m s] / [m s / kg]

      RE = [ ]

Reynolds number is a scalar

Let's evaluate for the given point

Where the data for methane are:

viscosity       μ = 11.2 10⁻⁶ Pa s

the density  ρ = 0.656 kg / m³

       D = 2 in (2.54 10⁻² m / 1 in) = 5.08 10⁻² m

       Re = 0.656 4 2 5.08 10⁻² /11.2 10⁻⁶

       Re = 1.19 10⁴

The Reynolds number is a key parameter in fluid mechanics that signals laminar or turbulent flow. It is a dimensionless quantity determined by fluid properties and flow characteristics.

The Reynolds number, a dimensionless parameter, indicates whether flow is laminar or turbulent in fluid mechanics. It is defined as Re = rho VD/mu, where rho is the fluid density, V is the velocity, D is the diameter, and mu is the fluid viscosity. The value of Reynolds number for methane flowing at 4 m/s through a 2-in-diameter pipe is calculated using this formula.

You place 100 grams of ice, with a temperature of −10∘C, in a styrofoam cup. Then you add an unknown mass of water, with a temperature of +10∘C, and allow the system to come to thermal equilibrium. For the calculations below, use the following approximations. The specific heat of solid water is 2 joules / gram, and the specific heat of liquid water is 4 joules / gram. The latent heat of fusion of water is 300 joules / gram. Assume no heat is exchanged with the surroundings. Express your answers in grams, using two significant digits.

If the final temperature of the system is -5∘C , how much water was added? ______________ grams

Answers

Answer:

Mass of water 2.9g

Explanation:

Ice

[tex]m_{ice}=100g[/tex]

[tex]c_{ice}=2J/g.K[/tex]

[tex]T_{ice,initial}=-10\°C[/tex]

[tex]T_{ice,final}=T_{equilibrium}=-5\°C[/tex]

Water

[tex]c_{water}=4J/g.K[/tex]

[tex]T_{water,initial}=10\°C[/tex]

[tex]T_{water,final}=0\°C[/tex]

[tex]T_{equilibrium}=-5\°C[/tex]

[tex]l_{water}=300J/g[/tex]

[tex]m_{water}=?g[/tex]

Step 1: Determine heat gained by ice

[tex]Q_{ice}=m_{ice}c_{ice}(T_{ice,final}-T_{ice,initial})[/tex]

[tex]Q_{ice}=100*2*(-5--10)[/tex]

[tex]Q_{ice}=1000J[/tex]

Step 2; Determine heat lost by water

[tex]Q_{water}=m_{water}c_{water}(T_{water,initial}-T_{water,final})+m_{water}l_{water}[/tex]

[tex]Q_{water}=m_{water}*4*(10-0)+m_{water}*300[/tex]

[tex]Q_{water}=40m_{water}+300m_{water}[/tex]

[tex]Q_{water}=340m_{water}[/tex]

Step 3: Heat gained by ice is equivalent to heat lost by water

[tex]Q_{ice}=Q_{water}[/tex]

[tex]1000=340m_{water}[/tex]

[tex]m_{water}=2.9g[/tex]

A device plugged into a wall outlet experiences an alternating current. The RMS current value is 2 A. The peak current is ___.

Answers

Answer:

The peak value of current will be 2.828 A.

Explanation:

Given that

Value of RMS current I(rms) = 2 A

Lets take peak current = I(p)

As we know that relationship between RMS and peak current given as

[tex]I(rms)=\dfrac{I(p)}{\sqrt2}[/tex]

Now by putting the values in the above equation

[tex]I(rms)=\dfrac{I(p)}{\sqrt2}[/tex]

[tex]2\ A=\dfrac{I(p)}{\sqrt2}[/tex]

[tex]I(p)=2\sqrt2[/tex]

As we know that

[tex]\sqrt{2}=1.414[/tex]

Therefore

I(p)=2.828 A

The peak value of current will be 2.828 A.

A resistor with R=300 ? and an inductor are connected in series across an ac source that has voltage amplitude 500 V. The rate at which electrical energy is dissipated in the resistor is 216 W. (a) What is the impedance Z of the circuit? (b) What is the amplitude of the voltage across the inductor? (c) What is the power factor?

Answers

Answer:

impedance Z = 416.66 ohm

voltage across inducance V = 346.99 V

power factor = 0.720

Explanation:

given data

resistor R = 300

voltage amplitude = 500 V

resistor = 216 W

to find out

impedance and amplitude of the voltage and power factor

solution

we apply here average power formula that is

average power = I²×R     ............1

I = [tex]\frac{Vrms}{Z}[/tex]

so

average power =  ([tex]\frac{Vrms}{Z}[/tex])²×R

Vrms = [tex]\frac{1}{\sqrt{2} }[/tex] × Vmax

Z = V × [tex]\sqrt{\frac{R}{2P} }[/tex]

Z = 500 × [tex]\sqrt{\frac{300}{2*216} }[/tex]

impedance Z = 416.66 ohm

and

we know voltage across inductor is here express as

V = I × X     .............2

so here X will be by inductance

Z² = R² + (X)²  

(X)²  = 416.66² - 300²  

X = 289.15 ohm

and I = [tex]\frac{V}{Z}[/tex]

I = [tex]\frac{500}{416.66}[/tex]

I = 1.20 A

so from equation 2

V = 1.20 × 289.15

voltage across inducance V = 346.99 V

and

average power = Vmax × Imax  ×  cos∅

tan∅ = [tex]\frac{289.15}{300}[/tex]

tan∅ = 43.95°

so power factor is

power factor = cos43.95°

power factor = 0.720

Final answer:

The problem involves the use of Ohm's law, impedance, power, and a power factor in an AC circuit. The impedance of the circuit, the voltage across the inductor, and the power factor can be calculated using given values, the principle of impedance and the relationships among AC voltage, current, resistance and power.

Explanation:

The question involves an AC circuit composed of a resistor and inductor in series connected to an AC voltage source. We have a resistor with a resistance (R) of 300 Ω and a dissipated power (P) of 216W. The voltage amplitude (Vo) of the AC source is 500 V. It is important to remember that in this context, impedance (Z), which has a unit of ohms, represents the total resistance in an AC circuit and can be calculated using Ohm's law for AC circuits.

(a) To find the impedance Z of the circuit, we consider that the power P is given by the relation P = Vo^2/R, substituting for P, and R, we can solve for Vo, which will be sqrt(P*R). Then, the rms voltage (V) is given by Vo/sqrt(2). Our current I would be P/V. Finally, applying Ohm's law, Z=V/I would give us the impedance.

(b) The voltage across the inductor can be found by using Pythagoras' Theorem in the context of an AC circuit, VL = sqrt(Vo^2 - VR^2), where VR is the voltage across the resistor (equal to I* R).

(c) Lastly, the power factor can be found as the cosine of the phase angle θ, which can also be defined as R/Z. We'd first calculate θ = arccos(R/Z), and then find the power factor as cos(θ).

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Stars form from clouds of gas and dust. As a protostar gravitationally contracts within its parent cloud, "conservation of angular momentum" says thatA) the protostar rotates more slowly.b) the protostar\'s rotation does not change.c) the protostar\'s axis of rotation will change direction.d) the protostar rotates more quickly.

Answers

Answer:

D) True. the protostar rotates more quickly.

Explanation:

If the system is isolated, the angular momentum must be retained.

Initial

        L₀ = I w₀

Final

       [tex]L_{f}[/tex] = [tex]I_{f}[/tex]  [tex]w_{f}[/tex]

      L₀ = [tex]L_{f}[/tex]

      I w₀ = [tex]I_{f}[/tex][tex]w_{f}[/tex]

     [tex]w_{f}[/tex]  = I /[tex]I_{f}[/tex] w₀

In general, the radius of the cloud decreases significantly to form the star, the moment of inertia must decrease, so the angular velocity must increase

Let's examine the answers

A) False. The opposite happens

B) False. Speed ​​changes

C) False. For this there must be an external force, which does not exist

D) True. You agree with the above

Final answer:

The correct answer to the question of what occurs to a protostar as it contracts due to the 'conservation of angular momentum' is that the protostar rotates more quickly (d). This is analogous to a figure skater pulling their arms in to spin faster and is supported by observations of star-forming regions like the Orion Nebula.

Explanation:

Stars form from clouds of gas and dust. As these clouds collapse under their own gravity, they form protostars, which spin due to the conservation of angular momentum. Conservation of angular momentum dictates that as the radius of a spinning object decreases, its rotation speed must increase to conserve angular momentum. This concept is similar to a figure skater who spins faster when they pull their arms in. Therefore, when a protostar gravitationally contracts within its parent cloud, it rotates more quickly because as it shrinks, the protostar's rate of spin increases to conserve angular momentum.

This increase in rotation speed results in the formation of a spinning accretion disk around the equator, which is easier to observe in some regions, such as the Orion Nebula or the Taurus star-forming region. Observations from telescopes like the Hubble Space Telescope support this understanding of the role of angular momentum in star formation. In conclusion, the correct answer to the question is (d) the protostar rotates more quickly.

If a camera lens gives the proper exposure for a photograph at a shutter speed of 1/200 s at an f-stop of f/2.80, the proper shutter speed at f/7.92 is

Answers

Answer:

[tex]\dfrac{1}{28.15432}\ s[/tex]

Explanation:

The factor by which the shutter speed is increased is [tex]\dfrac{7.92}{2.8}[/tex]

Exposure time will be increased by

[tex]f=2^{\dfrac{7.92}{2.8}}=7.1037\ s[/tex]

The proper shutter speed is given by

[tex]\dfrac{1}{T}f=\dfrac{1}{200}\times 2^{\dfrac{7.92}{2.8}}=\dfrac{1}{\dfrac{200}{2^{\dfrac{7.92}{2.8}}}}\\ =\dfrac{1}{28.15432}\ s[/tex]

The proper shutter speed is [tex]\dfrac{1}{28.15432}\ s[/tex]

The earth has radius R. A satellite of mass 100 kg is in orbit at an altitude of 3R above the earth's surface. What is the satellite's weight at the altitude of its orbit

Answers

Answer:

W= 61.3 N

Explanation:

The only force acting on the satellite is the one due to the attraction from Earth, which obeys the Newton's Universal Law of Gravitation, as follows:

Fg =G*ms*me / (res)²

This force, also obeys the Newton's 2nd Law, so we can write the following equation:

G*ms*me*/ (res)² = ms* a = ms*g

We call to the product of the mass times the acceleration caused by gravity (g), the weight of this mass, so we can write as follows:

G*ms*me / (res)² = ms*g = W (1)

where G = 6.67*10⁻11 N*m²/kg², ms= 100 kg, me= 5.97*10²⁴ kg, and

res=  4 *re = 4*6.37*10⁶ m.

Replacing all these known values in (1), we get the value of W:

W =(( 6.67*5.97/(4*6.37)²) *( 10⁻¹¹ * 10²⁴ /10¹²) )* 100 N = 61.3 N

The satellite's weight at the altitude of its orbit is;

61.31 N

Formula for gravitational force is;

F = GMm/R²

Where;

G is gravitational constant = 6.67 × 10⁻¹¹ N.m²/kg²

m is mass of earth = 5.97 × 10²⁴ kg

M is mass of satellite = 100 kg

Now, we are told that the altitude is 3R above the Earth's surface.

At the Earth's surface, the distance from the Earth's center is R where R is radius of earth.

Thus, total altitude from the Earth's center to the satellite it (3R + R) = 4R

Thus;

F = GMm/(4R)²

Where R is radius of earth = 6371 × 10⁶ m

Thus;

F = (6.67 × 10⁻¹¹ × 100 × 5.97 × 10²⁴)/(4 × 6371 × 10⁶)

F = 61.31 N

Now, from Newton's second law of motion, we know that the force is equal to the weight.

Thus;

Weight = 61.31 N

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Part A If the accuracy in measuring the position of a particle increases, what happens to the accuracy in measuring its velocity? If the accuracy in measuring the position of a particle increases, what happens to the accuracy in measuring its velocity? The accuracy in measuring its velocity also increases. The accuracy in measuring its velocity decreases. The accuracy in measuring its velocity remains the same. The accuracy in measuring its velocity becomes uncertain.

Answers

Answer:

The correct answer is the speed accuracy decreases

Explanation:

Before examining the final statements, let's review the uncertainty principle

       Δx  Δp> = h ’/ 2

       h ’= h / 2π

This is because the process of measuring one quantity affects the measurement of the other.

Let's review the claims

The speed of the particle is proportional to the moment

           p = mv

Therefore, if the position is measured more accurately (x) the accuracy of p must decrease

        Δp = h ’/ 2 Δx

The correct answer is the speed accuracy decreases

Final answer:

According to the Heisenberg uncertainty principle in physics, if the accuracy in measuring a particle's position increases, the accuracy in measuring its velocity decreases. This is a fundamental characteristic of the quantum world due to the wave-particle duality of matter.

Explanation:

The question refers to the Heisenberg uncertainty principle in physics, particularly quantum mechanics. This principle states that the more precisely the position of a particle is known, the less precisely its velocity (or momentum, to be more specific) can be known, and vice versa. Therefore, if the accuracy of measuring a particle's position increases, then the accuracy in measuring its velocity decreases.

This phenomenon isn't due to measurement techniques or technology. It's a fundamental limit defined by the nature of the quantum world. It derives from the wave-particle duality of matter, which means small particles like electrons behave both as particles and waves. If we measure the position very precisely, the particle acts more like a wave with an uncertain speed. Conversely, if we measure the speed very precisely, the particle acts more like a particle with an uncertain position.

For example, if we use an extremely short-wavelength electron probe to measure an electron's position, we'd be very accurate but massively disturb the electron's velocity in the process. Hence, increasing the precision in position measurement increases the uncertainty in velocity measurement.

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One of the greatest terrorism-related nuclear threats is from Select one: a. nuclear power plants. b. dirty bombs. c. nuclear warheads. d. None of these are correct.

Answers

Answer:

Dirty bomb

Explanation:

Among the nuclear bomb One type is a "dirty bomb." It combines a conventional explosive such as the dynamite with radioactive material which can spread when the system explodes.  The explosion is releasing "dirty" bits of radioactive particles which are extremely harmful and can cause loss equivalent to a nuclear attack.

Two protons, with equal kinetic energy, collide head-on. What is the minimum kinetic energy Kp of one of these protons necessary to make a pion-antipion pair? The rest energy of a pion is 139.6MeV.

Answers

Answer:

[tex]K_p=139.6\ MeV[/tex]

Explanation:

It is given that,

The rest energy of a pion is 139.6 MeV. Here, two protons having equal kinetic energy collides elastically. We need to find the minimum kinetic energy of one of these protons necessary to make a pion-antipion pair.

It can be calculated using conservation of energy and momentum, the total energy of the particles gets converted into rest mass energy of new particles. So,

[tex]2K_p=2\times E_{\pi^+}[/tex]

[tex]K_p=\times E_{\pi^+}[/tex]

[tex]K_p=139.6\ MeV[/tex]

So, the minimum kinetic energy of one of these protons necessary to make a pion-antipion pair is 139.6 MeV. Hence, this is the required solution.

An object initially at rest falls from a height H until it reaches the ground. Two of the following energy bar charts represent the kinetic energy K and gravitational potential energy Ug of the object-Earth system at two positions. The first position is when the object is initially released, and the second position is when the object is halfway between its release point and the ground. Which two charts could represent the mechanical energy of the object-Earth system? Select two answers. by Ug K Ug к Mechanical Energy at Halfway Point Mechanical Energy at Release Point Mechanical Energy at Release Point Mechanical Energy at Halfway Point d. c. к Ug Ug K Mechanical Energy at Release Point Mechanical Energy at Halfway Point Mechanical Energy at Release Point Mechanical Energy at Halfway Point

Answers

Answer:

Initial:   bar  power U₀

Final:    bar  power U = U₀ / 2

             bar Kinetic energy  K = U₀ / 2

These two bars are half the height of the initial bar

Explanation:

To know which graph is correct, let's discuss the solution to the problem

Initial mechanical energy

      Em₀ = U₀ = m g H

The mechanical energy at the midpoint

     Em₂ = K + U₂

As there is no friction, mechanical energy is conserved

     Em₀ = Em₂

     U₀ = K + U₂

     K = U₀ - U₂

     K = m g (H - y₂)

Indicates that position 2 corresponds to y₂ = H / 2

     K = m g (H –H / 2)

     K = ½ m g H

     K = ½ Uo

Therefore the graph must be

Initial:   bar  power U₀

Final:    bar  power U = U₀ / 2

             bar Kinetic energy  K = U₀ / 2

These two bars are half the height of the initial bar

Answer: B&C

Explanation:

Release point: Ug bar graph only K none

Halfway point: Ug and K are equal bar graph

second option,

Release point: empty graph

Halfway point: Ug down half K up half

A 1.0 kg piece of copper with a specific heat of cCu=390J/(kg⋅K) is placed in 1.0 kg of water with a specific heat of cw=4190J/(kg⋅K). The copper and water are initially at different temperatures. After a sufficiently long time, the copper and water come to a final equilibrium temperature. Part A Which of the following statements is correct concerning the temperature changes of both substances? (Ignore the signs of the temperature changes in your answer.) Which of the following statements is correct concerning the temperature changes of both substances? (Ignore the signs of the temperature changes in your answer.) The temperature change of the copper is equal to the temperature change of the water. The temperature change of the water is greater than the temperature change of the copper. The temperature change of the copper is greater than the temperature change of the water.

Answers

Answer:

The temperature change of the copper is greater than the temperature change of the water.

Explanation:

deltaQ = mc(deltaT)

Where,

delta T = change in the temperature

m =mass

c = heat capacity

[tex]\frac{(deltaT)_{Cu}}{(deltaT)_{w}}=\frac{4190J/kg.K}{390J/kg.K}\\ \\(deltaT)_{Cu}=10.74(deltaT)_{w}[/tex]

The temperature change in the copper is nearly 11 times the temperature change in the water.

So, the correct option is,

The temperature change of the copper is greater than the temperature change of the water.

Hope this helps!

Final answer:

The temperature change of the copper is equal to the temperature change of the water.

Explanation:

The temperature change of the copper is equal to the temperature change of the water.

When two substances come into contact, heat is transferred between them until they reach thermal equilibrium, where their temperatures are equal. According to the Law of Conservation of Energy, the heat lost by one substance is equal to the heat gained by the other substance. In this case, since all heat transfer occurs between the copper and water, the temperature change of the copper is equal to the temperature change of the water.

A cylinder with a piston contains 0.300 mol of oxygen at 2.50×105 Pa and 360 K . The oxygen may be treated as an ideal gas. The gas first expands isobarically to twice its original volume. It is then compressed isothermally back to its original volume, and finally it is cooled isochorically to its original pressure.Find the work done by the gas during the initial expansion.Find the heat added to the gas during the initial expansion.Find internal-energy change of the gas during the initial expansion.Find the work done during the final cooling;Find the heat added during the final cooling;Find the internal-energy change during the final cooling;Find the internal-energy change during the isothermal compression

Answers

Answer:

a) W =  900   J.  b) Q =  3142.8   J . c) ΔU =  2242.8   J. d) W = 0. e) Q =   2244.78   J.  g) Δ U  =  0.

Explanation:

(a) Work done by the gas during the initial expansion:

The work done W for a thermodynamic constant pressure process is given as;

W  =  p Δ V

where  

p  is the pressure and  Δ V  is the change in volume.

Here, Given;

P 1 = i n i t i a l  p r e s s u r e  =  2.5 × 10^ 5   P a

T 1 = i n i t i a l   t e m p e r a t u r e  =  360   K

n = n u m b er   o f   m o l e s  =  0.300  m o l  

The ideal gas equation is given by  

P V = nRT

where ,

p  =  absolute pressure of the gas  

V =  volume of the gas  

n  =  number of moles of the gas  

R  =  universal gas constant  =  8.314   K J / m o l   K

T  =  absolute temperature of the gas  

Now we will Calculate the initial volume of the gas using the above equation as follows;

PV  =  n R T

2.5 × 10 ^5 × V 1  =  0.3 × 8.314 × 360

V1 = 897.91 / 250000

V 1  =  0.0036   m ^3  = 3.6×10^-3 m^3

We are also given that

V 2  =  2× V 1

V2 =  2 × 0.0036

V2 =  0.0072   m^3  

Thus, work done is calculated as;

W  =  p Δ V  = p×(V2 - V1)

W =  ( 2.5 × 10 ^5 ) ×( 0.0072  −  0.0036 )

W =  900   J.

(b) Heat added to the gas during the initial expansion:

For a diatomic gas,

C p  =  7 /2 ×R

Cp =  7 /2 × 8.314

Cp =  29.1  J / mo l K  

For a constant pressure process,  

T 2 /T 1  =  V 2 /V 1

T 2  =  V 2 /V 1 × T 1

T 2  =  2 × T 1  = 2×360

T 2  =  720  K

Heat added (Q) can be calculated as;  

Q  =  n C p Δ T  = nC×(T2 - T1)

Q =  0.3 × 29.1 × ( 720  −  360 )

Q =  3142.8   J .

(c) Internal-energy change of the gas during the initial expansion:

From first law of thermodynamics ;

Q  =  Δ U + W

where ,

Q is the heat added or extracted,

Δ U  is the change in internal energy,

W is the work done on or by the system.

Put the previously calculated values of Q and W in the above formula to calculate  Δ U  as;

Δ U  =  Q  −  W

ΔU =  3142.8  −  900

ΔU =  2242.8   J.

(d) The work done during the final cooling:

The final cooling is a constant volume or isochoric process. There is no change in volume and thus the work done is zero.

(e) Heat added during the final cooling:

The final process is a isochoric process and for this, the first law equation becomes ,

Q  =  Δ U  

The molar specific heat at constant volume is given as;

C v  =  5 /2 ×R

Cv =  5 /2 × 8.314

Cv =  20.785  J / m o l   K

The change in internal energy and thus the heat added can be calculated as;  

Q  = Δ U  =  n C v Δ T

Q =  0.3 × 20.785 × ( 720 - 360 )

Q =   2244.78   J.

(f) Internal-energy change during the final cooling:

Internal-energy change during the final cooling  is equal to the heat added during the final cooling Q  =  Δ U  .

(g) The internal-energy change during the isothermal compression:

For isothermal compression,

Δ U  =  n C v Δ T

As their is no change in temperature for isothermal compression,  

Δ T = 0 ,  then,

Δ U  =  0.

During the initial expansion, the gas performed 900 J of work, absorbed 3142.8 J of heat, resulting in a change in internal energy of 2242.8 J. In subsequent processes, work and internal energy changes were determined, culminating in an isothermal compression with no internal energy change.

(a) Work done by the gas during the initial expansion:

The work done (W) for a thermodynamic constant pressure process is given by W = P ΔV, where P is the pressure and ΔV is the change in volume.

Given:

P₁ = initial pressure = 2.5 × 10^5 Pa

T₁ = initial temperature = 360 K

n = number of moles = 0.300 mol

The ideal gas equation is PV = nRT, where P is the absolute pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the universal gas constant (8.314 kJ/mol·K), and T is the absolute temperature of the gas.

Calculate the initial volume of the gas:

P₁V₁ = nRT₁

(2.5 × 10^5) × V₁ = 0.3 × 8.314 × 360

V₁ = (0.3 × 8.314 × 360) / (2.5 × 10^5)

V₁ = 0.0036 m³ = 3.6 × 10⁻³ m³

Given V₂ = 2 × V₁:

V₂ = 2 × 0.0036

V₂ = 0.0072 m³

Now, calculate the work done:

W = P ΔV = (2.5 × 10^5) × (0.0072 - 0.0036)

W = 900 J

(b) Heat added to the gas during the initial expansion:

For a diatomic gas, Cp = (7/2)R, where Cp is the molar heat capacity at constant pressure.

Cp = (7/2) × 8.314

Cp = 29.1 J/mol·K

For a constant pressure process, T₂/T₁ = V₂/V₁:

T₂ = (V₂/V₁) × T₁

T₂ = 2 × T₁ = 2 × 360 = 720 K

Heat added (Q) can be calculated as:

Q = nCpΔT = 0.3 × 29.1 × (720 - 360)

Q = 3142.8 J

(c) Internal-energy change of the gas during the initial expansion:

From the first law of thermodynamics Q = ΔU + W, where Q is the heat added or extracted, ΔU is the change in internal energy, and W is the work done on or by the system.

ΔU = Q - W = 3142.8 - 900

ΔU = 2242.8 J

(d) The work done during the final cooling:

The final cooling is a constant volume (isochoric) process, so there is no change in volume (ΔV = 0), and thus the work done is zero.

(e) Heat added during the final cooling:

For an isochoric process, Q = ΔU. The molar specific heat at constant volume is Cv = (5/2)R.

Cv = (5/2) × 8.314 = 20.785 J/mol·K

The change in internal energy and thus the heat added can be calculated as:

Q = ΔU = nCvΔT = 0.3 × 20.785 × (720 - 360)

Q = 2244.78 J

(f) Internal-energy change during the final cooling:

The internal-energy change during the final cooling is equal to the heat added during the final cooling, so ΔU = Q.

(g) The internal-energy change during the isothermal compression:

For isothermal compression, ΔU = 0 since there is no change in temperature (ΔT = 0).

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One might be tempted to say that exothermic processes are always spontaneous since the system is emitting energy (heat) in order to reach a (preferred) lower energy state. However, as we have just investigated, the spontaneous process for polymers is endothermic. This reveals that we must consider entropy changes when determining the nature of spontaneity. The most probable configuration of a system and its surroundings, naturally, is the one that will be observed. The condition for spontaneity can be recast using the concept of the free energy of the system, where a change in free energy results both from changes in the enthalpy (which includes internal potential and kinetic energies) and the entropy (the number of states accessible to the system). Δ G = Δ H − T Δ S.

An unknown chemical reaction undergoes an enthalpy change of Δ H =17 kJ/mol while the entropy increases by Δ S =50 J/(mol * K).

Above what temperature (in Kelvin) does this reaction occur spontaneously?

Answers

Answer : This reaction occur spontaneously at temperature above in kelvins is, 340 K

Explanation : Given,

[tex]\Delta H[/tex] = 17 KJ/mole = 17000 J/mole

[tex]\Delta S[/tex] = 50 J/mole.K

Gibbs–Helmholtz equation is :

[tex]\Delta G=\Delta H-T\Delta S[/tex]

As per question the reaction is spontaneous that means the value of [tex]\Delta G[/tex] is negative or we can say that the value of [tex]\Delta G[/tex] is less than zero.

[tex]\Delta G<0[/tex]

The above expression will be:

[tex]0>\Delta H-T\Delta S[/tex]

[tex]T\Delta S>\Delta H[/tex]

[tex]T>\frac{\Delta H}{\Delta S}[/tex]

Now put all the given values in this expression, we get :

[tex]T>\frac{17000J/mole}{50J/mole.K}[/tex]

[tex]T>340K[/tex]

Therefore, this reaction occur spontaneously at temperature above in kelvins is, 340 K

Final answer:

The reaction occurs spontaneously above 340 Kelvin, determined by applying the Gibbs free energy formula (ΔG = ΔH - TΔS) and ensuring the values for ΔH and ΔS are in compatible units.

Explanation:

The question requires us to determine above what temperature a reaction occurs spontaneously, given an enthalpy change (ΔH) of 17 kJ/mol and an entropy change (ΔS) of 50 J/(mol·K). To find this, we use the formula for Gibbs free energy change (ΔG = ΔH - TΔS), where a reaction is spontaneous when ΔG < 0.

First, we need to ensure that both values are in the same units, so we convert ΔH from kJ to J: 17 kJ/mol = 17000 J/mol. Then, we solve for T in the equation ΔG < 0, substituting ΔH and ΔS into the equation:

0 > 17000 J/mol - T(50 J/(mol·K)),

implying T > 17000 J/mol / 50 J/(mol·K) = 340 K.

Therefore, the reaction occurs spontaneously above 340 K.

A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a slit 0.107 mm wide. At the point in the pattern which is an angular distance of 3.09° from the center of the central maximum, the total phase difference between wavelets from the top and bottom of the slit is 56.0 rad.
a) What is the wavelenght of the radiation?
b) What is the intensity at this point, if the intensity at the center of the central maximum is I/O?

Answers

Answer:

a. λ = 647.2 nm

b. I₀  9.36 x 10⁻⁵

Explanation:

Given:

β = 56.0 rad , θ = 3.09 ° , γ = 0.170 mm = 0.170 x 10⁻³ m

a.

The wavelength of the radiation can be find using

β = 2 π / γ * sin θ

λ = [ 2π * γ * sin θ ] / β

λ = [ 2π * 0.107 x 10⁻³m * sin (3.09°) ] / 56.0 rad

λ = 647.14 x 10⁻⁹ m  ⇒  λ = 647.2 nm

b.

The intensity of the central maximum I₀

I = I₀ (4 / β² ) * sin ( β / 2)²

I = I₀ (4 / 56.0²) * [ sin (56.0 /2) ]²

I = I₀  9.36 x 10⁻⁵

Final answer:

To find the wavelength of the radiation and the intensity at a specific point in a single-slit diffraction pattern, we can use formulas and calculations based on the given information.

Explanation:

To find the wavelength of the radiation, we can use the formula for single-slit diffraction:

sin(θ) = mλ/w

Where θ is the angle from the center of the central maximum, m is the order of the bright fringe (m = 1 for the first bright fringe), λ is the wavelength, and w is the width of the slit. Rearranging the formula, we get:

λ = wsin(θ)/m

Plugging in the values, we have:

λ = (0.107 mm)(sin(3.09°))/(1)

Calculating this gives us the wavelength of the radiation.

To find the intensity at the given point, we can use the formula for intensity in single-slit diffraction:

I/I0 = (sin(θ)/θ)2

The given point is at an angular distance of 3.09°, so we can use this formula to calculate the intensity.

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An airplane flies at 150 km/hr. (a) The airplane is towing a banner that is b = 0.8 m tall and l = 25 m long. If the drag coef- ficient based on area bl is CD = 0.06, estimate the power required to tow the banner. (b) For comparison, determine the power required if the airplane was instead able to tow a rigid flat plate of the same size. (c) Explain why one had a larger power requirement (and larger drag) than the other. (d) Finally, determine the power required if the airplane was towing a smooth spherical balloon with a diameter of 2 m.

Answers

Final answer:

We can estimate the power required to tow a banner by using the drag force formula. If the airplane was towing a rigid flat plate, the power required would be different due to a different drag coefficient. The power requirement and drag are larger for the banner due to its higher drag coefficient compared to a flat plate. No calculation can be made for a smooth spherical balloon without the drag coefficient.

Explanation:

(a) To estimate the power required to tow the banner, we can use the formula: Power = Drag force * velocity. The drag force can be calculated using the drag coefficient and the area of the banner. The area is given by multiplying the height (0.8 m) by the length (25 m). The drag force can then be multiplied by the velocity of the plane (150 km/hr converted to m/s) to obtain the power required.

(b) If the plane was towing a rigid flat plate of the same size, the drag coefficient would be different. The power required can be calculated using the new drag coefficient and the same formula as in part (a).

(c) The power requirement and drag are larger for the banner because the drag coefficient for the banner is higher compared to that of a rigid flat plate. This means that the banner experiences more air resistance, requiring more power to tow.

(d) To determine the power required to tow a smooth spherical balloon with a diameter of 2 m, we would need the drag coefficient associated with the balloon. Since it is not provided in the question, we cannot calculate the power required.

An empty capacitor is capable of storing 1.0 × 10-4 J of energy when connected to a certain battery. If the distance between the plates is halved and then filled with a dielectric (κ = 4.3), how much energy could this modified capacitor store when connected to the same battery?

Answers

Answer:

Explanation:

Energy stored in a capacitor

= 1/2 C₁V²

capacity of a capacitor

c = εK A / d

k is dielectric and d is distance between plates .

When  the distance between the plates is halved and then filled with a dielectric (κ = 4.3)

capacity becomes 4.3  x 2 times

New capacity

C₂ = 8.6 C₁

Energy of modified capacitor

1/2 C₂ V²= 1/2 x 8.6 c x V²

Energy becomes

8.6 times.

Energy stored = 8.6 x 10⁻⁴ J

A velocity selector can be used to measure the speed of a charged particle. A beam of particles is directed along the axis of the instrument. The plates are separated by 3 mm and the magnitude of the magnetic field is 0.3 T. What voltage between the plates will allow particles of speed 5.0 x 10^5 m/sa. 1200 V b. 3800 V c. 7500 V d. 190 V e. 380 V

Answers

Answer:

Voltage, V = 450 volts

Explanation:

It is given that,

Separation between plates, d = 3 mm = 0.003 m        

magnitude of magnetic field, B = 0.3 T

Speed of the particle, [tex]v=5\times 10^5\ m/s[/tex]

The relation between the magnetic field, electric field and the velocity of the particle is given by :

[tex]v=\dfrac{E}{B}[/tex]

Also, [tex]E=\dfrac{V}{d}[/tex]

[tex]v=\dfrac{V}{Bd}[/tex]

[tex]V=vBd[/tex]

[tex]V=5\times 10^5\times 0.3\times 0.003[/tex]

V = 450 volts

So, the voltage between the plates will be 450 V. Hence, this is the required solution.

An observer stands on the side of the front of a stationary train. When the train starts moving with constant acceleration, the observer manages to measure the time it.takes for the second cart to pass him to be 5.0 seconds How long will it take for the 10th car to pass him? Assume all cars to be of the same length.
a. 2.8 s
b. 2.4 s
c. 2.0 s
d. 1.5 s
e. 1.1 s

Answers

To solve this problem we will use the linear motion description kinematic equations. We will proceed to analyze the general case by which the analysis is taken for the second car and the tenth. So we have to:

[tex]x = v_0 t \frac{1}{2} at^2[/tex]

Where,

x= Displacement

[tex]v_0[/tex] = Initial velocity

a = Acceleration

t = time

Since there is no initial velocity, the same equation can be transformed in terms of length and time as:

[tex]L = \frac{1}{2} a t_1 ^2[/tex]

For the second cart

[tex]2L \frac{1}{2} at_2^2[/tex]

When the tenth car is aligned the length will be 9 times the initial therefore:

[tex]9L = \frac{1}{2} at_3^2[/tex]

When the tenth car has passed the length will be 10 times the initial therefore:

[tex]10L = \frac{1}{2}at_4^2[/tex]

The difference in time taken from the second car to pass it is 5 seconds, therefore:

[tex]t_2-t_1 = 5s[/tex]

From the first equation replacing it in the second one we will have that the relationship of the two times is equivalent to:

[tex]\frac{1}{2} = (\frac{t_1}{t_2})^2[/tex]

[tex]t_1 = \frac{t_2}{\sqrt{2}}[/tex]

From the relationship when the car has passed and the time difference we will have to:

[tex](t_2-\frac{t_2}{\sqrt{2}}) = 5[/tex]

[tex]t_2 (\sqrt{2}-1) = 3\sqrt{2}[/tex]

[tex]t_2= (\frac{5\sqrt{2}}{\sqrt{2}-1})^2[/tex]

Replacing the value found in the equation given for the second car equation we have to:

[tex]\frac{L}{a} = \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2[/tex]

Finally we will have the time when the cars are aligned is

[tex]18 \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_3^2[/tex]

[tex]t_3 = 36.213s[/tex]

The time when you have passed it would be:

[tex]20\frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_4^2[/tex]

[tex]t_4 = 38.172[/tex]

The difference between the two times would be:

[tex]t_4-t_3 = 38.172-36.213 \approx 2s[/tex]

Therefore the correct answer is C.

Final answer:

The passage time for the 10th car in a uniformly accelerating train would be less than 5.0 seconds but not as low as any of the options provided (a to e), as each car passes more quickly than the previous one due to constant acceleration.

Explanation:

The question involves an observer timing how long it takes for individual cars of constant acceleration of a train to pass by. If it took 5.0 seconds for the second car to pass the observer, then the time it takes for the 10th car to pass him is merely the time it took for one car to pass because the train is accelerating at a constant rate, and all cars are of the same length. Considering the train is accelerating uniformly, the passage of each car happens more quickly over time.

Giving this knowledge, for the first car to pass, the train would have started at rest, so it would have taken longer than the second car which had the benefit of the train already moving at a certain velocity. Therefore, we can derive that the time for each subsequent car to pass will be less than 5.0 seconds, which questions the multiple-choice options. Based on this logic, we don't have to calculate the exact passing time for the 10th car, as there are no values given to calculate with, but we can infer that none of the times listed would be correct. Each car would be passing quicker than the last, but it would not be instantaneous; it would be a fraction of the 5 seconds but not as low as any of the times listed from a to e.

The size (radius) of an oxygen molecule is about 2.0 ×10−10 m. Make a rough estimate of the pressure at which the finite volume of the molecules should cause noticeable deviations from ideal-gas behavior at ordinary temperatures (T= 300K ). Assume that deviatons would be noticeable when volume of the gas per molecule equals the volume of the molecule itself.

Express your answer using one significant figure.

P = ? Pa

Answers

Answer:

P = 1 x 10⁸ Pa

Explanation:

given,

radius = 2.0 ×10⁻¹⁰ m

Temperature

T = 300 K

Volume of gas molecule =

[tex]V = \dfrac{4}{3}\pi r^3[/tex]

[tex]V = \dfrac{4}{3}\pi (2\times 10^{-10})^3[/tex]

 V = 33.51 x 10⁻³⁰ m³

we know,

P  V = 1 . k T

k = 1.38  x 10⁻²³ J/K

P(33.51 x 10⁻³⁰) = 1 . (1.38  x 10⁻²³) x 300

P = 1.235 x 10⁸ Pa

for 1 significant figure

P = 1 x 10⁸ Pa

A sinusoidally-varying voltage V(t)=V0sin(2pift) with amplitude V0 = 10 V and frequency f = 100 Hz is impressed across the plates of a circular-shaped parallel plate air-gap capacitor of radius a = 1.0 cm and plate separation d = 0.01 mm. The amplitude of Maxwell's displacement current ID flowing across the gap between the plates of this capacitor is?

Answers

Final answer:

The amplitude of Maxwell's displacement current flowing across the gap between the plates of the capacitor is 1.753 x 10^-7 A.

Explanation:

The amplitude of Maxwell's displacement current flowing across the gap between the plates of this capacitor can be determined using the formula for the displacement current, which is given by ID = ε0AdV/dt, where ε0 is the permittivity of free space, A is the area of the capacitor plates, and dV/dt is the rate of change of voltage with respect to time.

In this case, the area A of the circular plates is equal to πa^2, where a is the radius of the plates. Therefore, A = π(0.01 m)^2 = 0.000314 m^2. The rate of change of voltage can be obtained from the equation V(t) = V0sin(2πft), where V0 is the amplitude of the voltage, f is the frequency, and t is the time.

Substituting the given values V0 = 10 V and f = 100 Hz into the equation, we have V(t) = 10sin(2π(100)t). Differentiating this equation with respect to time gives dV/dt = 2000πcos(2π(100)t). Plugging in the value of t = 0 (for maximum displacement current) into the equation, we find that cos(2π(100)(0)) = cos(0) = 1.

Therefore, the amplitude of the Maxwell's displacement current ID is given by ID = (8.85 x 10^-12 F/m)(0.000314 m^2)(2000π)(1) = 1.753 x 10^-7 A.

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A 3.0-kg cylinder falls vertically from rest in a very tall, evacuated tube near the surface of the earth. What is its speed after the cylinder has fallen 6.0 m?

Answers

Answer:v=10.84 m/s

Explanation:

Given

mass of Cylinder [tex]m=3 kg[/tex]

height of cylinder [tex]h=6 m[/tex]

It is given that tube is evacuated so we can neglect air resistance so friction provided by the air is zero.

Since Energy cannot be destroyed but can be transformed from one form to another therefore Potential Energy of Cylinder is converted to Kinetic Energy of Cylinder

Potential Energy[tex]=mgh=3\times 9.8\times 6=176.4 J[/tex]

Kinetic Energy [tex]=\frac{1}{2}mv^2=\frac{1}{2}\times 3\times (v)^2[/tex]

[tex]176.4=\frac{1}{2}\times 3\times (v)^2[/tex]

[tex]v=\sqrt{117.6}[/tex]

[tex]v=10.84 m/s[/tex]

Speed of cylinder after the cylinder has fallen 6 m is 11 meter/second.

Energy conservation based problem

What information do we have?

Mass of cylinder = 3 kg

Height = 6 m

Using energy conservation theroy

mgh = (1/2)mv²

gh = (1/2)v²

(9.8)(6) = (1/2)v²

Velocity = 11 m/s (Approx.)

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