A car of mass 900 kg is traveling at 20 m/s when the brakes are applied. The car then comes to a complete stop in 5 s. What is the average power that the brakes produce in stopping the car?

A. 36,000 W
B. 7200 W
C. 3600 W
D. 1800 W

Answer:

c. is negative

Explanation:

When the magnitude of force is multiplied with the force vector's projection along the direction of the vector of displacement which is negative as it is a resistive force we get the work done.

As the wind is acting in opposite direction of the force which is being applied by the plane the work done will be negative. Also, the net work will be the sum of many smaller negative quantities.

Hence, the answer here is negative.

The time taken for the stone to hit the ground can be calculated using vertical motion equations, and the stone's speed at impact can be determined using the vertical velocity equation.

Time taken for the stone to hit the ground can be calculated using the vertical motion equation: t = sqrt(2h/g). Substituting the values, t = √(2×45/9.81) ≈ 3.0 seconds.

Stone's speed at impact can be calculated using the vertical velocity equation: vf = u + gt, where vf is the final velocity (0 m/s at impact), u is the initial vertical speed, and g is the acceleration due to gravity.

With the given information, the stone's speed at impact is approximately 29.4 m/s.

Answer:

The velocity is  [tex]4.6 m/s^2[/tex]

Explanation:

Given:

Force = 500N

Distance  s= 0

To find :

Its velocity at s = 0.5 m

Solution:

[tex]\sum F_{x}=m a[/tex]

[tex]F\left(\frac{4}{5}\right)-F_{S}=13 a[/tex]

[tex]500\left(\frac{4}{5}\right)-\left(k_{s}\right)=13 a[/tex]

[tex]400-(500 s)=13 a[/tex]

[tex]a = \frac{400 -(500s)}{13}[/tex]

[tex]a = (30.77 -38.46s) m/s^2[/tex]

Using the relation,

[tex]a=\frac{d v}{d t}=\frac{d v}{d s} \times \frac{d s}{d t}[/tex]

[tex]a=v \frac{d v}{d s}[/tex]

[tex]v d v=a d s[/tex]

Now integrating on both sides

[tex]\int_{0}^{v} v d v=\int_{0}^{0.5} a d s[/tex]

[tex]\int_{0}^{v} v d v=\int_{0}^{0.5}(30.77-38.46 s) d s[/tex]

[tex]\left[\frac{v^{2}}{2}\right]_{0}^{2}=\left[\left(30.77 s-19.23 s^{2}\right)\right]_{0}^{0.5}[/tex]

[tex]\left[\frac{v^{2}}{2}\right]=\left[\left(30.77(0.5)-19.23(0.5)^{2}\right)\right][/tex]

[tex]\left[\frac{v^{2}}{2}\right]=[15.385-4.807][/tex]

[tex]\left[\frac{v^{2}}{2}\right]=10.578[/tex]

[tex]v^{2}=10.578 \times 2[/tex]

[tex]v^{2}=21.15[/tex]

[tex]v = \sqrt{21.15}[/tex]

[tex]v = 4.6 m/s^2[/tex]

To solve this problem we will use the kinematic equations of angular motion in relation to those of linear / tangential motion.

We will proceed to find the centripetal acceleration (From the ratio of the radius and angular velocity to the linear velocity) and the tangential acceleration to finally find the total acceleration of the body.

Our data is given as:

[tex]\omega = 1.25 rad/s \rightarrow[/tex] The angular speed

[tex]\alpha = 0.745 rad/s2 \rightarrow[/tex] The angular acceleration

[tex]r = 4.65 m \rightarrow[/tex] The distance

The relation between the linear velocity and angular velocity is

[tex]v = r\omega[/tex]

Where,

r = Radius

[tex]\omega =[/tex] Angular velocity

At the same time we have that the centripetal acceleration is

[tex]a_c = \frac{v^2}{r}[/tex]

[tex]a_c = \frac{(r\omega)^2}{r}[/tex]

[tex]a_c = \frac{r^2\omega^2}{r}[/tex]

[tex]a_c = r \omega^2[/tex]

[tex]a_c = (4.65 )(1.25 rad/s)^2[/tex]

[tex]a_c = 7.265625 m/s^2[/tex]

Now the tangential acceleration is given as,

[tex]a_t = \alpha r[/tex]

Here,

[tex]\alpha =[/tex] Angular acceleration

r = Radius

[tex]\alpha = (0.745)(4.65)[/tex]

[tex]\alpha = 3.46425 m/s^2[/tex]

Finally using the properties of the vectors, we will have that the resulting component of the acceleration would be

[tex]|a| = \sqrt{a_c^2+a_t^2}[/tex]

[tex]|a| = \sqrt{(7.265625)^2+(3.46425)^2}[/tex]

[tex]|a| = 8.049 m/s^2 \approx 8.05 m/s2[/tex]

Therefore the correct answer is C.

Answer:

No

Explanation:

Let the reference origin be location of ship B in the beginning. We can then create the equation of motion for ship A and ship B in term of time t (hour):

A = 12 - 12t

B = 9t

Since the 2 ship motions are perpendicular with each other, we can calculate the distance between 2 ships in term of t

[tex]d = \sqrt{A^2 + B^2} = \sqrt{(12 - 12t)^2 + (9t)^2}[/tex]

For the ships to sight each other, distance must be 5 or smaller

[tex] d \leq 5[/tex]

[tex]\sqrt{(12 - 12t)^2 + (9t)^2} \leq 5[/tex]

[tex](12 - 12t)^2 + (9t)^2 \leq 25[/tex]

[tex]144t^2 - 288t + 144 + 81t^2 - 25 \leq 0[/tex]

[tex]225t^2 - 288t + 119 \leq 0[/tex]

[tex](15t)^2 - (2*15*9.6)t + 9.6^2 + 26.84 \leq 0[/tex]

[tex](15t^2 - 9.6)^2 + 26.84 \leq 0[/tex]

Since [tex](15t^2 - 9.6)^2 \geq 0[/tex] then

[tex](15t^2 - 9.6)^2 + 26.84 > 0[/tex]

So our equation has no solution, the answer is no, the 2 ships never sight each other.

Answer:

As point B is located inside the copper block so net electric field at point B is j.

Explanation:

Consider the figure attached below. The net electric field at location B,that is inside the copper block is zero because when a conductor is charged or placed in an electric field of  external charges, net charge lies on the surface of conductor and there is no electric field inside the conductor. As point B is located inside the copper block so net electric field at point B is zero as well direction of net electric field at point B is zero.

Answer:

option D.  

Explanation:

The correct answer is option D.                

Diffusion is the particle movement from high concentration to low concentration, such as air, water, etc.                      

Example: if you spray perfume spread throughout the room at one part of the fragrance scent.                                                        

The gas molecule thus spreads out in a gradient of concentration

Diffusion is equivalent to gas molecules spreading out in a concentration gradient. Option D.

Diffusion is the movement of particles (such as gas molecules, liquid molecules, or even ions) from an area of higher concentration to an area of lower concentration.

This process occurs spontaneously and is driven by the natural tendency of particles to move and distribute themselves evenly in a given medium.

Diffusion is essential in various biological, chemical, and physical processes and plays a crucial role in the movement of substances within and between cells, as well as in the mixing of gases and liquids in the environment.

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Answer:

5.5 × 10^14 Hz or s^-1

no orange light has less frequency so no photoelectric effect

Explanation:

hf = hf0 + K.E

HERE h is Planck 's constant having value 6.63 × 10 ^-34 J s

f is frequency of incident photon and f0 is threshold frequency

hf0 = hf- k.E

6.63 × 10 ^-34 × f0 = 6.63 × 10 ^-34× 6.66 × 10^14 - 7.74× 10^-20

6.63 × 10 ^-34 × f0 = 3.64158×10^-19

                           f0 = 3.64158×10^-19/ 6.63 × 10 ^-34

                           f0 = 5.4925 × 10^14

                            f0 =5.5 × 10^14 Hz or s^-1

frequency of orange light is 4.82 × 10^14 Hz which is less than threshold frequency hence photo electric effect will not be observed for orange light

Answer:

(D) Va > Vb. The forces between the atoms in a block and the atoms in a surface oppose the motion of the block and are greater on average for block B.

Explanation:

Given that, Two identical blocks, block A and block B, are placed on different horizontal surfaces. Block A is initially at rest on a smooth surface, while block B is initially at rest on a rough surface.

And now, a constant force F₀ is exerted on each block for a time Δt.

Now, speeds of blocks A and B are vA and vB respectively.

In case of block A the surface is smooth, so there is no opposing force.

Whereas as, in case of block B, the surface is rough which means friction opposes the motion

So, some of the applied force is used to overcome this friction which makes the speed of block B vB to be less than that of block A vA.

⇒ Va > Vb.

And completely eliminating friction is not possible.

so, even smooth surface has friction which is very very little.

so, The forces between the atoms in a block and the atoms in a surface oppose the motion of the block and are greater on average for block B.

The correct relation between vA and vB is vA > vB because block A encounters less friction on the smooth surface as compared to block B on the rough surface. Consequently, block A will have a larger final velocity vA when the same force is applied to both blocks for the same amount of time.

The correct relation is vA > vB. This is because block A is on a smooth surface, meaning there is less friction to oppose its motion when a force is applied, compared to block B which is on a rough surface. Friction is a force that opposes the motion of an object when it moves across a surface. In this case, the frictional force on block B is greater on average than the force on block A because block B rests on a higher-friction, or rougher, surface.

When we apply a constant force to both blocks, block A, encountering less frictional resistance, moves more easily than block B. Therefore, after force F0 is applied for Δt amount of time, block A on the smooth surface will have moved faster, hence having a larger final velocity vA, than block B on the rough surface (whose final velocity is vB). This falls under the domain of Newton's second law of motion, which states that the acceleration of a body is directly proportional to, and in the same direction as, the net force acting on the body, and inversely proportional to its mass.

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The magnitude of the induced emf in the second coil can be expressed using the mutual inductance, maximum current in the first coil, and angular frequency.

(a) The magnitude of the induced emf in the second coil, ε2, can be expressed as ε2 = M * (dI2/dt), where M is the mutual inductance and dI2/dt is the rate of change of current in the second coil.

(b) The magnitude of ε2 can also be expressed as ε2 = M * I0 * ω * cos(ωt), where I0 is the maximum current in the first coil and ω is the angular frequency.

(c) The maximum value of |ε2|, εmax, can be calculated by taking the maximum value of the function ε2 = M * I0 * ω * cos(ωt) over one period.

(d) To calculate the numerical value of εmax, substitute the values of M, I0, and ω into the equation ε2 = M * I0 * ω * cos(ωt) and evaluate it at the maximum value of cos(ωt), which is 1.

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Answer:

q = 1,297 10⁻¹⁹ C , n=1

Explanation:

For this problem we will use Newton's second law in the case of equilibrium

        [tex]F_{e}[/tex] + B - W = 0

Where [tex]F_{e}[/tex] is the electrical force up, B the thrust and W the weight of the drop.

Let's look for weight and thrust

oil

      ρ = m / V

      m =  ρ V  

Air

       B =   [tex]\rho _{air}[/tex] g V

Electric force

       [tex]F_{e}[/tex] = qE

       E = V / d

       [tex]F_{e}[/tex] = q V/d

Let's replace

      q V / d + [tex]\rho _{air}[/tex] g V - ρ V g = 0

      qV / d = (4/3 π r³) g (ρ –[tex]\rho _{air}[/tex])

      q = 4/3 π r³ (ρ –[tex]\rho _{air}[/tex])  d / V

Reduce to SI units

      d = 1.87 cm (1m / 100cm) = 1.87 10⁻² m

      ρ= 0.816 r / cm3 (1kg / 1000g) (102cm / 1m)³ = 816 kg / m³

      [tex]\rho_{air}[/tex] = 1.28 kg / m³

Let's calculate the charge

       r = d / 2 = 3.3 10⁻⁶ m

      r = 1.65 10⁻⁶ m

     q = 4/3 π (1.65 10⁻⁶)³  (816 - 1.28)   0.0187 / 2210

     q = 12.9717 10⁻²⁰ C

     q = 1,297 10⁻¹⁹ C

If we assume that the load is

     q = n e

In this case n = 1

Answer:

a) -1.97 rad/sec² b) -8.09*10⁻³ N.m

Explanation:

a) Assuming a constant angular acceleration, we can apply the definition of angular acceleration, as follows:

γ = (ωf -ω₀) / t

We know that the final state of the grindstone is at rest, so ωf =0

In order to be consistent in terms of units, we can convert ω₀ from rev/min to rad/sec:

ω₀ = 730 rev/min* (1 min/60 sec)* (2*π rad / 1 rev) = 73/3*π

⇒ γ = (0-73/3*π) / 38.8 sec = - 1.97 rad/sec²

b) In order to get the value of the frictional torque exerted on the grindstone, that caused it to stop, we can apply the rotational equivalent of the Newton's 2nd law, as follows:

τ = I * γ (1)

As the grindstone can be approximated by a solid disk, the rotational inertia I can be expressed as follows:

I = m*r² / 2, where m=1.5 kg and r = 0.074 m.

Replacing in (1) , m. r and γ (the one we calculated in a)), we get:

τ = (1.5 kg* (0.074)² m² / 2) * -1.97 rad/sec = -8.09*10⁻³ N.m

(The negative sign implies that the frictional torque opposes to the rotation of the grindstone).

Answer

given,

frequency of sound (f)= 120 Hz

speed of sound (v)= 343 m/s

a) speed of approaching = 3 % of speed of sound

                          = 0.03 x 343 = 10.3 m/s

     frequency you would hear

     [tex]f' = f\dfrac{v+v_0}{v}[/tex]

     [tex]f' = 120 \times \dfrac{343+10.3}{343}[/tex]

            f' = 123.60 Hz

b)  speed of going away = 3 % of speed of sound

                          = 0.03 x 343 = 10.3 m/s

     frequency you would hear

     [tex]f' = f\dfrac{v-v_0}{v}[/tex]

     [tex]f' = 120 \times \dfrac{343-10.3}{343}[/tex]

            f' = 116.39 Hz

Answer:

Energy stored in inductor will be 20.797 J

Explanation:

We have given inductance L = 3.54 H

And resistance R = 7.76 ohm

Battery voltage V = 26.6 VOLT

After very long time means at steady state inductor behaves as short circuit

So  current [tex]i=\frac{V}{R}=\frac{26.6}{7.76}=3.427Amp[/tex]

Now energy stored in inductor [tex]E=\frac{1}{2}Li^2=\frac{1}{2}\times 3.54\times 3.427^2=20.797J[/tex]

So energy stored in inductor will be 20.797 J

To find the energy stored in an inductor in an RL circuit at equilibrium, use Ohm's law to calculate the steady current, and then apply the energy formula E = (1/2)LI^2.

When the switch in an RL series circuit is closed for a long time, the current in the circuit reaches an equilibrium value due to the voltage provided by the battery. The current, I, can be calculated using Ohm's law, I = V/R, where V is the voltage of the battery and R is the resistance in the circuit. In this case, with a 26.6 V battery and a 7.76 Ω resistor, the current would be I = 26.6 V / 7.76 Ω. After calculating the current, the energy stored in the inductor at equilibrium can be found using the formula for energy stored in an inductor, which is E = (1/2)LI^2, where L is the inductance and I is the current.

The corresponding energy stored in the inductor after a long time can be determined using these calculations.

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Answer:

Difference in Length of steel is 0.000246m

Difference in Length of invar is 0.00001845m

Difference in Length of steel surveyor's tape is 0.00738m

Difference in Length of invar surveyor's tape is 0.0005535m

Explanation:

Linear expansivity of steel is volume expansivity ÷ 3

Linear expansivity of steel = 0.000036/°C ÷ 3 = 0.000012/°C

Difference in Length of steel= Linear expansivity × initial length × temperature change

= 0.000012 × 1 × (20.5-0)

= 0.000012×1×20.5 = 0.000246m

Linear expansivity of invar = volume expansivity of invar ÷ 3

Linear expansivity of invar= 0.0000027/°C ÷ 3= 0.0000009/°C

Difference in Length of invar = 0.0000009×1×(20.5-0) = 0.00001845m

Difference in Length of steel surveyor's tape= 0.000012×30×(20.5-0) = 0.00738m

Difference in Length of invar surveyor's tape = 0.0000009×30×(20.5-0) = 0.0005535m

The difference in length between a steel and invar meter stick at 20.5°C is 6.83 mm, and for two 30-m-long surveyor's tapes, the difference at the same temperature is 20.5 cm.

When a steel meter stick and an invar meter stick both expand due to an increase in temperature, we can determine their difference in length using the coefficients of linear expansion for each material.

The difference in length at 20.5°C can be determined using the formula:

ΔL = αLΔT

For steel, the coefficient of linear expansion (α) is 3.6 × 10^-5 /°C, and for invar it is 2.7 × 10^-6 /°C. Plugging in the values:

The difference in length is therefore 7.38 × 10^-3 m - 5.54 × 10^-4 m = 6.83 × 10^-3 m or 6.83 mm.

For two 30.00-m-long surveyor’s tapes:

The difference in length for the surveyor's tapes is 2.21 × 10^-1 m - 1.66 × 10^-2 m = 2.05 × 10^-1 m or 20.5 cm.

The tensile strength and ductility cannot be directly calculated from the provided information, but the proportion of cold work (or percentage deformation) can be computed based on the diameter change.

Despite the detailed context provided, the question lacks sufficient data or formulas to directly compute the tensile strength nor the ductility of a copper rod from its diameter change due to cold working. Usually, the tensile strength and ductility mean the ultimate tensile strength (the maximum stress that a material can withstand while being stretched or pulled before failing or breaking) and the percent elongation after a material specimen has been pulled and rupture occurs. To get these values, experimenting with the material and measuring would be necessary.

However, we can calculate the percentage of cold work using the geometric deformation change. The percentage of cold work (%CW) based on change in diameter can be calculated with the formula: %CW = [(initial area - final area)/initial area] x 100%. Here, the area is that of the rod cross-section, which for a cylinder is pi*(d/2)². Thus you can substitute and calculate for %CW with the given diameters.

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Answer:

[tex]2\times 10^{-3}\ C[/tex]

6000

1.2 J

[tex]3.33\times 10^{-6}\ F[/tex]

Explanation:

I = Current = 1 A

t = Time = 2 ms

n = Number of electrocyte

V = Voltage = 100 mV

Charge is given by

[tex]Q=It\\\Rightarrow Q=1\times 2\times 10^{-3}\\\Rightarrow Q=2\times 10^{-3}\ C[/tex]

The charge flowing through the electrocytes in that amount of time is [tex]2\times 10^{-3}\ C[/tex]

The maximum potential is given by

[tex]V_m=nV\\\Rightarrow n=\dfrac{V_m}{V}\\\Rightarrow n=\dfrac{600}{100\times 10^{-3}}\\\Rightarrow n=6000[/tex]

The number of electrolytes is 6000

Energy is given by

[tex]E=Pt\\\Rightarrow E=V_mIt\\\Rightarrow E=600\times 1\times 2\times 10^{-3}\\\Rightarrow E=1.2\ J[/tex]

The energy released when the electric eel delivers a shock is 1.2 J

Equivalent capacitance is given by

[tex]C_e=\dfrac{Q}{V_m}\\\Rightarrow C_e=\dfrac{2\times 10^{-3}}{600}\\\Rightarrow C_e=3.33\times 10^{-6}\ F[/tex]

The equivalent capacitance of all the electrocyte cells in the electric eel is [tex]3.33\times 10^{-6}\ F[/tex]

a) The charge that flows through the electrocytes is [tex]\( 2 \text{ mC} \).[/tex]

b) 6000 electrolytes must be in series to produce the maximum shock. c)  the energy released when the electric eel delivers a shock is [tex]\( 0.6 \text{ J} \).[/tex]

d) the equivalent capacitance of all the electrocyte cells in the electric eel is approximately [tex]\( 3.33 \mu\text{F} \)[/tex]

To answer the questions about the electric eel's shock, we'll use the provided information and relevant physics formulas.

Part (a): We know:

- Current [tex]\( I = 1 \text{ A} \)[/tex]

- Time [tex]\( t = 2 \text{ ms} = 2 \times 10^{-3} \text{ s} \)[/tex]

The charge Q that flows can be found using the relation:

[tex]\[ Q = I \times t \][/tex]

Plugging in the values:

[tex]\[ Q = 1 \text{ A} \times 2 \times 10^{-3} \text{ s} \][/tex]

[tex]\[ Q = 2 \times 10^{-3} \text{ C} \][/tex]

[tex]\[ Q = 2 \text{ mC} \][/tex]

Part (b): We know:

- Maximum potential [tex]\( V = 600 \text{ V} \)[/tex]

- Potential of each electrocyte [tex]\( V_{\text{single}} = 100 \text{ mV} = 0.1 \text{ V} \)[/tex]

The number of electrocytes n in series required to produce 600 V can be found by:

[tex]\[ n = \frac{V}{V_{\text{single}}} \][/tex]

Plugging in the values:

[tex]\[ n = \frac{600 \text{ V}}{0.1 \text{ V}} \][/tex]

[tex]\[ n = 6000 \][/tex]

Part (c): We know:

- Voltage [tex]\( V = 600 \text{ V} \)[/tex]

- Charge [tex]\( Q = 2 \text{ mC} = 2 \times 10^{-3} \text{ C} \)[/tex]

The energy E released can be found using the relation:

[tex]\[ E = \frac{1}{2} Q V \][/tex]

Plugging in the values:

[tex]\[ E = \frac{1}{2} \times 2 \times 10^{-3} \text{ C} \times 600 \text{ V} \][/tex]

[tex]\[ E = \frac{1}{2} \times 1.2 \text{ J} \][/tex]

[tex]\[ E = 0.6 \text{ J} \][/tex]

Part (d):  Given:

- Total voltage [tex]\( V = 600 \text{ V} \)[/tex]

- Charge [tex]\( Q = 2 \times 10^{-3} \text{ C} \)[/tex]

The capacitance C can be found using the relation:

[tex]\[ C = \frac{Q}{V} \][/tex]

Plugging in the values:

[tex]\[ C = \frac{2 \times 10^{-3} \text{ C}}{600 \text{ V}} \][/tex]

[tex]\[ C = \frac{2 \times 10^{-3}}{600} \text{ F} \][/tex]

[tex]\[ C = \frac{1}{300} \times 10^{-3} \text{ F} \][/tex]

[tex]\[ C \approx 3.33 \times 10^{-6} \text{ F} \][/tex]

[tex]\[ C = 3.33 \mu\text{F} \][/tex]

Answer:

289.714 times bigger

Explanation:

[tex]\Delta x[/tex] = Uncertainty in position

[tex]\Delta p[/tex] = Uncertainty in momentum = [tex]\Delta v m[/tex]

[tex]\Delta v[/tex] = Uncertainty in velocity = [tex]2\times 10^3\ m/s[/tex]

h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]

m = Mass of electron = [tex]9.1\times 10^{-31}\ kg[/tex]

From the Heisenberg uncertainty principle we have

[tex]\Delta x\Delta p=\dfrac{h}{4\pi}\\\Rightarrow \Delta x\Delta v m=\dfrac{h}{4\pi}\\\Rightarrow \Delta x=\dfrac{h}{4\pi\Delta v m}\\\Rightarrow \Delta x=\dfrac{6.626\times 10^{-34}}{4\pi \times 2\times 10^3\times 9.1\times 10^{-31}}\\\Rightarrow \Delta x=2.89714\times 10^{-8}\ m[/tex]

Comparing with 0.1 nm size atom

[tex]\dfrac{\Delta x}{x}=\dfrac{2.89714\times 10^{-8}}{0.1\times 10^{-9}}\\\Rightarrow \dfrac{\Delta x}{x}=289.714[/tex]

So, the electron’s minimum uncertainty in position is 289.714 times bigger than an atom of size 0.1 nm

The Heisenberg Uncertainty Principle in quantum physics states a limit to how precisely the position and momentum of a particle, such as an electron, can be known simultaneously. In the given context, though there's a large uncertainty in the electron's velocity, the uncertainty in its position within the atom remains significantly small, thus reflecting the inverse relationship between the precision of these two measurements.

The concept involved in this question is called the Heisenberg Uncertainty Principle which is a fundamental theory in quantum mechanics. This theory describes a limit to how precisely we can know both the simultaneous position of an object (such as an electron) and its momentum.

Using the principles of uncertainty (ΔxΔp ≥ h/4π), where h is the Planck constant, the uncertainty in position (Δx) is given. If the given velocity (v) of the electron is 2.0×103m/s with accuracy or uncertainty (Δv), the minimum uncertainty in position can be calculated, where the product of the uncertainties in position and velocity equal to or greater than Planck's constant divided by 4π, i.e. Δx ≥ h / (4πmΔv).

After calculation, it shows that the uncertainty in the electron's position within the atom is very small. Even though the uncertainty in velocity is large, the uncertainty in position remains smaller compared to the size of the atom. Hence, this represents the principle that increasing precision in measuring one quantity leads to greater uncertainty in the measurement of the other quantity.

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Sound intensity (energy) falls inversely proportional to the square of the distance  from the sound:

[tex]I \propto \frac{1}{r^2}[/tex]

Therefore if we have two values of intensities we have

[tex]\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}[/tex]

As we have that

[tex]r_1 = 3 r_2[/tex]

Then we have that

[tex]\frac{I_1}{I_2} = \frac{r_2^2}{(3r_2)^2}[/tex]

[tex]\frac{I_1}{I_2} = \frac{1}{9}[/tex]

Therefore the correct answer is D. The sound intensity drops to 1 / 9 of its original value.

Entropy is the energy that cannot be used to do useful work.

The correct answer is b. entropy. Entropy is a measure of the disorder or randomness in a system. It is a form of energy that cannot be converted into useful work. For example, when energy is transferred from one form to another, such as from chemical energy to thermal energy, some of the energy is lost as heat, which is a form of entropy.

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Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

Answer:

d) larger than Lily's

Explanation:

If the merry-go-round is rotating at a constant angular speed, this means that any two points, located at different positions along a radius, rotate at the same angular speed, which means that they sweep the same angle at a given time interval.

In order to both points keep aligned along the same radius, we have a single choice (assuming that we are talking about a rigid  body) to meet this premise:

The point farther of the center (Bob) must have a linear speed greater than a point closer to the center (Lily).

Mathematically, we can explain this result as follows:

ω = Δθ / Δt (by definition of angular velocity) (1)

but, by definition of angle, we can say the following:

θ = s/r , where s is the arc along the circumference, and r, the radius.

⇒Δθ = Δs /r

Replacing in (1) we have:

ω = (Δs /Δt) / r

By definition, Δs/Δt = v, so, arranging terms, we get:

v = ω*r

If ω=constant, if r increases, v increases.

So, as Bob is at a distance r from the center larger than Lily's, Bob's speed must be larger than Lily's.

Regardless of their position, Bob and Lily will rotate at the same angular speed. However, due to being further from the center of rotation, Bob's linear speed will be greater than Lily's.

The subject matter of the question revolves around rotational motion, specifically concerning angular speed, and how location on a rotating reference frame, such as a merry-go-round, affects linear speed. Both Bob and Lily, being on the same merry-go-round, share the same angular speed (w), regardless of their location on the platform.

The linear speed of an object in rotational motion is given by: Speed = radius x angular speed. Given that Bob is on the outer edge (larger radius) while Lily is near the center (smaller radius), it's evident that Bob's linear or tangential speed (how fast he's actually moving along a path) will be larger than Lily's for the same angular speed. Therefore, the answer is d) Bob's speed is larger than Lily's.

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Answer:

I think d is the answer haha

Galaxy collisions were more common in the past as C. the universe was smaller, hence galaxies were closer together.

Galaxy collisions were more common in the past than they are today primarily because galaxies were closer together in the past due to the smaller size of the universe. As the universe expanded, it carried galaxies along with it, increasing the distances between them and making collisions less frequent. When the universe was younger, galaxy interactions and collisions were frequent events that significantly influenced galaxy evolution, increased star-formation rates, and contributed to the prevalence of quasars during that time.

Observations indicate that when the universe was about 20% of its current age, interactions such as galaxy mergers happened most frequently, likely accounting for the active quasar populations observed from that epoch. These collisions often resulted in starburst galaxies, and the debris from these encounters could fuel the supermassive black holes at the centers of galaxies.

The distance between your eye and the image of the spider in the mirror is 85 cm. This is found by adding the distance from the spider to the mirror and from your eyes to the mirror.

The situation regards the properties of reflection in a mirror. In a plane mirror, the image of an object appears to be the same distance behind the mirror as the object is in front of the mirror. So the image of the spider in the mirror is 20cm behind the mirror. The eye's distance from the mirror is 65cm. Therefore, the distance from your eye to the image of the spider in the mirror is the sum of these two distances: 20cm (distance from the spider to the mirror) + 65cm (distance from your eye to the mirror) = 85cm.

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Answer:

0.5 mph in the opposite direction

Explanation:

[tex]m_1[/tex] = Mass of Cadillac = 1000 kg

[tex]v_1[/tex] = Velocity of Cadillac = 1 mph

[tex]m_2[/tex] = Mass of Volkswagen = 2000 kg

[tex]v_2[/tex] = Velocity of Volkswagen

In order to know the speed the system must have the momentum exchange

As the linear momentum of the system is conserved

[tex]m_1v_1+m_2v_2=0\\\Rightarrow v_2=-\dfrac{m_1v_1}{m_2}\\\Rightarrow v_2=\dfrac{1000\times 1}{2000}\\\Rightarrow v_2=-0.5\ mph[/tex]

The speed of the impact is given by 0.5 mph in the opposite direction

Final answer:

To bring the Cadillac to a halt using a 2000 kg Volkswagen, we can utilize the conservation of momentum. The necessary impact speed for the Volkswagen is calculated to be approximately 0.22352 m/s, opposing the direction of the Cadillac's motion.

Explanation:

To stop the Cadillac using a head-on collision with the Volkswagen, we must apply the principle of conservation of momentum which states that the total momentum of a system remains constant if no external forces act upon it. Given that both vehicles will come to a halt after the collision, we can set their combined momentum to zero.

Calculating Impact Speed

The Cadillac's momentum is its mass times its velocity (mass of Cadillac ×speed of Cadillac). Converting 1 mph to meters per second (approximately 0.44704 m/s), we get the Cadillac's momentum as 1000 kg × 0.44704 m/s.

The Volkswagen's mass is 2000 kg and we want to find out with what speed it should hit the Cadillac to bring both to a halt. Let's denote this unknown speed as v. The momentum of the Volkswagen right before the impact is 2000 kg × v.

Using conservation of momentum:

Total momentum before collision = Total momentum after collision

(1000 kg × 0.44704 m/s) + (2000 kg × v) = 0

447.04 kg× m/s + 2000 kg × v = 0

To find the speed v, we'll solve the equation:

2000 kg × v = -447.04 kg× m/s

v = -447.04 kg× m/s / 2000 kg

v = -0.22352 m/s

The negative sign indicates that the Volkswagen must be traveling in the opposite direction of the Cadillac's motion, which we already know. Thus, the required speed for the Volkswagen for a head-on impact to bring the Cadillac to a halt is approximately 0.22352 m/s.

Answer:

0.243

Explanation:

Step 1: Identify the given parameters

Force (f) = 5kN, length of pitch (L) = 5mm, diameter (d) = 25mm,

collar coefficient of friction = 0.06 and thread coefficient of friction = 0.09

Frictional diameter =45mm

Step 2: calculate the torque required to raise the load

[tex]T_{R} = \frac{5(25)}{2} [\frac{5+\pi(0.09)(25)}{\pi(25)-0.09(5)}]+\frac{5(0.06)(45)}{2}[/tex]

[tex]T_{R}[/tex] = (9.66 + 6.75)N.m

[tex]T_{R}[/tex] = 16.41 N.m

Step 3: calculate the torque required to lower the load

[tex]T_{L} = \frac{5(25)}{2} [\frac{\pi(0.09)(25) -5}{\pi(25)+0.09(5)}]+\frac{5(0.06)(45)}{2}[/tex]

[tex]T_{L}[/tex] = (1.64 + 6.75)N.m

[tex]T_{L}[/tex] = 8.39 N.m

Since the torque required to lower the thread is positive, the thread is self-locking.

The overall efficiency = [tex]\frac{F(L)}{2\pi(T_{R})}[/tex]

                        = [tex]\frac{5(5)}{2\pi(16.41})}[/tex]

                        = 0.243

The overall efficiency of the power screw is approximately 38.6%. The torque required to raise the load is 46.8 N·m, while the torque to lower the load is about 22.53 N·m.

Calculating Lead and Mean Diameter:Lead (L) = Pitch (p) = 5 mm

Mean diameter (dm) = 25 mm

Calculating Torque to Raise the Load ([tex]T_{up[/tex]):

Using the formula: [tex]T_{up[/tex] = (W dm/2) (L + π μt dm)/(π dm - μt L)

Where:

W = 5 kN = 5000 N

μt = 0.09

dm = 25 mm = 0.025 m

L = 5 mm = 0.005 m

Substituting the values:

[tex]T_{up[/tex] = (5000 × 0.025/2) (0.005 + π × 0.09 × 0.025)/(π × 0.025 - 0.09 × 0.005)

[tex]T_{up[/tex] ≈ 46.8 N·m

Calculating Torque to Lower the Load ([tex]T_{down[/tex]):

Using the formula: [tex]T_{down[/tex] = (W dm/2) (L - π μt dm)/(π dm + μt L)

Substituting the values:

[tex]T_{down[/tex] = (5000 × 0.025/2) (0.005 - π × 0.09 × 0.025)/(π × 0.025 + 0.09 × 0.005)

[tex]T_{down[/tex] ≈ 22.53 N·m

Calculating Overall Efficiency (η):

Efficiency η = (tan θ / (tan(θ + φ)))

Where:

θ = tan-1(L/π dm)

φ = tan-1(μt)

θ ≈ 0.064

φ ≈ 0.09

η ≈ (tan 0.064) / (tan(0.064 + 0.09))

η ≈ 38.6%

Therefore, the overall efficiency of the power screw is approximately 38.6%, the torque to raise the load is 46.8 N·m, and the torque to lower the load is around 22.53 N·m.

Answer:

a)  v = 160.9 Km / h, b) K = 9,988 10⁵ J , c) d = 335.2 m , d) a = - 2.9796 m / s²

Explanation:

a) The average speed is defined as the distance traveled in the time interval

      v = d / t

Let's reduce the magnitudes to the SI system

      t = 30 min (60 s / 1min) = 1800 s

      d = 50 mile (1609 m / 1mile) = 80450 m

      v = 80450/1800

      v = 44.6944 m / s

      v = 44.6944 m / s (1 km / 1000m) (3600 s / 1h)

      v = 160.9 Km / h

b)   W = 2 ton (1000 kg / 1 ton) = 1000 kg

      K = ½ m v²

      K = ½ 1000  44.694²

      K = 9,988 10⁵ J

c) take t = 15 s to stop

    v = v₀ - at

    v = 0

    a = v₀ / t

    a = 44,694 / 15

    a = 2.9796 m / s²

    v² = v₀² - 2 a d

    v = 0

    d = v₀² / 2 a

    d = 44.694² / (2 2.9796)

    d = 335.2 m

d) the average acceleration is the change of speed in the time interval

   a = (v - v₀) / (t -t₀)

   a = (0 - 44.694) / 15

   a = - 2.9796 m / s²

To solve this problem we will use the mathematical definition of the light years in metric terms, from there, through the kinematic equations of motion we will find the distance traveled as a function of the speed in proportion to the elapsed time. Therefore we have to

[tex]1Ly =9.4605284*10^{15}m \rightarrow 'Ly'[/tex]means  Light Year

Then

[tex]14.4Ly = 1.36231609*10^{17} m[/tex]

If we have that

[tex]v= \frac{x}{t} \rightarrow t = \frac{x}{t}[/tex]

Where,

v = Velocity

x = Displacement

t = Time

We have that

[tex]t = \frac{1.36231609*10^{17}}{0.96c} \rightarrow c[/tex]= Speed of light

[tex]t = \frac{1.36231609*10^{17}}{0.96(3*10^8)}[/tex]

[tex]t= 454105363 s (\frac{1hour}{3600s})[/tex]

[tex]t= 126140 hours(\frac{1day}{24hours})[/tex]

[tex]t= 5255.85 days(\frac{1 year}{365days})[/tex]

[tex]t = 14.399 years[/tex]

Therefore will take 14.399 years

It takes Rob approximately 15.0 years to complete the trip relative to a frame of reference fixed with respect to Earth.

  To solve this problem, we can use the concept of time dilation from the theory of special relativity. According to this concept, time measured in a frame of reference moving at a high velocity relative to a stationary observer (in this case, Earth) will be dilated, or ""slowed down,"" compared to the time measured by the stationary observer.

 The time dilation formula is given by:

 [tex]\[ t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}} \][/tex]

 where:

- [tex]\( t' \)[/tex]is the time interval measured by the moving observer (Rob),

- [tex]\( t \)[/tex]is the time interval measured by the stationary observer (Earth),

-[tex]\( v \)[/tex] is the relative velocity between the two frames of reference, and

- [tex]\( c \)[/tex] is the speed of light in a vacuum.

 Given that Rob's speed is [tex]\( 0.960c \)[/tex]and the distance to the star system is [tex]\( 14.4 \)[/tex] light-years, we can calculate the time it takes for Rob to complete the trip according to Earth's frame of reference.

 First, we calculate the Lorentz factor [tex]\( \gamma \),[/tex] which is the factor by which time is dilated:

[tex]\[ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \][/tex]

 Substituting \( v = 0.960c \):

 [tex]\[ \gamma = \frac{1}{\sqrt{1 - (0.960)^2}} \][/tex]

[tex]\[ \gamma = \frac{1}{\sqrt{1 - 0.9216}} \][/tex]

[tex]\[ \gamma = \frac{1}{\sqrt{0.0784}} \][/tex]

[tex]\[ \gamma = \frac{1}{0.280} \][/tex]

[tex]\[ \gamma \approx 3.5714 \][/tex]

 Now, we calculate the time [tex]\( t \)[/tex] it takes for Rob to travel 14.4 light-years at a speed of [tex]\( 0.960c \):[/tex]

[tex]\[ t = \frac{d}{v} \][/tex]

[tex]\[ t = \frac{14.4 \text{ light-years}}{0.960c} \][/tex]

[tex]\[ t = \frac{14.4}{0.960} \text{ years} \][/tex]

[tex]\[ t = 15 \text{ years} \][/tex]

 Since the Lorentz factor [tex]\( \gamma \)[/tex] is approximately 3.5714, the time measured on Earth for Rob's journey, taking into account time dilation, would be:

[tex]\[ t' = \gamma \cdot t \][/tex]

[tex]\[ t' \approx 3.5714 \cdot 15 \text{ years} \][/tex]

[tex]\[ t' \approx 53.571 \text{ years} \][/tex]

 However, since the question asks for the time according to the frame of reference fixed with respect to Earth, we do not need to multiply by the Lorentz factor. The time [tex]\( t \)[/tex] is already the time measured by the stationary observer on Earth.

 Therefore, the time it takes Rob to complete the trip relative to Earth is approximately 15.0 years.

Answer:

74 N

Explanation:

T = Tension in the string = 74 N

[tex]\rho[/tex] = Density of the steel = 7800 kg/m³

A = Area = [tex]\pi r^2[/tex]

r = Radius = [tex]8.5\times 10^{-4}\ m[/tex]

Linear density is given by

[tex]\mu=\rho A\\\Rightarrow \mu=7800\times \pi (8.5\times 10^{-4})^2\\\Rightarrow \mu=0.0177\ kg/m[/tex]

The linear density is 0.0177 kg/m

Velocity is given by

[tex]v=\sqrt{\dfrac{T}{\mu}}\\\Rightarrow v=\sqrt{\dfrac{74}{0.0177}}\\\Rightarrow v=64.65903\ m/s[/tex]

The velocity of the wave on the guitar string is 64.65903 m/s

Answer:

Explanation:

Fission means breaking or splitting i.e. Nuclear fission involves breaking up of unstable nucleus into smaller pieces and simultaneously producing energy.  

Fusion means joining of different elements to form a unified whole. In Nuclear fusion smaller nuclei combined to form a new heavy nucleus associated with a large amount of energy release.

Thus nuclear Fusion and fission can release Energy.