A heat engine uses two containers held at different temperatures. One container is at 294 K 294 K , while the other is kept at 552 K 552 K . What is the maximum possible efficiency for this engine?

Answer:

(a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

Explanation:

Given that,

Tensile strength = 196 MPa

Maximum strain = 0.380

Diameter = 50.0 μm

Length = 12.0 cm

We need to calculate the area

Using formula of area

[tex]A=\dfrac{\pi}{4}\times d^2[/tex]

Put the value into the formula

[tex]A=\dfrac{\pi}{4}\times(50.0\times10^{-6})^2[/tex]

[tex]A=1.96\times10^{-9}\ m^2[/tex]

We need to calculate the magnitude of the force

Using formula of force

[tex]F=\sigma A[/tex]

Put the value into the formula

[tex]F=196\times10^{6}\times1.96\times10^{-9}[/tex]

[tex]F=0.38416\ N[/tex]

(b). If the length of a strand of the hair is 12.0 cm at its breaking point

We need to calculate the unstressed length

Using formula of strain

[tex]strain=\dfrac{\Delta l}{l_{0}}[/tex]

[tex]\Delta l=strain\times l_{0}[/tex]

Put the value into the formula

[tex]\Delta l=0.380\times l_{0}[/tex]

Length after expansion is 12 cm

We need to calculate the original length

Using formula of length

[tex]l=l_{0}+\Delta l[/tex]

Put the value into the formula

[tex]I=l_{0}+0.380\times l_{0}[/tex]

[tex]l=1.38l_{0}[/tex]

[tex]l_{0}=\dfrac{l}{1.38}[/tex]

[tex]l_{0}=\dfrac{12\times10^{-2}}{1.38}[/tex]

[tex]l_{0}=0.0869\ m[/tex]

Hence, (a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

Answer:

The distance covered by the lighting flash is 1700 meters.

Explanation:

Given that,

The speed of sound travels in air at a speed of about 340 meter second, v = 340 m/s

The sound of thunder is heard 5 seconds after the flash of lightning is seen, t = 5 s

To find,

The distance far away was the lightning flash.

Solution,

The distance covered by the lighting flash is given by the product of speed and time. It is given by :

[tex]d=v\times t[/tex]

[tex]d=340\ m/s\times 5\ s[/tex]

d = 1700 meters

So, the distance covered by the lighting flash is 1700 meters. Therefore, this is the required solution.

a) The equation 1/2 mv² = 1/2 mv0² + mgh is not dimensionally consistent; b)  The equation v = v0 + at² is not dimensionally consistent; c) The equation ma = v² is not dimensionally consistent.

(a) The left-hand side of the equation has dimensions of energy (M L² T⁻²), while the right-hand side has dimensions of energy plus length (M L² T⁻² + ML).

This is because the term mgh represents the potential energy gained by an object of mass m when it is lifted to a height h above the ground. Therefore, the dimensions of the two sides of the equation do not match, and the equation cannot be correct.

(b)The left-hand side of the equation has dimensions of velocity (L T⁻¹), while the right-hand side has dimensions of velocity plus time squared (L T⁻¹ + T²).

This is because the term at² represents the change in velocity over time due to acceleration. Therefore, the dimensions of the two sides of the equation do not match, and the equation cannot be correct.

(c) The left-hand side of the equation has dimensions of force (M L T⁻²), while the right-hand side has dimensions of velocity squared (L² T⁻²).

This is because the term v² represents the square of the velocity of an object. Therefore, the dimensions of the two sides of the equation do not match, and the equation cannot be correct.

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The dimensional analysis shows that equation (a) is correct while equations (b) and (c) are incorrect. Equation (b) is missing a time component, while equation (c) incorrectly equates force to velocity squared.

Let's analyze each equation in terms of their dimensions:

By comparing dimensions on both sides of an equation, we can spot if the equations are incorrect or not.

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To solve this problem we will apply the concepts given by the kinematic equations of motion. For this purpose it will be necessary with the given data to obtain the deceleration. With this it will be possible again to apply one of the kinematic equations of motion that does not depend on time, but on distance, to find how far the block would slide with the quadruplicate velocity

Our values are given as,

[tex]\text{Initial speed} =V_i = 2.3 m/s[/tex]

[tex]\text{Final speed}= V_f = 0 m/s[/tex]

[tex]\text{Stopping distance = }d = 1.9 m[/tex]

[tex]a = acceleration[/tex]

[tex]\text{mass} = m = 3.1kg[/tex]

Using the kinematic equation of motion we have

[tex]V_f^2 = V_i^2 + 2 a d[/tex]

[tex]0^2 = 2.3^2 + 2 a (1.9)[/tex]

[tex]a = -1.39211 m/s^2[/tex]

Now if the initial velocity is quadrupled we have that,

[tex]\text{Initial speed} =V_i' = 2.3*4 m/s = 9.2m/s[/tex]

[tex]\text{Final speed}= V_f' = 0 m/s[/tex]

[tex]\text{Stopping distance = }d'[/tex]

[tex]V_f^2' = V_i^2' + 2 a d'[/tex]

Replacing the values

[tex]0^2 = 9.2^2+ 2 (-1.39211) d'[/tex]

[tex]d' = 30.44m[/tex]

Therefore the block would have slipped around 30.44 if its initial velocity quadrupled.

The question is about calculating the electric force on a point charge near an infinitely long wire using the concepts of electric fields and Coulomb's law within the domain of Physics.

The question involves calculating the magnitude of the electric force that an infinitely long, thin wire exerts on a point charge placed at a certain distance from it. This falls under the subject of Physics, specifically dealing with electrostatics and the concept of electric fields and forces.

First, we need to find the electric field due to the wire at the location of the point charge. The electric field E from an infinitely long wire is given by the formula E = (2kλ)/r, where k is Coulomb's constant (8.99 × 109 Nm2/C2), λ is the linear charge density of the wire, and r is the distance from the wire to the point where the electric field is being calculated.

After calculating E, we can find the force exerted on the charge q using F = qE. Since we're given q = -4.00 nC and λ = 3.00×10−9C/m, with r = 9.00 cm, we can substitute these values into the formulas to calculate the electric field and then the force acting on the point charge.

The magnitude of the electric force exerted by the wire on the point charge is 2.40 × 10⁻⁶ N.

To find the magnitude of the electric force that an infinitely long, thin wire with linear charge density λ = 3.00×10⁻⁹ C/m exerts on a point charge q = -4.00 nC placed a distance of 9.00 cm from the wire, we use the principle that an electric field E is created by a line of charge.

The electric field created by an infinitely long, thin wire is given by:

E =  (2kλ)/r

where  k = 8.99 × 10⁹ Nm²/C²

            λ = 3.00×10⁻⁹ C/m , charge density

             r = 9 × 10⁻² m

Substitute the given values:

E = (2 ×  8.99 × 10⁹ Nm²/C² × 3.00×10⁻⁹ C/m ) / 9 × 10⁻² m

Calculating this, we get:

E ≈ 600 N/C

The force F on the point charge q due to the electric field E is given by:

F = q × E

Since q = -4.00 nC = -4.00 × 10⁻⁹ C:

F = (-4.00 × 10⁻⁹ C) × 600 N/C ≈ -2.40 × 10⁻⁶ N

Therefore, the magnitude of the electric force is 2.40 × 10⁻⁶N.

Answer:

Electric field magnitude

E = K/qd

Where

K = kinetic energy of electron

d = electron distance

q = charge

Explanation:

Given the relationship between workdone and energy

Work-energy theorem:

Net workdone = Energy change

W = ∆E

In this case

W = ∆K.E

And,

∆K.E = K(final) - K(initial)

To stop the kinetic energy | K(final) = 0

K(initial) = K (given)

∆K.E = 0 - K = -K

Let the electric force on the electron has magnitude F.

And

W = -Fd = ∆K.E = -K

-Fd = -K

F = K/d .....1

The magnitude of the electric field E that can stop these electron in a distance d:

E = F/q ......2

Where q is the charge on electron.

substituting equation 1 to 2

E = (K/d)/q = K/qd

E = K/qd

Explanation:

(a) Velocity is given by :

[tex]v=\dfrac{ds}{dt}[/tex]

s is the length of the distance

t is the time

The dimension of v will be, [tex][v]=[LT^-1][/tex]      

(b) The acceleration is given by :

[tex]a=\dfrac{dv}{dt}[/tex]

v is the velocity

t is the time

The dimension of a will be, [tex][a]=[LT^{-2}][/tex]

(c) Since, [tex]d=\int\limits{v{\cdot}dt} =[LT^{-1}][T]=[L][/tex]

(d) Since, [tex]v=\int\limits{a{\cdot}dt} =[LT^{-2}][T]=[LT^{-1}][/tex]

(e)

[tex]\dfrac{da}{dt}=\dfrac{[LT^{-2}]}{[T]}[/tex]

[tex]\dfrac{da}{dt}=[LT^{-3}]}[/tex]

Hence, this is the required solution.

Answer:

a) p=0, b) p=0, c) p= ∞

Explanation:

In quantum mechanics the moment operator is given by

              p = - i h’  d φ / dx

             h’= h / 2π

We apply this equation to the given wave functions

a)  φ = [tex]e^{ikx}[/tex]

        .d φ dx = i k [tex]e^{ikx}[/tex]

We replace

        p = h’ k [tex]e^{ikx}[/tex]

        i i = -1

The exponential is a sine and cosine function, so its measured value is zero, so the average moment is zero

            p = 0

b) φ = cos kx

           p = h’ k sen kx

The average sine function is zero,

          p = 0

c) φ = [tex]e^{-ax^{2} }[/tex]

         d φ / dx = -a 2x  [tex]e^{-ax^{2} }[/tex]

         .p = i a g ’2x  [tex]e^{-ax^{2} }[/tex]

       The average moment is

         p = (p₂ + p₁) / 2

         p = i a h ’(-∞ + ∞)

         p = ∞

Answer:

a)   v_average = 11 m / s, b)  t = 0.0627 s

, c)    F = 7.37 10⁵ N

, d)   F / W = 35.8

Explanation:

a) truck speed can be found with kinematics

         v² = v₀² - 2 a x

The fine speed zeroes them

           a = v₀² / 2x

           a = 22²/2 0.69

           a = 350.72 m / s²

The average speed is

           v_average = (v + v₀) / 2

           v_average = (22 + 0) / 2

           v_average = 11 m / s

b) The average time

          v = v₀ - a t

          t = v₀ / a

          t = 22 / 350.72

          t = 0.0627 s

c) The force can be found with Newton's second law

             F = m a

             F = 2100 350.72

             F = 7.37 10⁵ N

.d) the ratio of this force to weight

             F / W = 7.37 10⁵ / (2100 9.8)

             F / W = 35.8

.e) Several approaches will be made:

- the resistance of air and tires is neglected

- It is despised that the force is not constant in time

- Depreciation of materials deformation during the crash

Answer:

Development in science:

There a number of individuals in the world around us who has improved the field of science and made it easy for us to know this universe more easily then ever. This universe is still unexplored and there are going to be more then billions of stars which are still not studied and has an immense amount of information regarding the universe.

Stephen Hawking: The scientist who made it easy for each of us to know the universe and more over the cosmos in a more detailed form, as before this no one knew about the different patterns of entities and there properties.

A genius who was handicapped:

He was a genius in cosmology, mathematics, and other subjects of science as he was diagnosed by the amyotrophic lateral sclerosis(ALS), which made it unable for him to live a normal life.But, he communicated through a computer which detected his nerve signals and shared his thoughts about any thing or subject.

Why him?

Stephen hawking was just tremendous in making it more understandable about the black holes, time worms, and space etc.As it was known to very small number of people before he took the stage.

Answer:

34.9 kg

Explanation:

Since there are no net external forces on the system, the center of gravity does not move.

Let's say that m is the mass of the man, M is the mass of the boat, and L is the length of the boat.

When the man is at the stern, the distance between the center of gravity and the pier is:

CG = (m L + M (L/2)) / (m + M)

When the man reaches the prow, the distance between the center of gravity and the pier is:

CG = (m x + M (x + L/2)) / (m + M)

Since these are equal:

(m L + M (L/2)) / (m + M) = (m x + M (x + L/2)) / (m + M)

m L + M (L/2) = m x + M (x + L/2)

m L + M (L/2) = m x + M x + M (L/2)

m L = m x + M x

m L − m x = M x

m (L − x) = M x

M = m (L − x) / x

Plugging in values:

M = 89.3 kg (5.8 m − 4.17 m) / 4.17 m

M = 34.9 kg

The required mass of boat is 34.9 kg.

The given problem is based on the concept of the center of mass. The point of analysis where the entire mass of the system is known to be concentrated is known as the center of mass.

Given data:

The mass of man is, m = 89.3 kg.

The length of boat is, L = 5.8 m.

The distance away from the pier is, d = 4.17 m.

Since there are no net external forces on the system, the center of gravity does not move. Let's say that m is the mass of the man, M is the mass of the boat

When the man is at the stern, the distance between the center of gravity and the pier is:

CG = (m L + M (L/2)) / (m + M)

When the man reaches the prow, the distance between the center of gravity and the pier is:

CG = (m d + M (d + L/2)) / (m + M)

Since these are equal:

(m L + M (L/2)) / (m + M) = (m d + M (d + L/2)) / (m + M)

m L + M (L/2) = m d + M (d + L/2)

m L + M (L/2) = m d + M d + M (L/2)

Further solving as,

m L = m d + M d

m L − m d = M d

m (L − x) = M x

M = m (L − x) / x

M = 89.3 kg (5.8 m − 4.17 m) / 4.17 m

M = 34.9 kg

Thus, we can conclude that the required mass of boat is 34.9 kg.

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Answer:

c. 1.2×10¹⁵ Hz

Explanation:

Work Function: This is the minimum amount of energy a photon requires to liberate an electron from the surface of a metal.

Mathematically, it can be represented as

E' = hf' ................................... Equation 1

Where E' = work function of gold, f' = minimum frequency ( threshold frequency), h = Planck's constant

Making f' the subject of the equation,

f' = E'/h................................. Equation 2

Given: E' = 4.8 ev = 4.8×1.6×10⁻¹⁹ J = 7.68×1.6×10⁻¹⁹ J, h = 6.626×10⁻³⁴Js.

Substituting into equation 2

f' = 7.68×10⁻¹⁹/ 6.626×10⁻³⁴

f' = 1.16×10¹⁵ Hz.

f' ≈ 1.2×10¹⁵ Hz

Thus the minimum frequency =  1.2×10¹⁵ Hz

The right option is c. 1.2×10¹⁵ Hz

Using Planck's equation, the minimum frequency of photons required to remove an electron from gold, given a work function of 4.8 eV, is approximately 1.15 * 10^15 Hz or 1.15 PHz.

The question is asking for the minimum frequency of photons capable of removing electrons from gold. This is related to the photoelectric effect, where electrons can be ejected from a metal surface by incident light. In this process, energy of the incoming light is absorbed by the electron, which can then be ejected if the light's energy is greater than the binding energy of the electron, which is also known as the work function.

The work function for gold is given as 4.8 eV. To calculate the minimum frequency, we need to employ Planck's equation, E = hf, which implies frequency, f = E/h. Given that E is the energy of the photon which must be equal to the work function (4.8 eV or 4.8 * 1.6 x 10^-19 J) and h is Planck’s constant (6.626 x 10^-34 J s), we can solve for f. The calculation gives approximately 1.15 * 10^15 Hz or 1.15 PHz, which is not listed among the provided answer choices.

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The car accelerates from 0 m/s to 19.1 m/s in 101 m, with an acceleration of approximately 1.81 m/s².

Given  that,

The car is stopped at a traffic light on a straight road.

When the light turns green, the car accelerates.

The car reaches a speed of 19.1 m/s (68.8 km/h).

The distance covered during acceleration is 101 m

To calculate the acceleration of the car,

Use the kinematic equation:

v² = u² + 2as

Where,

v is the final velocity (19.1 m/s in this case),

u is the initial velocity (0 m/s since the car is stopped),

a is the acceleration that we're trying to find,

s is the distance covered (101 m).

Plugging in the known values, we have:

19.1² = 0² + 2a(101)

Simplifying further:

365.81 = 202a

Now, let's solve for a:

a = 365.81 / 202

Calculating that, we find:

a ≈ 1.81 m/s²

Hence,

The acceleration of the car is approximately 1.81 m/s².

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The acceleration of the car is 1.85 m/s^2.

To calculate the acceleration of the car, we can use the equation:

v^2 = u^2 + 2as

Where:

Plugging in the given values, we have:

19.1^2 = 0 + 2a(101)

Simplifying the equation gives us:

a = (19.1^2) / (2 * 101) = 1.85 m/s^2

Therefore, the acceleration of the car is 1.85 m/s^2.

Answer:

No

Explanation:

Although the direction of position, velocity or acceleration of an object in marry-go-round problem changes continuously,however the magnitude of the position, velocity and acceleration do remains the same. Marry-go-round is nothing but a machine found in fairs that turn round in circular motion. So, the laws of circular motion are applicable in it.

By employing Coulomb's Law and additional physical principles like Newton's third law, it becomes possible to determine the characteristics of an unknown electrical charge from the force it experiences and exerts.

The question concerns two charges, one of which is negative and known (-0.510 µC), and the other is unknown. You're asked to find the value of this unknown charge and the force it exerts on the known charge. To do this, we would use Coulomb's Law, which states that the electric force between two charges is directly proportional to the product of their charges, and inversely proportional to the square of the distance separating them.

So, using the formula F = k*q1*q2/r^2, where F is the force, k is Coulomb's constant (approximately 9 * 10^9 N.m^2/C^2), q1 and q2 are the charges, and r is the distance between them, we can solve for the unknown charge(q2) - which gives us q2 = Fr^2/(kq1). Substituting the provided values in the question, we should be able to solve it.

For part (b), the sign of the force that the unknown charge exerts on the known charge would be exactly opposite in direction to the force it experiences, as per Newton's third law (equal and opposite reactions). However, we'd need to know the magnitude of the unknown charge to calculate the actual force.

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the magnitude of the charge of object A is[tex]\( 2.78 \times 10^{-10} \, \text{C} \).[/tex]

We can use the formula for electric field intensity ( E ) to calculate the charge of object A. The electric field intensity is given by:

[tex]\[ E = \frac{k \cdot |q|}{r^2} \][/tex]

Where:

- ( E ) is the electric field intensity (40.0 N/C),

- ( k) is Coulomb's constant [tex](\( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)),[/tex]

- [tex]\( |q| \)[/tex] is the magnitude of the charge on object A (what we're trying to find), and

- ( r ) is the distance from object A to point P (0.250 m).

First, rearrange the formula to solve for[tex]\( |q| \)[/tex]:

[tex]\[ |q| = \frac{E \cdot r^2}{k} \][/tex]

Now, substitute the given values into the formula:

[tex]\[ |q| = \frac{40.0 \, \text{N/C} \cdot (0.250 \, \text{m})^2}{8.99 \times 10^9 \, \text{N m}^2/\text{C}^2} \]\[ |q| = \frac{40.0 \times 0.0625}{8.99 \times 10^9} \, \text{C} \]\[ |q| = \frac{2.5}{8.99 \times 10^9} \, \text{C} \]\[ |q| = 2.78 \times 10^{-10} \, \text{C} \][/tex]

So, the magnitude of the charge of object A is[tex]\( 2.78 \times 10^{-10} \, \text{C} \).[/tex]

To solve this problem we will start by differentiating the values in each of the states of matter. Subsequently through the thermodynamic tables we will look for the values related to the entropy, enthalpy and respective specific volumes. Through the relationship of Power defined as the product between mass and enthalpy and mass, specific volume and pressure, we will find the energetic values in the two states investigated. We will start defining the states

State 1

[tex]T_1 = 700\°C[/tex]

[tex]P_1 = 4 Mpa[/tex]

From steam table

[tex]h_1 =3906.41 KJ/Kg[/tex]

[tex]s_1 = 7.62 KJ/Kg.K[/tex]

Now

[tex]s_1 = s_2 = 7.62 KJ/Kg.K[/tex] As 1-2 is isentropic

State 2

[tex]P_2 = 20 Kpa[/tex]

[tex]s_2 = 7.62 KJ/Kg \cdot K[/tex]

From steam table

[tex]h_2 = 2513.33 KJ/Kg[/tex]

PART A) The power produced by turbine is the product between the mass and the enthalpy difference, then

[tex]Power = m \times (h_1-h_2)[/tex]

[tex]P = (50)(3906.41 - 2513.33)[/tex]

[tex]P = 69654kW[/tex]

b) Pump Work

State 3

[tex]P_3 = 20 Kpa[/tex]

[tex]\upsilon= 0.001 m^3/kg[/tex]

The Work done by the pump is

[tex]W= m\upsilon \Delta P[/tex]

[tex]W = (50)(0.001)(4000-20)[/tex]

[tex]W = 199kJ[/tex]

Answer:

[tex]8.90392\times 10^{-9}\ W[/tex]

[tex]5.36518\times 10^{-9}\ W[/tex]

Explanation:

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

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

t = Time taken = 3.8 ms

[tex]\lambda[/tex] = Wavelength

n = Number of protons = [tex]8\times 10^7[/tex]

Power is given by

[tex]P=\dfrac{E}{t}\\\Rightarrow P=\dfrac{nh\dfrac{c}{\lambda}}{t}\\\Rightarrow P=\dfrac{8\times 10^7\times 6.626\times 10^{-34}\times \dfrac{3\times 10^8}{470\times 10^{-9}}}{3.8\times 10^{-3}}\\\Rightarrow P=8.90392\times 10^{-9}\ W[/tex]

The power is [tex]8.90392\times 10^{-9}\ W[/tex]

[tex]P=\dfrac{nh\dfrac{c}{\lambda}}{t}\\\Rightarrow P=\dfrac{8\times 10^7\times 6.626\times 10^{-34}\times \dfrac{3\times 10^8}{780\times 10^{-9}}}{3.8\times 10^{-3}}\\\Rightarrow P=5.36518\times 10^{-9}\ W[/tex]

The power is [tex]5.36518\times 10^{-9}\ W[/tex]

To solve this problem we will proceed to use Newton's second law for which the mass (m) multiplied by the acceleration (a) is defined as the Force (F) applied on a body, mathematically that is,

[tex]F = ma[/tex]

According to the statement for the first object, the acceleration is,

[tex]a = \frac{F}{m} \rightarrow 1^{st} Object[/tex]

For the second object the acceleration is,

[tex]8a = \frac{3F}{m_2} \rightarrow 2^{nd} Object[/tex]

Solving for the mass of the second object,

[tex]m_2 = \frac{3F}{8a}[/tex]

[tex]m_2 = \frac{3F}{8(F/m)}[/tex]

[tex]m_2 = \frac{3}{8}m[/tex]

Therefore the correct answer is D.

To find the distance between the first-order red and blue fringes formed by a diffraction grating, we calculate the diffraction angles using the grating equation and then use trigonometry to determine the positions of the fringes on the screen. The distance apart is the absolute difference between these positions.

The student is asking about the distance between the first-order red and blue fringes that appear on a screen as a result of a diffraction pattern formed by light from a hydrogen discharge lamp passing through a diffraction grating. To answer this question, we apply the formula for diffraction maxima, d sin( heta) = m ext{ extlambda}, where d is the distance between the grating lines,  ext{ extlambda} is the wavelength of light,  ext{ exttheta} is the diffraction angle, and m is the order of the maximum.

Given that the grating has 500 lines per mm or 500,000 lines per meter, we can calculate d as 1/500,000 meters. Since we are looking for the first-order maxima (m=1), we can find the angles for each color using the formula sin( heta) =  ext{ extlambda}/d. We can then use basic trigonometry to find the positions of the red and blue fringes on the screen, and their distance apart.

For the 656 nm red light:

ext{ exttheta}_{red} = sin^{-1}( ext{ extlambda}_{red}/d)
ext{ exttheta}_{red} = sin^{-1}(656 ext{ exttimes}10^{-9} m / (1/500,000 m))

For the 486 nm blue light:

ext{ exttheta}_{blue} = sin^{-1}( ext{ extlambda}_{blue}/d)
ext{ exttheta}_{blue} = sin^{-1}(486 ext{ exttimes}10^{-9} m / (1/500,000 m))

After calculating the angles, the positions on the screen for first-order maxima are found by L tan( heta), where L is the distance from the grating to the screen (1.50 m in this case).

The final distance between the red and blue fringes is the absolute difference between their positions on the screen.

Answer:

The electrostatic force between and electron and a proton is [tex]F=2.30\times 10^{-20}\ N[/tex]

Explanation:

It is given that, charge [tex]q_1[/tex] is placed at a distance [tex]r_o[/tex] from charge [tex]q_2[/tex]. The force acting between charges is given by :

[tex]F=\dfrac{kq_1q_2}{r_o^2}[/tex]

We need to find the force if the distance between them is reduced to [tex]r_o/4[/tex]. It is given by :

[tex]F'=\dfrac{kq_1q_2}{(r_o/4)^2}[/tex]

[tex]F'=16\times \dfrac{kq_1q_2}{r_o^2}[/tex]

[tex]F'=16\times F[/tex]

So, if the the distance between them is reduced to [tex]r_o/4[/tex], the new force becomes 16 times of the previous force.

The electrostatic force between and electron and a proton separated by 0.1 mm or [tex]10^{-4}\ m[/tex] is :

[tex]F=\dfrac{kq_1q_2}{r_o^2}[/tex]

[tex]F=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{(10^{-4})^2}[/tex]

[tex]F=2.30\times 10^{-20}\ N[/tex]

So, the electrostatic force between and electron and a proton is [tex]F=2.30\times 10^{-20}\ N[/tex]. Hence, this is the required solution.

When the distance between two charges is reduced, the magnitude of the force on one charge due to the other increases. This is because the electrostatic force is inversely proportional to the square of the distance between the charges.

When the distance between two charges, q1 and q2, is reduced from r0 to r0/4, the magnitude of the force on q1 due to q2 increases. This is because the electrostatic force is inversely proportional to the square of the distance between the charges.

For example, if we have charges q1 = 2 C and q2 = 4 C, and the original distance r0 = 2 m, the force between them would be F = k × (q1 × q2) / r0^2 = k × (2 × 4) / (2^2) = k × 4, where k is the Coulomb's constant.

If we reduce the distance to r0/4 = 0.5 m, the force would become F' = k × q1 × q2) / (r0/4)^2 = k × 4) / (0.5^2) = k ×64, which is 16 times greater than the original force.

Answer:

Bulk modulus = 1.35 × [tex]10^{10}[/tex] Pa

Explanation:

given data

density = 1400 kg/m³

frequency = 370 Hz

wavelength = 8.40 m

solution

we get here bulk modulus of the liquid that is

we know Bulk Modulus = [tex]v^2*\rho[/tex]   ...............

here [tex]\rho[/tex] is density i.e 1400 kg/m³

and v is =  frequency × wavelength

v = 370 × 8.40 = 3108 m/s

so here bulk modulus will be as

Bulk modulus = 3108² × 1400

Bulk modulus = 1.35 × [tex]10^{10}[/tex] Pa

Answer:

The time taken by the rotating object to speed up from 15.0 s to 33.3 rad/s is 5.30 seconds.        

Explanation:

It is given that,

Initial angular speed of the object, [tex]\omega_o=15\ rad/s[/tex]

Final angular speed of the object, [tex]\omega_f=33.3\ rad/s[/tex]

Angular acceleration of the object, [tex]\alpha =3.45\ rad/s^2[/tex]

Angular acceleration of an object is object is defined as the change in angular velocity per unit time. It is given by :

[tex]\alpha =\dfrac{\omega_f-\omega_i}{t}[/tex]

[tex]t =\dfrac{\omega_f-\omega_i}{\alpha}[/tex]

[tex]t =\dfrac{33.3-15}{3.45}[/tex]

t = 5.30 seconds

So, the time taken by the rotating object to speed up from 15.0 s to 33.3 rad/s is 5.30 seconds. Hence, this is the required solution.

The time taken will be "5.30 seconds".

According to the question,

Initial angular speed,

Final angular speed,

Angular acceleration,

As we know,

→ [tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]

or,

→ [tex]t = \frac{\omega_f -\omega_i}{\alpha}[/tex]

By substituting the values, we get

     [tex]= \frac{33.3-15}{3.45}[/tex]

     [tex]= \frac{18.3}{3.45}[/tex]

     [tex]= 5.30 \ seconds[/tex]

Thus the above answer i.e., "option b" is right.

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Answer:

For Part A:

ΔX=11.8813 m

Part B:

[tex]t=0.8194 s[/tex]

Explanation:

Note: In order to find Part A we first have to find Part B i.e time

Data given:

[tex]V_i=0[/tex]

a=-9.8m/s^2 (-ve is because ranch is falling down)

Δy=-3.29m (-ve is because ranch is falling down)

Second equation of Motion:

Δy=[tex]V_i*t+\frac{1}{2}g*t^2[/tex]

V_i=0, Equation will become

Δy=[tex]\frac{1}{2}g*t^2[/tex]

[tex]t=\sqrt{2Y/g} \\t=\sqrt{2*-3.29/-9.8} \\t=0.8194 s[/tex]

For Part A:

Again:

Second equation of Motion:

ΔX=[tex]V_i*t+\frac{1}{2}a*t^2[/tex]

Since velocity is constant a=0

V_i=14.5m/s, t=0.8194 sec

ΔX=[tex]V_i*t[/tex]

ΔX=[tex]14.5*0.8194[/tex]

ΔX=11.8813 m

Part B: (Calculated above)

Data given:

[tex]V_i=0[/tex]

a=-9.8m/s^2 (-ve is because ranch is falling down)

Δy=-3.29m (-ve is because ranch is falling down)

Second equation of Motion:

Δy=[tex]V_i*t+\frac{1}{2}g*t^2[/tex]

V_i=0, Equation will become

Δy=[tex]\frac{1}{2}g*t^2[/tex]

[tex]t=\sqrt{2Y/g} \\t=\sqrt{2*-3.29/-9.8} \\t=0.8194 s[/tex]

Answer:

A. the indices of refraction matched

Explanation:

The index of refraction, or refractive

index, is a measure of how fast light

rays travel through a given medium.

Alternatively, it could be said that

the refractive index is the measure of

the bending of a light ray when

passing from one medium to

another. Mathematically, it can be

represented as a ratio between two

different velocities – the velocity of

light in vacuum and the velocity of

light in a given medium.

For example, try putting a pencil in a jar full of water. If you look at the pencil from above, it would look as though the pencil has bent in the water. That happens due to the refraction of light. It occurs because as light rays enter water, they slow down, as the speed of light in water is lower than the speed of light in air. The magnitude of how much a medium refracts a light ray is determined by the index

If you place a glass cylinder in Wessin Oil, the view of the cylinder nearly disappeared when the eyedropper was full of Wessin Oil because the indices of refraction matched here. Thus, the correct option is A.

Refraction is the phenomena of bending of light, which also happens with the rays of sound, water and other waves as it passes from one transparent substance into another substance or medium. This phenomena of bending of light waves by the refraction makes it possible for the people to have lenses, magnifying glasses, prisms, and rainbows. Even, our eyes depend upon this bending of light for the visibility and other effects of light.

If we place a glass cylinder in Wessin Oil, the view of the cylinder nearly disappeared when the cylinder nearly disappeared when the eyedropper was full of Wessin Oil because the indices of refraction matched.

Therefore, the correct option is A.

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Answer:

D = 72.68 m

Explanation:

given,

R = 100 m

angular speed = 0.1 rad/s

distance she can be pulled before the centripetal acceleration reaches 5g = 49 m/s².

using conservation of Angular momentum

[tex]I_i\omega_i= I_f\omega_f[/tex]

[tex]mr_i^2\omega_i=m r_f^2\omega_f[/tex]

[tex]\omega_f = \dfrac{r_i^2}{r_f^2}\times \omega_i[/tex]

[tex]\omega_f = \dfrac{r_i^2}{r_f^2}\times \omega_i[/tex]

we know,

centripetal acceleration

[tex]a = \dfrac{v^2}{r}[/tex]

v = r ω

[tex]a =\omega_f^2 r_f [/tex]

[tex]a =(\dfrac{r_i^2}{r_f^2}\times \omega_i)^2 r_f [/tex]

[tex]a =\dfrac{r_i^4\times \omega_i^2}{r_f^3}[/tex]

[tex]r_f^3=\dfrac{100^4\times 0.1^2}{5\times 9.8}[/tex]

[tex]r_f^3=20408.1632[/tex]

[tex]r_f = 27.32\ m[/tex]

distance she has reached inward is equal to

D = 100 - 27.23

D = 72.68 m

Answer:

Option A is correct (0) ( The electric field at the center of circular charged ring is zero)

Explanation:

Option A is correct (0) ( The electric field at the center of circular charged ring is zero)

The reason why electric field at the center of circular charged ring is zero because the fields at the center of the circular ring due to any point are cancelled by electric fields of another point which is at 180 degree to that point i.e opposite to that charge.

Answer: a) 0

The electric field from opposite directions at the centre of the circular ring cancel each other out and give a resultant of zero

Explanation:

Given that the ring is perfectly circular with radius b which has a uniform distributed charge q around it.

The electric field experienced at the centre of the ring from opposite directions are given as

E1 = kₑq/b²

E2 = -kₑq/b²

It experience the two electric field E1 and E2 from opposite directions at the centre. So the resultant electric field is given by:

E = E1 + E2

E = kₑq/b² - kₑq/b²

E = 0

The electric field from opposite directions at the centre of the circular ring cancel each other out and give a resultant of zero

To solve this problem we will apply the concept related to the conservation of the Momentum. We will then start considering that the amount of initial momentum must be equal to the amount of final momentum. Considering that all the objects at the initial moment have the same initial velocity (Zero, since they start from rest) the final moment will be equivalent to the multiplication of the mass of each object by the velocity of each object, so

Initial Momentum = Final Momentum

[tex](m_B+m_1+m_2)v_i = m_1v_1+m_2v_2+m_Bv_B[/tex]

Here,

[tex]m_B[/tex] =  mass of Raft

[tex]m_1[/tex] = Mass of swimmers 1

[tex]m_2[/tex] = Mass of swimmers 2

[tex]v_i[/tex] = Initial velocity (of the three objects)

[tex]v_B[/tex] = Velocity of Raft

Replacing,

[tex](199+52+70)*0 = (52)(4)+(70)(-4)+199v_B[/tex]

Solving for [tex]v_B[/tex]

[tex]vB = \frac{72}{199}[/tex]

[tex]v_B = 0.3618m/s[/tex]

Therefore the velocity the rarft start to move is 0.3618m/s

Answer:

The velocity of the pin is opposite its acceleration on the way up.

(d) option is correct.

Explanation:

when the juggler throws a bowling pin straight in the air, the acceleration working on the pin is in the downward direction due to the gravitational force of the earth.

According to Newton's Universal Law of Gravitation

''The gravitational force is a force that attracts any objects with mass''

Final answer:

The acceleration due to gravity is always downward, so when the pin is thrown up, its velocity is opposite to its acceleration. At the peak, the velocity is zero and then aligns with the direction of gravity on the descent.

Explanation:

When a juggler throws a bowling pin straight up in the air, the acceleration due to gravity is always directed towards the ground, which means it is downwards. As the pin moves upwards, its velocity is in the opposite direction of the acceleration. At the peak of its motion, the velocity of the pin is zero, after which it starts to fall back down, and its velocity is then in the same direction as acceleration. Thus, the correct statement is (d) The velocity of the pin is opposite its acceleration on the way up.

To solve this problem we will apply the concepts related to the potential, defined from the Coulomb laws for which it is defined as the product between the Coulomb constant and the load, over the distance that separates the two objects. Mathematically this is

[tex]V = \frac{kq}{r}[/tex]

k = Coulomb's constant

q = Charge

r = Distance between them

[tex]q = 18 pC \rightarrow q = 1.8*10^-11 C[/tex]

[tex]d = 2.4mm \rightarrow r = 1.2 mm = 1.2*10^-3 m[/tex]

Replacing,

[tex]V = \frac{kq}{r}[/tex]

[tex]V = \frac{ (9*10^9)*(1.8*10^{-11})}{(1.2*10^{-3})}[/tex]

[tex]V = 135 V[/tex]

Therefore the potential at the surface of the raindrop is 135 V