A flat circular coil having a diameter of 25 cm is to produce a magnetic field at its center of magnitude, 1.0 mT. If the coil has 100 turns how much current must pass through the coil?

Answer:

Electrical charge, q = 75 C

Explanation:

It is given that,

Voltage, V = 4 V

Energy stored, E = 300 J

Energy stored in the battery is given by :

[tex]E=q\times V[/tex]

q is the electrical charge

[tex]q=\dfrac{E}{V}[/tex]

[tex]q=\dfrac{300\ J}{4\ V}[/tex]

q = 75 C

So, the electrical charge of 75 C must flow between the battery's terminals to completely drain the battery if it is fully charged. Hence, this is the required solution.

Solution:

To calculate the average distance between the given parabola and the x-axis

y = 77x(1616 - x)

x ∈ [0, 16]

avg distance = [tex]\int_{0}^{16}\frac{77x(1616 - x)dx}{\int_{0}^{16} x dx}[/tex]

                     = [tex]2\int_{0}^{16}(\frac{(124432x - 77x^{2})dx}{[x^{2}]_{0}^{16}}[/tex]

                      =[tex] 2\int_{0}^{16}\frac{\frac{124432x^{2}}{2}- \frac{77x^{3}}{3}}{[x^{2}]_{0}^{16}}[/tex]

                      = [tex]16^{2}[\frac{62216 - 25.67\times 16}{16^{2}}][/tex]

avg distance = 61805 unit

Answer:

68.445 cm³/s

Explanation:

Given:

Volume, V = [tex]\frac{4}{3}\pi r^3[/tex]

radius = 5.85 cm

Growth rate of radius = 0.5 cm/week

now

differentiating the volume with respect to time 't', we get:

[tex]\frac{dV}{dt}=\frac{d(\frac{4}{3}\pi r^3)}{dt}[/tex]

or

[tex]\frac{dV}{dt}=(\frac{4}{3}\pi )3r^2\frac{dr}{dt}[/tex]

now, substituting the value of r (i.e at r = 5.85cm) in the above equation, we get:

[tex]\frac{dV}{dt}=4\pi 5.85^2\times 0.5[/tex]

or

[tex]\frac{dV}{dt}=68.445cm^3/s[/tex]

hence, the rate of change of volume at r = 5.85cm is 68.445 cm³/s

Answer:

2.08 x 10⁻⁶ T

Explanation:

[tex]v[/tex] = velocity of electron = 5.0 x 10⁷ m/s

q = charge on electron

B = magnetic field = ?

E = electric field = 104 V/m

Magnetic force on the electron is given as

[tex]F_{B} = qvB[/tex]

Electric force on the electron is given as

[tex]F_{E} = qE[/tex]

For the electron to pass without being deflected, we must have

[tex]F_{B} = F_{E}[/tex]

[tex]qvB = qE[/tex]

[tex]vB = E[/tex]

(5.0 x 10⁷) B = 104

B = 2.08 x 10⁻⁶ T

It takes approximately 2.80 seconds for the second-place cyclist to catch up with the leader.

To find the time it takes for the second-place cyclist to catch up with the leader, we can set up an equation based on their positions.

Let's assume that the time it takes for the second-place cyclist to catch up is "t" seconds.

For the leader:

Distance = Velocity × Time

Distance(leader) = Velocity(leader) × t

For the second-place cyclist:

Distance = Initial Distance + Velocity × Time + 0.5 × Acceleration × Time²

Distance(second) = Initial Distance(second) + Velocity(second) × t + 0.5 × Acceleration(second) × t²

Given:

Velocity(leader) = 11.9 m/s

Initial Distance(second) = 10.6 m

Velocity(second) = 9.80 m/s

Acceleration(second) = 1.20 m/s²

Equating the two distances:

Distance(leader) = Distance(second)

Velocity(leader) × t = Initial Distance(second) + Velocity(second) × t + 0.5 × Acceleration(second) × t²

Now, plug in the values and solve for "t":

11.9t = 10.6 + 9.80t + 0.5 × 1.20 × t²

Simplify the equation:

0.6t² + 2.1t - 10.6 = 0

solving the equation we get,

t = 2.80292945

t = -6.2898

Since time can not be negative so considering t = 2.80292945.

So, it takes approximately 2.80 seconds for the second-place cyclist to catch up with the leader.

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To determine the time elapsed before the second-place cyclist catches the leader in a bicycle race, we use the equations of motion to calculate when their positions will be equal, accounting for the leader's constant velocity and the second-place cyclist's acceleration.

The problem at hand involves the second-place cyclist catching up to the leader in a bicycle race. Given that the leader has a constant velocity of 11.9 m/s and the gap between them is 10.6 m, while the second-place cyclist has a velocity of 9.80 m/s and an acceleration of 1.20 m/s2, we can find out the time it takes for the second-place cyclist to catch up by using the equations of motion.

The equation for the leader's position as a function of time will be:
x1(t) = 11.9t + 10.6 ,
and for the second-place cyclist, considering initial velocity (u), acceleration (a), and the initial position being 0, the position as a function of time is:
x2(t) = 0 + 9.80t + 0.5(1.20)t2.

To find the time when the second-place cyclist catches up to the leader, we need to set x1(t) = x2(t) and solve for t:

11.9t + 10.6 = 9.80t + 0.5(1.20)t2

Solving this quadratic equation will give us the time elapsed before the second-place cyclist catches the leader.

Answer:

The acceleration of each of the two blocks and the tension in the rope are 5.92 m/s and 147.44 N.

Explanation:

Given that,

Mass = 12.0 kg

Angle = 31.0°

Friction coefficient = 0.158

Mass of second block = 38.0 kg

Using formula of frictional force

[tex]f_{\mu} = \mu N[/tex]....(I)

Where, N = normal force

[tex]N = mg\cos\theta[/tex]

Put the value of N into the formula

[tex]N =12\times9.8\times\cos 31^{\circ}[/tex]

[tex]N=100.80\ N[/tex]

Put the value of N in equation (I)

[tex]f_{mu}=0.158\times100.80[/tex]

[tex]f_{mu}=15.9264\ N[/tex]

Now, Weight of second block

[tex]W = mg[/tex]

[tex]W=38.0\times9.8[/tex]

[tex]W=372.4\ N[/tex]

The horizontal force is

[tex]F = mg\sintheta[/tex]

[tex]F=12\times9.8\times\sin 31^{\circ}[/tex]

[tex]F=60.5684\ N[/tex]....(II)

(I). We need to calculate the acceleration

[tex]a=m_{2}g-\dfrac{f_{\mu}+mg\sin\theta}{m_{1}+m_{2}}[/tex]

[tex]a=\dfrac{372.4-(15.9264+60.5684)}{12+38}[/tex]

[tex]a=5.92\ m/s^2[/tex]

(II). We need to calculate the tension in the rope

[tex]m_{2}g-T=m_{2}a[/tex]

[tex]-T=38\times5.92-38\times9.8[/tex]

[tex]T=147.44\ N[/tex]

Hence, The acceleration of each of the two blocks and the tension in the rope are 5.92 m/s and 147.44 N.

Answer:

The distance is 5 m.

Explanation:

Given that,

Horizontal velocity = 20.0 m/s

Time t = 0.250 s

We need to calculate the horizontal distance

The distance traveled by the baseball with horizontal velocity in time t is given by

Using formula for horizontal distance

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

[tex]d = 20.0\times0.250[/tex]

[tex]d=5\ m[/tex]

Hence, The distance is 5 m.

To find the force exerted by the ground on each wheel of the automobile, we can analyze the forces acting on the car and use the principle of equilibrium. By considering the weight of the car and the distribution of weight between the front and rear axles, we can determine the force exerted by the ground on each wheel.

The force exerted by the ground on each wheel of the automobile can be determined by considering the forces acting on the car. Since the car is in equilibrium, the sum of the vertical forces must be zero. The weight of the car is distributed between the front and rear axles according to their distances from the center of mass. Using this information, we can calculate the force exerted by the ground on each wheel.

First, we find the weight of the car by multiplying its mass by the acceleration due to gravity: W = (mass of the car) x (acceleration due to gravity). In this case, the acceleration due to gravity is 9.8 m/s².

Next, we find the force exerted by the ground on the rear wheels. Since the car is in equilibrium, the sum of the torques about any point must be zero. Taking the point where the front wheels contact the ground as the pivot point, we can set up an equation using the weight of the car, the distances between the front and rear axles and the center of mass, and the force exerted by the ground on the rear wheels. Solving this equation allows us to find the force exerted by the ground on the rear wheels, and since the front and rear wheels share the weight of the car equally, the force exerted by the ground on each wheel is half of this value.

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Answer:

c

Explanation:

When a satellite is orbiting the earth , a constant force is being applied on it which means it must has acceleration. Also the direction of satellite is always being changed when it is orbitting to there is always change in the velocity vector which means acceleration.

You can view in the attached diagram to understand how the velocity is being changed.

Answer:

Final velocity is 181.61 m/s at angle 40.44° below horizontal.

Explanation:

Initial horizontal velocity = 170 cos 35.6 = 138.23 m/s

Final horizontal velocity = 138.23 m/s

Considering vertical motion of projectile:

Initial vertical velocity, u = 170 sin 35.6 = 98.96 m/s

Acceleration, a = -9.81 m/s²

Displacement, s = -208 m

We have v² = u² + 2as

Substituting

             v² = 98.96² + 2 x -9.81 x -208

             v = 117.79 m/s

Final velocity,

            [tex]v=\sqrt{138.23^2+117.79^2}=181.61m/s[/tex]

            [tex]\theta =tan^{-1}\left ( \frac{117.79}{138.23}\right )=40.44^0[/tex]

Final velocity is 181.61 m/s at angle 40.44° below horizontal.

Answer:

150 fils

Explanation:

Power = 125 Watt

Time 4 hours a day

Energy per day = Power x time per day = 125 x 4 = 500 watt hour

Energy for 30 days = 500 x 30 watt hour = 15000 watt hour = 15 KWH

Cost of 1 KWH = 10 fils

Cost of 15 KWH = 15 x 10 = 150 fils

Answer:

4.78 x 10^-11 J

Explanation:

A = 1.5 x 10^-4 m^2

d = 2 mm = 2 x 10^-3 m

V = 12 V

Let C be the capacitance of the capacitor

C = ε0 A / d

C = (8.854 x 10^-12 x 1.5 x 10^-4) / (2 x 10^-3)

C = 6.64 x 10^-13 F

Energy stored, U = 1/2 CV^2

U = 0.5 x 6.64 x 10^-13 x 12 x 12

U = 4.78 x 10^-11 J

The energy stored in a parallel-plate capacitor connected to a 12-V battery, with plate area of 1.5 x 10^-4 m^2 and separation of 2.0 mm, is calculated using the formula U = (1/2)CV^2 and is found to be 4.8 x 10^-11 J.

The capacitance C of a parallel-plate capacitor is given by the formula C = ε_0 * A / d, where ε_0 is the vacuum permittivity (ε_0 = 8.85 x 10^-12 F/m), A is the area of the plates, and d is the separation between the plates.

Given that the area A is 1.5 x 10^-4 m^2, the separation d is 2.0 mm = 2.0 x 10^-3 m, and the voltage V is 12 V, we can plug in these values to first determine capacitance and then calculate the energy stored.

First, calculate the capacitance:

C = ε_0 * A / d = (8.85 x 10^-12 F/m)(1.5 x 10^-4 m^2) / (2.0 x 10^-3 m) = 6.6 x 10^-12 F

Next, calculate the energy stored:

U = (1/2)CV^2 = (1/2)(6.6 x 10^-12 F)(12 V)^2 = 4.8 x 10^-11 J

Therefore, the energy stored in the capacitor is 4.8 x 10^-11 J, which corresponds to option (a).

Answer:

16.035 revolutions

Explanation:

Part 1:

t = 6 s, f0 = 0 , f = 5 rps,

Let the number of revolutions be n1.

Use first equation of motion for rotational motion

w = w0 + α t

2 x 3.14 x 5 = 0 + α x 6

α = 5.233 rad/s^2

Let θ1 be the angle turned.

Use second equation of motion for rotational motion

θ1 = w0 t + 12 x α x t^2

θ1 = 0 + 0.5 x 5.233 x 6 x 6 = 94.194 rad

n1 = θ1 / 2π = 94.194 / 2 x 3.14 = 15 revolutions

Part 2:

f0 = 5 rps, f = 0, t = 13 s

Let the number of revolutions be n2.

Use first equation of motion for rotational motion

w = w0 + α t

0 = 2 x 3.14 x 5 + α x 13

α = - 2.415 rad/s^2

Let θ2 be the angle turned.

Use third equation of motion for rotational motion

w^2 = w0^2 + 2 x α x θ2

0 = 2 x 3.14 x 5 - 2 x 2.415 x θ2

θ2 = 6.5 rad  

n2 = θ2 / 2π = 6.5 / 2 x 3.14 = 1.035 revolutions

total revolutions n = n1 + n2 = 15 + 1.035 = 16.035 revolutions

Answer:

Momentum, p = 5 kg-m/s

Explanation:

The magnitude of the momentum of an object is the product of its mass m and speed v i.e.

p = m v

Mass, m = 3 kg

Velocity, v = 1.5 m/s

So, momentum of this object is given by :

[tex]p=3\ kg\times 1.5\ m/s[/tex]

p = 4.5 kg-m/s

or

p = 5 kg-m/s

So, the magnitude of momentum is 5 kg-m/s. Hence, this is the required solution.

Answer:

The length of the rectangular key is 0.4244 m

Explanation:

Given that,

Power = 7.46 kW

Speed = 1200 rpm

Shearing stress of shaft = 30 MPa

Mini shearing stress of key = 240 MPa

We need to calculate the torque

Using formula of power

[tex]P=\dfrac{2\pi NT}{60}[/tex]

Where, P = power

N = number of turns

Put the value into the formula

[tex]7.46\times10^{3}=\dfrac{2\pi\times1200\times T}{60}[/tex]

[tex]T=\dfrac{7.46\times10^{3}\times60}{2\pi\times1200}[/tex]

[tex]T=59.36\ N-m[/tex]

We need to calculate the distance

[tex]\tau_{max}=\dfrac{16T}{\pi d^3}[/tex]

[tex]d^3=\dfrac{16\times59.36}{\pi\times30}[/tex]

[tex]d=(10.077)^{\dfrac{1}{3}}[/tex]

[tex]d=2.159\ m[/tex]

Width of key is one fourth of the shaft diameter

[tex]W=\dfrac{1}{4}\times2.159[/tex]

[tex]W=0.53975\ m[/tex]

The shear stress induced in key

[tex]\tau_{max}=\dfrac{F}{Wl}[/tex]

[tex]\tau_{max}=\dfrac{\dfrac{T}{\dfrac{d}{2}}}{wl}[/tex]

[tex]\tau_{max}=\dfrac{2T}{dWl}[/tex]

[tex]240=\dfrac{2\times59.36}{2.159\times0.53975\times l}[/tex]

[tex]l=\dfrac{2\times59.36}{2.159\times0.53975\times240}[/tex]

[tex]l=0.4244\ m[/tex]

Hence, The length of the rectangular key is 0.4244 m

1. Area of Circular End: [tex]A = π x (1.5)^2 = 7.07 sq ft.[/tex]

2. Weight of Gasoline: W = 133.5π lbs.

3. Fluid Force: F = 2043.45π lbs on half-full tank end.

let's break down the problem step by step.

1. Identify Relevant Parameters:

  - Diameter of the tank (D) = 3 feet

  - Radius of the tank (r) = D/2 = 1.5 feet

  - Density of gasoline (ρ) = 42 pounds per cubic foot

  - Depth of the gasoline (h) when the tank is half full = 1.5 feet

2. Calculate the Area of the Circular End:

  - The area (A) of a circle is given by the formula:[tex]A = π x r^2[/tex]

  - Substituting the value of radius (r = 1.5 feet), we get:

  - [tex]A = πx (1.5)^2 = π x 2.25 = 7.07[/tex] square feet

3. Determine the Weight of the Gasoline:

  - The weight (W) of the gasoline can be calculated using the formula: W = ρx V

  - Where V is the volume of the gasoline.

  - Since the tank is half full, the volume of the gasoline is half the volume of the cylinder.

  - The volume (V) of the cylinder is given by the formula: [tex]V = πxr^2 x h[/tex]

  - Substituting the values of radius (r = 1.5 feet) and height (h = 1.5 feet), we get:

[tex]- V = π x (1.5)^2 x 1.5 = π x 2.25x 1.5 = 3.375π cubic feet[/tex]

  - So, the weight of the gasoline (W) is: W = 42x3.375π = 133.5π pounds

4. Calculate the Fluid Force:

  - The fluid force (F) on a circular end of the tank can be calculated using the formula: F = ρ x g x V x h

  - Where g is the acceleration due to gravity.

  - Since we're working in pounds and feet, we'll use the value of acceleration due to gravity, [tex]g ≈ 32.2 ft/s^2.[/tex]

  - Substituting the values, we get:

  - F = 42x 32.2x3.375π x 1.5 = 2043.45π pounds

5.Final Calculation:

  - Therefore, the fluid force on a circular end of the tank when the tank is half full is approximately 2043.45π pounds.

So, the complete calculation and explanation show that the fluid force on a circular end of the tank when it is half full is about 2043.45π pounds.

The net work done by friction on the body of a snake slithering in a complete circle can be found by calculating the work done by the friction force. The friction force is equal to the coefficient of friction multiplied by the normal force. The net work done by friction is -2πr × friction force = -2π(1.04 m) × 19.5 N = -129.77 J.

The net work done by friction on the body of a snake slithering in a complete circle can be found by calculating the work done by the friction force. The friction force is equal to the coefficient of friction multiplied by the normal force. In this case, the friction force is opposing the motion of the snake, so the work done by friction is negative.

Considering a complete circle of radius 1.04 m, the distance traveled by the snake is equal to the circumference of the circle, which is 2πr. The normal force is equal to the weight of the snake, which is 78.0 N. The friction force is equal to the coefficient of friction (0.25) multiplied by the normal force (78.0 N), so the friction force is 19.5 N.

The net work done by friction is then calculated by multiplying the friction force by the distance traveled and taking into account the negative sign to indicate the opposing direction of friction. Therefore, the net work done by friction on the body of the snake is -2πr × friction force = -2π(1.04 m) × 19.5 N = -129.77 J.

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Answer:

final equilibrium temperature of the system is ZERO degree Celcius

Explanation:

Hear heat given by water + iron = heat absorbed by ice

so here first we will calculate the heat given by water + iron

[tex]Q_1 = m_1s_2\Delta T_1 + m_2 s_2 \Delta T_1[/tex]

[tex]Q_1 = (200)(0.450)(40 - T) + (50)(4.186)(40 - T)[/tex]

now the heat absorbed by ice so that it will melt and come to the final temperature

[tex]Q_2 = m s \Delta T + mL + m s_{water}\Delta T'[/tex]

[tex]Q_2 = 50(2.09)(0 + 6) + 50(335) + 50(4.186)(T - 0)[/tex]

now we will have

[tex]17377 + 209.3T = 3600 - 90T + 8372 - 209.3T[/tex]

[tex]17377 + 209.3T + 90T + 209.3T = 11972[/tex]

[tex]T = -10.6[/tex]

since it is coming out negative which is not possible so here the ice will not completely melt

so final equilibrium temperature of the system is ZERO degree Celcius

Answer:

Beats are the difference in frequency.

(a) is correct option.

Explanation:

Beat :

Beat is the difference of the frequency of two waves.

The difference in frequency is equal to the number of beat per second.

Amplitude :

Amplitude of the wave is the maximum displacement.

Frequency :

Frequency is the number oscillations of wave in per second.

Intensity :

Intensity is the power per unit area.

Hence, Beats are the difference in frequency.

Answer:

W = 86 J

Explanation:

Work done by friction force is given by the formula

[tex]W = F_f . d[/tex]

here since the wagon is pulled by horizontal force such that speed remains constant

so here we will have

[tex]F_f = F_{ext} = 22 N[/tex]

Now we will have

[tex]W = 22\times 3.9 [/tex]

[tex]W = 86 J[/tex]

so work done against friction force will be equal to 86 J

The correct answer is option c 86.

Calculating Work Done by the Child:

In this  problem, we need to determine the amount of work done by a child who is pulling a 12 kg wagon over a distance of 3.9 meters with a force of 22 Newtons, assuming there is no friction.

The formula to calculate work (W) is:

W = F * d

where W is work, F is the force applied, and d is the distance covered.

Let's plug in the values provided:

F = 22 N
d = 3.9 m

Thus, the work done is:

W = 22 N * 3.9 m

W = 85.8 J

Therefore, the correct answer is approximately 86 J.

Answer:

(a) 7.92 A

(b) 0.133 Nm

Explanation:

N = 216

r = 2.32 cm = 0.0232 m

(a) M = 2.89 Am^2

M = N i A

Where, A be the area of the coil and i be the current in the coil

2.89 = 216 x i x 3.14 x 0.0232 x 0.0232

i = 7.92 A

(b) B = 46 mT = 0.046 T

Torque, τ = M B Sin 90

τ = 2.89 x 0.046 x 1 = 0.133 Nm

The current required to produce a magnetic dipole moment of 2.89 A·m² in a coil of 216 turns and radius 2.32 cm is approximately 7.92 A. The maximum torque this coil can experience in a 46.0 mT magnetic field is about 0.133 N·m.

To solve these problems, we'll use the relationship between the magnetic dipole moment, current, and torque in a magnetic field.

Part (a): Calculate the Current

The magnetic dipole moment (μ) of a coil is given by:

μ = NIA

where N is the number of turns, I is the current, and A is the area of the coil.

Given:

N = 216 turns

μ = 2.89 A·m²

Radius (r) = 2.32 cm = 0.0232 m

The area (A) of the coil is:

A = πr² = π(0.0232 m)² ≈ 1.69 × 10⁻³ m²

To find the current (I), rearrange the formula:

I = μ / (N × A)

Substitute the values:

I = 2.89 A·m² / (216 × 1.69 × 10⁻³ m²) ≈ 7.92 A

Part (b): Calculate the Maximum Torque

The maximum torque (τ) experienced by the coil in a magnetic field (B) is given by:

τ = μB

Given the magnetic field (B) is 46.0 mT = 0.046 T, substitute the values:

τ = 2.89 A·m² × 0.046 T ≈ 0.133 N·m

Thus, the current required to achieve the given magnetic dipole moment is approximately 7.92 A, and the maximum torque experienced by the coil in a 46.0 mT magnetic field is approximately 0.133 N·m.

Answer:

(a) 3.82 x 10⁷ m/s

(b) 4.5 MV/m

Explanation:

(a)

ΔV = change in the electric potential as the proton moves = 7.60 x 10⁶ Volts

q = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C

v = speed gained by the proton

m = mass of proton = 1.67 x 10⁻²⁷ kg

Using conservation of energy

Kinetic energy gained by proton = Electric potential energy

(0.5) m v² = q ΔV

inserting the values

(0.5) (1.67 x 10⁻²⁷) v² = (1.6 x 10⁻¹⁹) (7.60 x 10⁶)

v = 3.82 x 10⁷ m/s

(b)

d = distance over which the potential change = 1.70 m

Electric field is given as

E = ΔV/d

E = 7.60 x 10⁶/1.70

E = 4.5 x 10⁶ V/m

E = 4.5 MV/m

Explanation:

It is given that,

Electric field, E = 554 N/C

Time, [tex]t=52\ ns=52\times 10^{-9}\ s[/tex]

Electric force, F = qE

For both electron and proton, [tex]F=1.6\times 10^{-19}\ C\times 554\ N/C[/tex]

[tex]F=8.86\times 10^{-17}\ N[/tex]

For electron, [tex]F=m_ea_e[/tex]

[tex]a_e=\dfrac{F}{m_e}[/tex]

[tex]a_e=\dfrac{8.86\times 10^{-17}\ N}{9.1\times 10^{-31}\ kg}[/tex]

[tex]a_e=9.73\times 10^{13}\ m/s^2[/tex]

Using first equation of motion as :

[tex]v=u+at[/tex]

u = 0

[tex]v=9.73\times 10^{13}\ m/s^2\times 52\times 10^{-9}\ s[/tex]

v = 5059600 m/s

or

v = 5.05 × 10⁶ m/s

For proton :

[tex]F=m_pa_p[/tex]

[tex]a_p=\dfrac{F}{m_e}[/tex]

[tex]a_p=\dfrac{8.86\times 10^{-17}\ N}{1.67\times 10^{-27}\ kg}[/tex]

[tex]a_p=5.3\times 10^{10}\ m/s^2[/tex]

Using first equation of motion as :

[tex]v=u+at[/tex]

u = 0

[tex]v=5.3\times 10^{10}\ m/s^2\times 52\times 10^{-9}\ s[/tex]

v = 2756 m/s

Hence, this is the required solution.

Answer:

Spring constant, k = 2.304 N/m

Explanation:

It is given that,

Mass of spring, m = 0.7 kg

Angular frequency, [tex]\omega=2.4\ rad/s[/tex]

We need to find the spring constant of the spring. It is a case of SHM. Its angular frequency is given by :

[tex]\omega=\sqrt{\dfrac{k}{m}}[/tex]

[tex]k=\omega^2\times m[/tex]

[tex]k=(2.4)^2\times 0.4\ kg[/tex]

k = 2.304 N/m

So, the spring constant of the spring is 2.304 N/m. Hence, this is the required solution.

Answer:

The pressure of the air is 499.91 kPa.

Explanation:

Given that,

Initial pressure = 430 kPa

Temperature = 13.0+273=286 K

Final temperature = 59.5+273=332.5 K

We need to calculate the final pressure

Using relation of pressure and temperature

At constant volume,

[tex]\dfrac{P'}{P}=\dfrac{T'}{T}[/tex]

[tex]\dfrac{P'}{430}=\dfrac{332.5}{286}[/tex]

[tex]P'=\dfrac{332.5}{286}\times430[/tex]

[tex]P'=499.91\ kPa[/tex]

Hence,The pressure of the air is 499.91 kPa.

Answer:

[tex]B_2 = 1.71 mT[/tex]

Explanation:

As we know that the magnetic field near the center of solenoid is given as

[tex]B = \frac{\mu_0 N i}{L}[/tex]

now we know that initially the length of the solenoid is L = 18 cm and N number of turns are wounded on it

So the magnetic field at the center of the solenoid is 2 mT

now we pulled the coils apart and the length of solenoid is increased as L = 21 cm

so we have

[tex]\frac{B_1}{B_2} = \frac{L_2}{L_1}[/tex]

now plug in all values in it

[tex]\frac{2.0 mT}{B_2} = \frac{21}{18}[/tex]

[tex]B_2 = 1.71 mT[/tex]

The magnetic field strength inside a solenoid is dependent on the density of the coils per unit length. If a solenoid is extended, spreading the same number of coils over a greater distance, the magnetic field strength at its center decreases proportionally.

The magnetic field strength (B) inside a solenoid is proportional to the number of coils per unit length (n) and the current flowing through the coils (I). We can express this mathematically as B = μonI. The key concept here is that the field strength, B, depends on the density of the coils (n); in other words, how tightly or loosely coiled the wires are.

When you initially have a solenoid 18 cm long, and you extend it to 21 cm without changing the current or the size of the coils, the coil density per unit length decreases. Therefore, the magnetic field strength should decrease proportionally. In effect, you are spreading the same number of coils (field lines) over a greater distance, which dilutes the strength of the magnetic field at the solenoid's center. The exact strength would require knowing more specifics about the solenoid, but this discussion gives the general principle.

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Answer:

5.3 x 10⁻⁹ C

Explanation:

r = radius of cylindrical shell = 10⁻⁵ m

L = length = 0.32 m

A = area

Area is given as

A = 2πrL

A = 2 (3.14) (10⁻⁵) (0.32)

A = 20.096 x 10⁻⁶ m²

d = separation = 10⁻⁸ m

[tex] k_{appa} [/tex] = dielectric constant = 4

Capacitance is given as

[tex]Q=\frac{k_{appa}\epsilon _{o}A}{d}[/tex]                               eq-1

V = Potential difference across the membrane = 74 mV = 0.074 Volts

Q = magnitude of charge on each side

Magnitude of charge on each side is given as

Q = CV

using eq-1

[tex]Q=\frac{k_{appa} \epsilon _{o}AV}{d}[/tex]

Inserting the values

[tex]Q=\frac{4 (8.85\times 10^{-12})(20.096\times 10^{-6})(0.074)}{10^{-8}}[/tex]

Q = 5.3 x 10⁻⁹ C

The membrane is parallel and act as a parallel plate capicitor.The magnitude of the charge on each side of the membrane will be 5.3×10⁻⁹C.

It is an type capacitor is an in which two metal plates arranged in such away so that they are connected in parallel andhaving some distance between them.

A dielectric medium is must in between these plates help to stop the flow of electric current through it due to its non-conductive nature .

The given data in the problem is;

r is the radius of cell of thin cyliner=10⁻⁵m

L is the length =0.32

d is the thickness =10⁻⁸m

K is the dielectric constant=4

v is the potential difference across the membrane=74 mv=0.074 v

q is the megnitude of charge=?

A is the area of capicitor=2πrl

[tex]\rm A = 2\pi rl\\\\\rm A= 2\times3.14\times10^{-5}\times0.32\\\\\rm A= 20.096\times10^{-6}[/tex]

The given formula for the parallel plate capicitor as ,

[tex]\rm Q=\frac{K\varepsilon _0AEr}{d} \\\\\rm Q=\frac{4\times(8.85\times10^{-12})(20.096\times10^{-6}\times0.074}{10^{-8}}\\\\\rm Q=5.3\TIMES10^{-9}\;C[/tex]

Hence the charge on each side of the membrane will be 5.3×10⁻⁹C.

To learn more about the parallel plate capicitor refer to the link;

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Answer:

option d)

Explanation:

The amount of heat required to convert the 1 g of ice at 0 degree C to 1 g water at 0 degree C is called latent heta of fusion.

H = m Lf

where,, h is the heat required, m is the mass and Lf is the latent heat of fusion

1140 = m x 79.7

m = 14.3 g

Answer:

the girl moves towards right with a velocity of 4.1m/s

Explanation:

Since the system is isolated the momentum of the system is conserved

Initial momentum = Final Momentum

Since initially the system is at rest thus [tex]\overrightarrow{p_{i}}=0[/tex]

Now the final momentum of boy = [tex]m_{boy}×velocity[/tex]

[tex]\overrightarrow{p_{boy}}=66.0\times -2.80m/s\\\\\overrightarrow{p_{boy}}=-184.8kgm/s[/tex]

Now for girl let the velocity = u hence her moumentum is 45[tex]\times u[/tex]

Thus equating final momentum to zero we have

[tex]-184.8kgm/s[/tex]+[tex]45\times u = 0[/tex]

[tex]u=\frac{184.8}{45}m/s[/tex]

hence [tex]u=4.1m/s[/tex]

Thus the girl moves towards right with a velocity of 4.1m/s

When the girl pushes her brother on roller blades, causing him to move backwards, the principle of conservation of momentum dictates that the girl will move in the opposite direction with a velocity of 4.16 m/s towards the east.

The scenario presented involves the conservation of momentum, which is a fundamental concept in physics. When the girl pushes her brother on roller blades, and he moves backward at a velocity of 2.80 m/s, by the principle of conservation of momentum, the girl will move in the opposite direction.

Since no external forces are acting on the system (assuming friction is ignored), the total momentum before and after the push remains constant. The combined momentum of the boy and girl before the push is zero because they are both initially at rest. After the push, the momentum of the boy is his mass multiplied by his velocity (66.0 kg × 2.80 m/s).

To find the subsequent motion of the girl, let's calculate:

Therefore, the girl will move towards the east at a velocity of 4.16 m/s as a result of the push.

Answer:

0.695 m

Explanation:

α = angular acceleration of the fan blade = 14.1 rad/s²

a = tangential acceleration at the point concerned = acceleration due to gravity = 9.8 m/s²

r = distance of the point from axis of rotation at which tangential acceleration is same as acceleration due to gravity

We know the relation between

a = r α

Inserting the values

9.8 = 14.1 r

r = 0.695 m