Two blocks of masses mA and mB are connected by a massless spring. The blocks are moved apart, stretching the spring, and subsequently released from rest. Find (a) the ratio of velocities 7of the blocks at any point of their ensuing motion (when their velocities are non-zero) and (b) the ratio of the kinetic energies of the blocks.

q = 1156363.6W/m².

To calculate the heat flux per unit area (W/m²) of a sheet made of metal:

q = -k(ΔT/Δx)

q = -k[(T₂ - T₁)/Δx]

Where, k is the thermal conductivity of the metal, ΔT is the temperature difference and Δx is the thick.

With Δx = 11 mm = 11x10⁻³m, T₂ = 350°C and T₁ = 110°C, and k = 53.0 W/m-K:

q = -53.0W/m-K[(110°C - 350°C)/11x10⁻³m

q = 1156363.6W/m²

Answer:

v = 11.3 m/s

Explanation:

Since there is no external force on the system of two cars

so here momentum of the system of two cars will remain constant

it is given here that two cars combined and move together at 41.5 degree North of East after collision

so here direction of final momentum is given as

[tex]tan\theta = \frac{P_y}{P_x}[/tex]

now we have initial momentum in the same direction

[tex]P_1 = m(10 m/s)\hat j[/tex]

[tex]P_2 = mv\hat i[/tex]

now we have

[tex]\frac{P_1}{P_2} = tan\theta[/tex]

[tex]\frac{m(10 m/s)}{mv} = tan41.5[/tex]

[tex]\frac{10}{v} = 0.88[/tex]

[tex]v = 11.3 m/s[/tex]

Using the principles of conservation of momentum and trigonometry, the initial velocity of car B can be calculated to be approximately 9.75 m/s.

The problem can be solved using the principles of conservation of momentum. In a two-dimensional collision, the components of the momentum in the X and Y directions are separately conserved. Since we're given that the cars have equal mass and they stick together after colliding to make a single object, the magnitude of the velocity of the combined object after the collision must be equal to the vector sum of their initial velocities.

Car A is moving northward and thus its velocity is contributing only to the Y component of the total momentum. Car B, on the other hand, is moving eastward, so its velocity is contributing to the X component. When the cars collide and stick together, they move at an angle of 41.5° north of east, meaning there is both an X and Y component to their velocity.

Using trigonometry, the X component (eastward direction, i.e., car B's direction) can be found by VB = VA * tan(Θ). Given that VA is 10 m/s and Θ is 41.5 degrees, VB can be calculated to be approximately 9.75 m/s.

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Answer:

Option B is the correct answer.

Explanation:

Speed of car = 70mph [tex]=70\times \frac{1609}{3600}=31.28m/s[/tex]

Deceleration of car = 2 m/s²

We have equation of motion

          v = u + at

          0 = 31.28 - 2 t

           t = 15.64 seconds.

We also have

          [tex]s=ut+\frac{1}{2}at^2=31.28\times 15.64-\frac{1}{2}\times 2\times 15.64^2=244.70m[/tex]

Option B is the correct answer.

Answer:

Thickness of magazine = 9.42 mm.

Explanation:

Considering vertical motion of spider:-

Initial velocity, u =  0.870 sin 35 = 0.5 m/s

Acceleration , a = -9.81 m/s²

Time, t = 0.077 s

We have equation of motion s= ut + 0.5 at²

Substituting

   s= 0.5 x 0.077 - 0.5 x 9.81 x 0.077²

    s = 9.42 x 10⁻³ m = 9.42 mm

Thickness of magazine = 9.42 mm.

The thickness of the magazine is determined using the vertical component of the spider's initial velocity and kinematic equations for constant acceleration. By calculating and converting the vertical displacement to millimeters, we can find the thickness of the magazine the spider lands on.

Given that the initial velocity in the vertical direction (Vy) can be found using the formula Vy = V * sin(θ), where V is the initial velocity of the spider and θ is the angle of projection, we can calculate Vy = (0.870 m/s) * sin(35.0°).

The vertical displacement (y) can then be calculated using the kinematic equation for constant acceleration, y = Vy * t + (1/2) * g * t^2, where t is the time of flight, and g is the acceleration due to gravity (approximately -9.81 m/s^2). Since we want to find the displacement in the upward direction and the spider lands on the magazine after 0.0770 s, we are looking for the magnitude of y when the spider lands.

Substituting the values we have: y = (0.870 m/s) * sin(35.0°) * 0.0770 s + (0.5) * (-9.81 m/s^2) * (0.0770 s)^2. The negative sign in the acceleration term accounts for the direction of gravity, which is downwards. After calculating this expression, we convert the result from meters to millimeters (by multiplying by 1000) to obtain the thickness of the magazine.

Answer:

A) 854.46 kPa

Explanation:

P₁ = initial pressure of the gas = 400 kPa

P₂ = final pressure of the gas = ?

T₁ = initial temperature of the gas = 110 K

T₂ = final temperature of the gas = 235 K

Using the equation

[tex]\frac{P_{1}}{T_{1}}=\frac{P_{2}}{T_{2}}[/tex]

Inserting the values

[tex]\frac{400}{110}=\frac{P_{2}}{235}[/tex]

P₂ = 854.46 kPa

Explanation:

0.05

momentum =Mv

then,

Mv=mv

v2=0.12*6/14

v2=0.05kgm/s

The change in the magnitude of the momentum of the ball is 2.4 kg·m/s.

To calculate the change in the magnitude of the momentum of the ball, we need to find the initial momentum and the final momentum of the ball. Momentum is calculated by multiplying the mass of an object by its velocity. Let's start by calculating the initial momentum:

Initial momentum = mass × initial velocity

Final momentum = mass × final velocity

Change in momentum = final momentum - initial momentum

Plugging in the given values:

Initial momentum = 0.12 kg × 6 m/s

Final momentum = 0.12 kg × (-14 m/s) (since the ball changes direction)

Change in momentum = (-1.68 kg·m/s) - (0.72 kg·m/s) = -2.4 kg·m/s.

The change in the magnitude of the momentum of the ball is 2.4 kg·m/s. However, since the question is asking for the magnitude, we take the absolute value of the change in momentum, which is 2.4 kg·m/s.

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Answer:

Power, P = 600 watts

Explanation:

It is given that,

Mass of sprinter, m = 54 kg

Speed, v = 10 m/s

Time taken, t = 3 s

We need to find the average power generated. The work done divided by time taken is called power generated by the sprinter i.e.

[tex]P=\dfrac{W}{t}[/tex]

Work done is equal to the change in kinetic energy of the sprinter.

[tex]P=\dfrac{\dfrac{1}{2}mv^2}{t}[/tex]

[tex]P=\dfrac{\dfrac{1}{2}\times 54\ kg\times (10\ m/s)^2}{3\ s}[/tex]

P = 900 watts

So, the average power generated by the sprinter is 900 watts. Hence, this is the required solution.

Final answer:

The average power generated is 900 Watts, which is approximately 1.21 horsepower.

Explanation:

The question involves calculating the average power generated by a sprinter during acceleration. Power is defined as the work done per unit time. To find the power, we need to first calculate the work done, which, in the case of the sprinter, is the kinetic energy gained as they accelerate to the final speed.

In this case, we can calculate the kinetic energy (KE) of the sprinter using the formula KE = 0.5 × mass ×[tex]velocity^2[/tex]. The mass of the sprinter is given as 54 kg and the final velocity is 10 m/s. Thus, KE = 0.5 × 54 kg × [tex](10 m/s)^2[/tex] = 2700 Joules.

Now, the power can be found by dividing the work done by the time it takes to do that work. The sprinter takes 3 seconds to achieve this final velocity. So, the average power P is P = work/time = 2700 Joules / 3 s = 900 Watts.

Converting this power to horsepower (1 horsepower = 746 Watts), we get approximately P = 900 W × 1 hp/746 W = 1.21 horsepower.

Answer:

1.53 x 10^-5 m^2

Explanation:

use the Stefan's law

Energy per unit time = σ x A x T^4

σ = 5.67 x 10 -8 W/m^2 K^4

96 = 5.67 x 10^-8 x A x (3242)^4

A = 1.53 x 10^-5 m^2

Answer:

23 cm

Explanation:

m = 64.3 kg, diameter = 0.910 cm

radius, r = 0.455 cm = 4.55 x 10^-3 m

A = 3.14 x (4.55 x 10^-3)^2 = 6.5 x 10^-5 m^2

L = 32.2 m

Y = 1.35 x 10^9 Pa

Let the stretched length is l.

Use the formula for Young's modulus

Y = mg L / A l

l = m g L / A Y

l = (64.3 x 9.8 x 32.2) / (6.5 x 10^-5 x 1.35 x 10^9)

l = 0.23 m

l = 23 cm

Final answer:

The self-inductance of the coil is calculated using the energy stored in its magnetic field and the current flowing through it. By rearranging the formula W = (1/2) * L * I², the self-inductance L is found to be 0.11875 H.

Explanation:

To calculate the self-inductance of the coil, we can use the formula for the energy stored in the magnetic field of an inductor:

Energy (W) = (1/2) * L * I²

Where:

W is the energy stored in the magnetic field

L is the self-inductance of the coil

I is the current flowing through the coil

From the question, we're given that the energy (W) is 9.5 x 10⁻³ J, and the current (I) is 0.40 A. We can rearrange the formula to solve for L:

L = 2W / I²

Substitute the given values:

L = 2 * (9.5 x 10⁻³ J) / (0.40 A)²

L = 2 * (9.5 x 10⁻³ J) / (0.16 A²)

L = (19 x 10⁻³ J) / (0.16 A²)

L = 0.11875 H

Therefore, the self-inductance of the coil is 0.11875 H (henrys).

Answer:

at the highest point of the path the acceleration of ball is same as acceleration due to gravity

Explanation:

At the highest point of the path of the ball the speed of the ball becomes zero as the acceleration due to gravity will decelerate the motion of ball due to which the speed of ball will keep on decreasing and finally it comes to rest

So here we will say that at the highest point of the path the speed of the ball comes to zero

now by the force diagram we can say that net force on the ball due to gravity is given by

[tex]F_g = mg[/tex]

now the acceleration of ball is given as

[tex]a = \frac{F_g}{m}[/tex]

[tex]a = \frac{mg}{m} = g[/tex]

so at the highest point of the path the acceleration of ball is same as acceleration due to gravity

Final answer:

The magnitude of the acceleration of a tennis ball at its highest point, in an evacuated chamber and under the influence of gravity alone, is 9.8 m/s². This remains constant regardless of the ball's position in its trajectory.

Explanation:

The question asks about the magnitude of the acceleration of a tennis ball at its highest point when it is shot vertically upward in an evacuated chamber with an initial speed. This scenario is governed by the principles of classical mechanics, specifically under the domain of gravity's influence on objects in motion. Regardless of whether the object is moving up or down or is at its peak height, the acceleration due to gravity on Earth remains constant.

Acceleration at the highest point, or at any point during its motion under gravity alone (assuming no air resistance), is -9.8 m/s2. It's essential to understand that the negative sign indicates the direction of acceleration, which is towards the center of the Earth but does not affect the magnitude of acceleration. Hence, the magnitude of acceleration is 9.8 m/s2, indicating that gravity is the only force acting on the ball throughout its trajectory.

Answer:

The emf is 0.574 volt.

Explanation:

Given that,

Magnetic field [tex]B =3.00\times10^{-5}\ T[/tex]

Length = 75.0 m

Velocity = 255 m/s

We need to calculate the emf

Using formula of emf

[tex]\epsilon=Blv[/tex]

Where, v = velocity

l = length

B = magnetic field

Put the value into the formula

[tex]\epsilon= 3.00\times10^{-5}\times75.0\times255[/tex]

[tex]\epsilon=0.574\ volt[/tex]

Hence, The emf is 0.574 volt.

The emf induced between the wingtips of a jet airplane flying at 255 m/s with a 75.0 m wingspan in Earth's magnetic field of strength 3.00 x 10^-5 T is 0.57 V. This is calculated using the formula for motional emf: E = Blv.

To calculate the emf induced between the airplane wings due to Earth's magnetic field, we use the formula for motional emf: E = Blv, where B is the magnetic field strength, l is the length of the conductor moving through the field (in this case, the wingspan of the airplane), and v is the velocity of the conductor.

Substituting the given values into the formula we get:
E = (3.00 x 10^-5 T) * (75.0 m) * (255 m/s) = 0.57 V

This means that a 0.57 Volt emf is induced between the wingtips of the airplane due to the Earth's magnetic field. This is a direct application of Faraday's law of electromagnetic induction, which states that a changing magnetic field will induce an emf in a conductor.

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Answer:

1.18 x 10^-3 m^3/s

Explanation:

η = 0.25 N s/m^2, radius, r = 5 mm = 0.005 m, l = 25 cm = 0.25 m

P = 300 kPa = 300,000 Pa, Volume of flow = ?

By use of Poiseuillie's equation

Volume of flow = π P r^4 / (8 η l)

Volume of flow = (3.14 x 300000 x 0.005^4) / (8 x 0.25 x 0.25)

Volume of flow = 1.18 x 10^-3 m^3/s

Using Poiseuille's Law for viscous flow, the volume of oil flowing through the tube per second is calculated to be 1.8 liters/sec.

The volume of oil flowing through the tube per second can be calculated using Poiseuille's Law, which is an equation for viscous flow. This Law states that the flow rate of fluid through a pipe (Q) is directly proportional to the fourth power of the radius (r) of the tube, the pressure difference (ΔP) and inversely proportional to the viscosity (η) of the fluid, and the length (L) of the tube.

Using Poiseuille's Law, we have:

Q = (π/8) * (ΔP/r^4L) * η

By substituting the provided values:

Q = (π/8) * (300,000 Pa/(0.005 m)^4 * 0.25 m) * 0.25 N ∙ s/m^2

This yields: Q = 0.0018 m^3/s, or 1.8 liters/sec.

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The average current passing through a device is given by:

I = Q/Δt

I is the average current

Q is the amount of charge that has passed through the device

Δt is the amount of elapsed time

Given values:

Q = 600C

Δt = 0.500hr = 1800s

Plug in the values and solve for I:

I = 600/1800

I = 0.333A

The average current  flowing in the flashlight bulb is 0.333 A

The given parameters:

quantity of the charge, Q = 600 C

time of the current flow, t = 0.5 hour

To find:

The average current is calculated from quantity of charge that flows in the given time period:

Q = It

where:

I is the average current

Q is the quantity of the charge

t is the time period

From the equation above, make the average current the subject of the formula:

[tex]I = \frac{Q}{t} \\\\I = \frac{600 \ C}{0.5 \ hr \times 3600 \ s} \\\\I = \frac{600 \ C}{1800 \ s} \\\\I = 0.333 \ A[/tex]

Thus, the average current is 0.333 A

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Explanation:

The kinetic energy of an object is associated with its motion. Mathematically, it is given by :

[tex]E_k=\dfrac{1}{2}mv^2[/tex]........(1)

Where

m is the mass of the car

v is the velocity of the car

And the momentum of an object is equal to the product of mass and velocity i.e.

[tex]momentum(p)=mass(m)\times velocity(v)[/tex]..........(2)

From equation (1) and (2) it is clear that the kinetic energy and the momentum is directly proportional to the velocity of the object. As a car increases velocity, its kinetic energy or momentum increase faster.

Explanation:

It is given that,

Diameter of the copper wire, d = 2.3 cm

Radius of copper wire, r = 1.15 cm = 0.0115 m

Distance between bird's feet, l = 3.9 cm = 0.039 m

The resistivity of the wire, [tex]\rho=1.7\times 10^{-8}\ \Omega-m[/tex]

We need to find the potential difference between bird's feet. The resistance of the wire is calculated by :

[tex]R=\rho\times \dfrac{l}{A}[/tex]

[tex]R=1.7\times 10^{-8}\ \Omega-m\times \dfrac{0.039\ m}{\pi(0.0115\ m)^2}[/tex]

R = 0.00000159 ohms

[tex]R=1.59\times 10^{-6}\ \Omega[/tex]

Let V is the potential difference between bird's feet. It can be calculated using Ohm's law as :

[tex]V=I\times R[/tex]

[tex]V=45\ A\times 1.59\times 10^{-6}\ \Omega[/tex]

V = 0.000071 volts

or

[tex]V=7.1\times 10^{-5}\ volts[/tex]

So, the potential difference between the bird's feet is [tex]7.1\times 10^{-5}\ volts[/tex]. Hence, this is the required solution.

Even with a large estimated resistance for the bird (say, R_bird = 1000 Ω), the potential difference (V_bird) would be very small (V_bird = 45 A * 1000 Ω ≈ 45000 V).  However, in reality, most of the current will flow through the much lower resistance wire, making the actual potential difference between the bird's feet much closer to zero.

Calculation (for illustration purposes only):

Even though the potential difference is negligible, we can estimate its upper bound (worst-case scenario) by assuming all the current flows through the bird's body.

Wire Resistance:

Diameter (d) = 2.3 cm = 0.023 m

Radius (r) = d/2 = 0.0115 m

Wire length between feet (l) = 3.9 cm = 0.039 m

Resistivity (ρ) = 1.7 x 10^-8 Ω · m

Wire Resistance (R_wire) = ρ * (l / π * r^2) = (1.7 x 10^-8 Ω · m) * (0.039 m / (π * (0.0115 m)^2)) ≈ 4.2 x 10^-7 Ω (very small)

Assuming All Current Through Bird:

Current (I) = 45 A

Bird's Body Resistance (R_bird) = V_bird / I (Since we don't have V_bird, this is just a placeholder)

Total Circuit Resistance (ignoring wire resistance):

R_total = R_bird + R_wire ≈ R_bird (because R_wire is negligible)

Potential Difference Across Bird (upper bound):

V_bird = I * R_total = I * R_bird (since R_wire is negligible)

Answer:

Gravitational potential energy, PE = 2.8 J

Explanation:

It is given that,

Mass of the bluebird, m = 34 g = 0.034 kg

It flies from the ground to the top of an 8.5-m tree, h = 8.5 m

We need to find the change in the bluebird's gravitational potential energy as it flies to the top of the tree. It can be calculated as :

[tex]PE=m\times g\times h[/tex]

[tex]PE=0.034\ kg\times 9.8\ m/s^2\times 8.5\ m[/tex]

PE = 2.83 J

or

PE = 2.8 J

So, the gravitational potential energy as it flies to the top of the tree is 2.8 J. Hence, this is the required solution.

Answer:

a)  Initial angular speed = 30 rad/s

b) Final angular speed = 70 rad/s        

Explanation:

a) We have equation of motion s = ut + 0.5at²

    Here s = 400 radians

              t = 8 s

              a = 5 rad/s²

    Substituting

             400 = u x 8 + 0.5 x 5 x 8²

              u = 30 rad/s

   Initial angular speed = 30 rad/s

b) We have equation of motion v = u + at

     Here u = 30 rad/s

               t = 8 s

              a = 5 rad/s²  

    Substituting

             v = 30 + 5 x 8 = 70 rad/s    

   Final angular speed = 70 rad/s        

Answer:

[tex]T_2[/tex] = [tex]90.667K[/tex]

Explanation:

Given:

For the first solenoid

Number of turns, n₁ = 1200 turns/m

Current, I₁ = 2.5 A

Paramagnetic material temperature, T₁ = 320 K

Now for the second solenoid

Number of turns, n₂ = 1000 turns/m

Current, I₂ = 0.85 A

Paramagnetic material temperature = T₂

The magnetic flux (B) is given as

[tex]B=\frac{c\mu_onI}{T}[/tex]

where,

c = curie's constant

μ₀ = arithmetic constant

also it is given that the magnetization in both the cases are same

therefore the magnetic flux will also be equal

thus,

[tex]\frac{c\mu_on_1I_1}{T_1}[/tex] = [tex]\frac{c\mu_on_2I_2}{T_2}[/tex]

or

[tex]\frac{n_1I_1}{T_1}[/tex] = [tex]\frac{n_2I_2}{T_2}[/tex]

or

[tex]\frac{1200\times 2.5}{320}[/tex] = [tex]\frac{1000\times 0.85}{T_2}[/tex]

or

[tex]9.375[/tex] = [tex]\frac{850}{T_2}[/tex]

or

[tex]T_2[/tex] = [tex]\frac{850}{9.375}[/tex]

or

[tex]T_2[/tex] = [tex]90.667K[/tex]

Answer:

y = 43.2 cm

Explanation:

As we know by the formula of diffraction we have

[tex]a sin\theta = N\lambda[/tex]

here we have

[tex]\theta = 17.4[/tex]

N = 4

[tex]\lambda = 602 nm[/tex]

now we have

[tex]a sin(17.4) = 4(602\times 10^{-9})[/tex]

[tex]a = 8.05 \times 10^{-6} m[/tex]

now the distance of screen is 138 cm

now we can say

[tex]\frac{y}{L} = tan\theta[/tex]

[tex]y = L tan(\theta)[/tex]

[tex]y = 138 tan(17.4) = 43.2 cm[/tex]

Answer:

1.90 x 10² cal

1.90 x 10² cal

0.133 cal/(g °C)

Explanation:

For water :

[tex]m_{w}[/tex] = mass of water = 112 g

[tex]c_{w}[/tex] = specific heat of water = 1 cal/(g °C)

[tex]T_{wi}[/tex] = initial temperature of water = 90.0 °C

[tex]T_{wf}[/tex] = final temperature of water = 88.3 °C

[tex]Q_{w}[/tex] = Heat lost by water

Heat lost by water is given as

[tex]Q_{w}= m_{w}c_{w}(T_{wi} - T_{wf})[/tex]

[tex]Q_{w}[/tex] = (112) (1) (90.0 - 88.3)

[tex]Q_{w}[/tex] = 1.90 x 10² cal

[tex]Q_{B}[/tex] = Heat gained by the block

As per conservation of energy

Heat gained by the block = Heat lost by water

[tex]Q_{B}[/tex] = [tex]Q_{w}[/tex]

[tex]Q_{B}[/tex] = 1.90 x 10² cal

For Block :

[tex]m_{B}[/tex] = mass of block = 21.0 g

[tex]c_{B}[/tex] = specific heat of block

[tex]T_{bi}[/tex] = initial temperature of block = 20.0 °C

[tex]T_{bf}[/tex] = final temperature of block = 88.3 °C

[tex]Q_{B}[/tex] = Heat gained by Block = 1.90 x 10² cal

Heat gained by water is given as

[tex]Q_{B}[/tex] = m_{B}c_{B}(T_{bf} - T_{bi})[/tex]

1.90 x 10²  = (21.0) (88.3 - 20.0) c_{B}

c_{B} = 0.133 cal/(g °C)

Answer:

The time required for the package to hit the ground = 14.63 seconds.

Explanation:

Considering vertical motion of care package:-

Initial velocity, u =  0 m/s

Acceleration , a = 9.81 m/s²

Displacement, s = 1050 m

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

    1050 = 0 x t + 0.5 x 9.81 x t²

    t = 14.63 seconds

The time required for the package to hit the ground = 14.63 seconds.

To determine the time required for the care package to hit the ground, we can separate the vertical and horizontal components of the motion. By using the equation h = (1/2)gt^2 and plugging in the values h = 1050 m and g = 9.8 m/s^2, we find that it will take approximately 14.63 seconds for the care package to hit the ground.

To determine the time required for the care package to hit the ground, we can separate the vertical and horizontal components of the motion. Since the horizontal velocity of the airplane does not change, it will not affect the time of flight. The vertical motion can be analyzed using the equation of motion for free fall, which is h = (1/2)gt^2, where h is the initial vertical displacement, g is the acceleration due to gravity, and t is the time of flight. In this case, the initial vertical displacement is 1050 meters and the acceleration due to gravity is approximately 9.8 m/s^2. Plugging these values into the equation, we can solve for t and find the time required for the care package to hit the ground.

By using the equation h = (1/2)gt^2 and plugging in the values h = 1050 m and g = 9.8 m/s^2, we have 1050 = (1/2)(9.8)t^2. Solving for t, we get t^2 = (1050) / (0.5)(9.8) = 214.29. Taking the square root of both sides, we find t ≈ 14.63 seconds. Therefore, it will take approximately 14.63 seconds for the care package to hit the ground.

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Explanation:

The velocity field o a steady flow is given by :

[tex]V=(U_o+bx)i[/tex]

Where

U₀ and b are constant

We know that, the acceleration is given by :

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

[tex]a=\dfrac{d((U_o+bx)i)}{dt}[/tex]

a = b i

a = 0.3 m/s²

So, acceleration corresponding to this velocity is 0.3 m/s². Hence, this is the required solution.

Answer:

Required charge [tex]q=2.6\times 10^{9}C[/tex].

[tex]n=1.622\times 10^{10}\ electrons[/tex]

Explanation:

Given:

Diameter of the isolated plastic sphere = 25.0 cm

Magnitude of the Electric field = 1500 N/C

now

Electric field (E) is given as:

[tex]E =\frac{kq}{r^2}[/tex]

where,

k = coulomb's constant = 9 × 10⁹ N

q = required charge

r = distance of the point from the charge where electric field is being measured

The value of r at the just outside of the sphere = [tex]\frac{25.0}{2}=12.5cm=0.125m[/tex]

thus, according to the given data

[tex]1500N/C=\frac{9\times 10^{9}N\times q}{(0.125m)^2}[/tex]

or

[tex]q=\frac{0.125^2\times 1500}{9\times 10^{9}}[/tex]

or

Required charge [tex]q=2.6\times 10^{9}C[/tex].

Now,

the number of electrons (n) required will be

[tex]n=\frac{required\ charge}{charge\ of\ electron}[/tex]

or

[tex]n=\frac{2.6\times 10^{-9}}{1.602\times 10^{-19}}[/tex]

or

[tex]n=1.622\times 10^{10}\ electrons[/tex]

To produce an electric field of 1500 N/C just outside the surface of the plastic sphere, approximately 337.5 nC of excess charge must be distributed uniformly within its volume.

To produce an electric field just outside the surface of the sphere, the excess charge on the surface of the sphere should be uniformly distributed within its volume. The electric field just outside a uniformly charged sphere is given by:

E = k * q / r2

Where E is the electric field, k is the Coulomb constant [tex](9 * 109 N m2/C^2)[/tex] q is the charge, and r is the distance from the center of the sphere to the point outside the surface. In this case, E is given as 1500 N/C and r is half the diameter of the sphere (12.5 cm).

Substituting the values into the equation, we can solve for q:

[tex]1500 N/C = (9 * 109 N m2/C2) * q / (0.125 m)^2[/tex]

Solving for q, we find that the excess charge on the surface of the sphere should be approximately [tex]3.375 x 10-7 C,[/tex] or about 337.5 nC.

Answer:

1.85 x 10^10 cycles per second

Explanation:

B = 0.66 T, theta = 90 degree, q = 1.6 x 10^-19 C,

The time period of electron is given by

T = 2 π m / B q

Frequency is teh reciprocal of time period.

f = 1 /T

f = B q / (2 π m)

f = (0.66 x 1.6 x 10^-19) / (2 x 3.14 x 9.1 x 10^-31)

f = 1.85 x 10^10 cycles per second

Answer:

(c) 3P/5

Explanation:

The formula to calculate the power is:

[tex]P=\frac{W}{T}[/tex]

where

W is the work done

T is the time required for the work to be done

In the second part of the problem, we have

Work done: 3W

Time interval: 5T

So the power required is

[tex]P=\frac{3W}{5T}=\frac{3}{5}\frac{W}{T}=\frac{3}{5}P[/tex]

Answer:

f = 91.1 Hz

Explanation:

As we know that it is given here that wavelength of sound in platinum is same as the wavelength of sound in air

so we can use the formula of wavelength in two mediums

[tex]\frac{speed}{frequency} = \frac{speed}{frequency}[/tex]

now it is given that

[tex]v_{air} = 340 m/s[/tex]

[tex]v_{pt} = 2800 m/s[/tex]

frequency of sound in platinum is 750 Hz

now frequency of sound in air = f

now from above formula

[tex]\frac{340}{f} = \frac{2800}{750}[/tex]

[tex]f = 91.1 Hz[/tex]

Answer:

26.8 m/s

Explanation:

[tex]v[/tex]  = constant speed of the car

[tex]V[/tex]  = speed of sound = 343 m/s

[tex]f[/tex] = actual frequency of the horn

[tex]f_{app}[/tex] = frequency heard as the car approach = 76 Hz

frequency heard as the car approach is given as

[tex]f_{app}=\frac{vf}{V - v}[/tex]

[tex]76 =\frac{vf}{343 - v}[/tex]                               eq-1

[tex]f_{rec}[/tex] = frequency heard as the car recedes = 65 Hz

frequency heard as the car goes away is given as

[tex]f_{rec}=\frac{vf}{V + v}[/tex]

[tex]65 =\frac{vf}{343 + v}[/tex]                                  eq-2

dividing eq-1 by eq-2

[tex]\frac{76}{65}=\frac{343+v}{343-v}[/tex]

[tex]v[/tex] = 26.8 m/s

Final answer:

To determine the car's speed using the Doppler Effect, we calculate the difference in observed sound frequencies as the car approaches and moves away. Applying formulas for Doppler Effect calculations, the speed of the car comes out to be approximately 14.6 m/s.

Explanation:

The question revolves around the phenomenon known as the Doppler Effect, which is observed when a sound source moves relative to an observer. To calculate the speed v of the car, we use the Doppler Effect equations for sound frequencies heard when the source is moving towards and then away from the observer:

For the source approaching:

f' = f * ((v + vo) / (v - vs))

, where:

f' is the observed frequency when the source is approaching (76 Hz)

f is the original frequency emitted by the source

v is the speed of sound (343 m/s)

vo is the speed of the observer (0 m/s, since the observer is stationary)

vs is the speed of the source (the car's speed, what we are solving for)

For the source receding:

f'' = f * ((v - vo) / (v + vs))

, where:

f'' is the observed frequency when the source is receding (65 Hz)

To find the car's speed, we need to solve for vs in both equations. By eliminating f (since it's the same for both equations), we can solve for vs. Using these equations, we determine that the speed of the car is approximately 14.6 m/s.

Answer:

angular speed of both the children will be same

Explanation:

Rate of revolution of the merry go round is given as

f = 4.04 rev/min

so here we have

[tex]f = \frac{4.04}{60} =0.067 rev/s[/tex]

here we know that angular frequency is given as

[tex]\omega = 2\pi f[/tex]

[tex]\omega = 2\pi(0.067)[/tex]

[tex]\omega = 0.42 rad/s[/tex]

now this is the angular speed of the disc and this speed will remain same for all points lying on the disc

Angular speed do not depends on the distance from the center but it will be same for all positions of the disc

Answer:

Part a)

L = 25.13 kg m^2/s

Part b)

N = 3200 rev

Part c)

torque = 0.837 Nm

Explanation:

Part a)

As we know that angular frequency is given as

[tex]\omega = 2\pi f[/tex]

[tex]\omega = 2\pi(800/60)[/tex]

[tex]\omega = 83.77 rad/s[/tex]

now the angular momentum is given as

[tex]L = I\omega[/tex]

[tex]L = (0.3)(83.77)[/tex]

[tex]L = 25.13 kg m^2/s[/tex]

Part b)

angular acceleration of fan is given as

[tex]\alpha = 2\pi\frac{800/60}{30}[/tex]

[tex]\alpha = 2.79 rad/s^2[/tex]

total number of revolutions are

[tex]N = \frac{1}{4\pi}\alpha t^2[/tex]

[tex]N = \frac{1}{4\pi}(2.79)(120^2)[/tex]

[tex]N = 3200 rev[/tex]

Part c)

Torque is given as

[tex]\tau = I\alpha[/tex]

[tex]\tau = (0.30)(2.79) = 0.837 Nm[/tex]