Photons with wavelength 1 pm are incident on electrons. What is the frequency of the Compton-scattered photons at an angle of 60°?

Explanation:

It is given that,

Mass of object, m = 2 kg

Initial velocity, u = 3 m/s (east)

Final velocity, v = - 7 m/s (west)

The Impulse can be calculated using the change in momentum of an object i.e.

J = m(v-u)

[tex]J=2\ kg(-7\ m/s-3\ m/s)[/tex]

J = -20 kg-m/s

So, the Impulse of this object is 20 kg-m/s but the direction is opposite. Hence, statement (1) is correct i.e. Student #1: "The impulse is equal to the change in momentum, which is (2 kg)(3 m/s + 7 m/s) = 20 kg m/s."

The correct student to agree with is Student #2. The impulse applied to an object is equal to the change in its momentum.

The change in momentum is calculated by multiplying the mass of the object by the change in its velocity.

First, let's calculate the change in velocity.

[tex]\[ \Delta v = v_f - v_i = (-7 \text{ m/s}) - (3 \text{ m/s}) = -10 \text{ m/s} \][/tex]

The negative sign indicates that the velocity has changed in the opposite direction. The magnitude of the change in velocity is 10 m/s.

Now, we multiply the mass of the object by the change in velocity to find the change in momentum:

[tex]\[ \Delta p = m \cdot \Delta v = (2 \text{ kg}) \cdot (-10 \text{ m/s}) = -20 \text{ kg m/s} \][/tex]

The impulse, which is equal to the change in momentum, is therefore:

[tex]\[ \text{Impulse} = \Delta p = -20 \text{ kg m/s} \][/tex]

The negative sign indicates that the impulse is in the opposite direction to the initial velocity. The magnitude of the impulse is 20 kg m/s.

Student #1 incorrectly adds the initial and final velocities without considering their directions. Student #3 incorrectly divides the change in momentum by the time interval, which is not necessary since impulse is the change in momentum, not the rate of change of momentum. Student #4 incorrectly states that we cannot find the impulse without knowing the force; however, we can find the impulse using the change in momentum, as we have done.

Answer:

1.5 m/s²

Explanation:

m = mass of the clarinet case = 3.450 kg

F = upward force on the clarinet through the air = 38.92 N

W = weight of the clarinet case in down direction

weight of the clarinet case is given as

W = mg

W = (3.450) (9.8)

W = 33.81 N

a = acceleration of the case

Force equation for the motion of the case is given as

F - W = ma

38.92 - 33.81 = (3.450) a

a = 1.5 m/s²

Answer:

5.43 x 10^-3 Nm

Explanation:

N = 52.5, radius, r = 5.35 cm = 0.0535 m, B = 0.455 T, I = 25.3 mA = 0.0253 A

Torque = N I A B Sin theta

Here, theta = 90 degree

Torque = 52.5 x 0.0253 x 3.14 x 0.0535 x 0.0535 x 0.455

Torque = 5.43 x 10^-3 Nm

Final answer:

The magnitude of the impulse delivered by the bat to the ball is 5.1 kg·m/s. The magnitude of the average force exerted by the bat on the ball is 3.4 × 10^3 N.

Explanation:

To find the magnitude of the impulse delivered by the bat to the ball, we need to use the formula for impulse:

Impulse = change in momentum

Momentum is given by the product of mass and velocity (momentum = mass × velocity). The change in momentum is the final momentum minus the initial momentum (change in momentum = final momentum - initial momentum).

Given that the mass of the baseball is 0.15 kg, the initial velocity is +17 m/s, and the final velocity is -17 m/s, we can calculate the magnitude of the impulse:

Impulse = (0.15 kg × -17 m/s) - (0.15 kg × 17 m/s) = -5.1 kg·m/s

So, the magnitude of the impulse delivered by the bat to the ball is 5.1 kg·m/s.

To find the magnitude of the average force exerted by the bat on the ball, we can use the formula for average force:

Average force = impulse / time

Given that the time the ball is in contact with the bat is 1.5 ms (1.5 × 10^-3 s) and the magnitude of the impulse is 5.1 kg·m/s, we can calculate the magnitude of the average force:

Average force = 5.1 kg·m/s / (1.5 × 10^-3 s) = 3.4 × 10^3 N

Therefore, the magnitude of the average force exerted by the bat on the ball is 3.4 × 10^3 N.

A 0.293-kg particle undergoes simple harmonic motion along the horizontal x-axis between the points x1 = -0.285 m and x2 = 0.395 m. The period of oscillation is 0.641 s. Find the frequency, f, the equilibrium position, xeq, the amplitude, A, the maximum speed, vmax, the maximum magnitude of acceleration, amax, the force constant, k, and the total mechanical energy, Etot.

That being said,

f=1/T=1/0.641=1.56 s

xeq=(x1+x2)/2=(-0.285+0.395)/2=0.055 m

A=(x2-x1)/2=(0.395+0.285)/2=0.340 m

x(t)=Acos(wt+ϕ)

v(t)=-wAsen(wt+ϕ)

w=2πf=2*3.14*1.56=9.8 rad/s

vmax=-wA=9.8*0.340=-3.3 m/s

a(t)=-w^2Acos(wt+ϕ)

amax=-w^2A=96.04*0.340=-32.6 m/s^2

w=(k/m)^1/2 => k=mw^2=0.293*96.04=28.1 N/m

Etot=1/2kA^2

Etot=0.5*28.1*0.116=1.63 J

The characteristics of the simple harmonic movement allow to find the results for the questions about the movement of the mass with the spring are:

     a) The frequency is: f = 1.72 Hz

     b) The equilibrium position is: x₀ = 0.088 m

     c) The amplitude is: A = 0.686 m

     d) The maximum speed is: v = 7.42 m / s

     e) the maximum acceleration is: a = 80.16 m / s²

     f) The mechanical energy is: E = 3.44 J

Given parameters

To find.

    a) Frequency.

    b) the equilibrium position.

    c) The amplitude.

    d) Maximum speed.

    e) Maximum acceleration.

    f) The total mechanical energy.

Simple harmonic motion is a periodic motion where the restoring force is proportional to the elongation. This movement is fully described by the expression

       x = A cos (wt + Ф)

       w² = [tex]\sqrt{ \frac{k}{m} }[/tex]  

Where x is the displacement, A the amplitude of the movement, w the angular velocity, t the time, Ф a phase constant that meets the initial conditions, k the spring constant and m the mass attached to the spring.

a) Frequency and period are related.

          [tex]f= \frac{1}{T}[/tex]  

Let's calculate.

        [tex]f = \frac{1}{0.581 }[/tex]  

         f = 1.72 Hz

b) The equilibrium position occurs at the midpoint of the movement

       [tex]x_o = \frac{x_f + x_o}{2}[/tex]  

       [tex]x_o = \frac{0.387 + (-0.299)}{2}[/tex]  

       x₀ = 0.088 m

c) The amplitude is the maximum elongation of the spring

        [tex]A= x_f - x_o[/tex]  

        A = 0.387 - (-0.299)

        A = 0.686 m

d) The speed is defined by the variation of the position with respect to time.

          [tex]v = \frac{dx}{dt}[/tex]

Let's make the derivatives

          v = - A w cos (wt + Ф)

The angular velocity is related to the period.

        [tex]w= \frac{\pi }{T} \\w= \frac{2 \pi }{0.581}[/tex]

        w = 10.81 rad / s

The speed is maximum when the cosine  is maximum ±1

        v = A w

        v = 0.686 10.81

        v = 7.42 m / s

e) Acceleration is defined as the change in velocity over time.

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

           a = -A w² sin (wt + Ф)

The acceleration is maximum when the sine function is maximum ±1

          a = A w²

          a = 0.686 10.81²

          a = 80.16 m / s²

f) The mechanical energy is calculate for the point of maximum elongation.

          E = ½ k A²  

          k = w² m

Let's replace.

          E = ½ w² m A²

Let's calculate.

          E = ½ 10.81² 0.125 0.686²

          E = 3.44 J

In conclusion, using the characteristics of simple harmonic motion we can find the results for the questions about the motion of the mass with the spring are:

     a) The frequency is: f = 1.72 Hz

     b) The equilibrium position is: x₀ = 0.088 m

     c) The amplitude is: A = 0.686 m

     d) The maximum speed is: v = 7.42 m / s

     e) the maximum acceleration is: a = 80.16 m / s²

     f) The mechanical energy is: E = 3.44 J

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Answer:

No answer is correct.

Explanation:

given data:

Q1 = 400 gpm

P1 = 28 psi

Q2 = ?

P2 = 30 psi

Change in pressure in a pipe is given as

[tex]\Delta P = \frac{32\mu vl}{D^{2}}[/tex]

where v is velocity and it is given as [tex]v = \frac{Q}{A}[/tex]

[tex]\Delta P = \frac{32\mu Ql}{AD^{2}}[/tex]

Therefore, change in pressure is directly proportional to flow

thus we have

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

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

[tex]Q_{2}=\frac{30}{28}*400 = 428.57 gpm[/tex]

[tex]Q_{2} =428.57 gpm[/tex]

no answer is correct

Answer:

428.5 gmp

Explanation:

Given that,

A piping system is operating at quantity = 400 gpm

Pressure = 28 psi

Increased pressure = 30 psi

We need to calculate the resultant flow rate

We know that,

The flow rate is directly proportional to pressure

[tex]Q\propto P[/tex]

Therefore,

[tex]\dfrac{Q_1}{Q_2}=\dfrac{P_1}{P_2}[/tex]

where,

[tex]Q_1\rightarrow 400\text{ gpm}[/tex]

[tex]Q_2\rightarrow x\text{ gpm}[/tex]

[tex]P_1\rightarrow 28\text{ psi}[/tex]

[tex]P_2\rightarrow 30\text{ psi}[/tex]

By substituting into formula

[tex]\dfrac{400}{x}=\dfrac{28}{30}[/tex]

[tex]x=\dfrac{12000}{28}\approx 428.5\text{ gpm}[/tex]

Hence, The resultant flow rate will be 428.5 gmp

Answer:

the lowest frequency is [tex]7.5\times 10^{14} Hz[/tex]

Explanation:

In the question it is given that wavelength(L) in the range of 200μm to 400μm.

let ν be frequency of wave v velocity = 3\times 10^8

velocity v= Lν

therefore ν=[tex]\frac{v}{L}[/tex]

frequency ν be lopwest when L will be heighest

ν(lowest)=[tex]\frac{3\times 10^8}{400\times 10^-9}[/tex]

ν=[tex]7.5\times 10^{14} Hz[/tex]

Answer:

2.16 x 10^7 years

Explanation:

R = 6.9 x 10^20 m

ω = 9.2 x 10^-15 rad/s

T = 2π / ω

T = ( 2 x 3.14 ) / ( 9.2 x 10^-15)

T = 6.826 x 10^14 second

T = 2.16 x 10^7 years

By using the star's given angular velocity and the formula for a revolution's time period, it can be calculated that it takes the star approximately 2.17 x 10^8 years to complete one revolution around the center of the galaxy.

The timeframe in which a star completes one revolution around the center of a galaxy is referred to as a galactic year. In this scenario, the star's time period for one revolution, T, can be calculated using the formula T = 2π/ω, where ω represents the star's angular speed. So, T = 2π / (9.2 x 10^-15 rad/s). The resulting value is in seconds since the angular velocity was in rad/s. This can then be converted to years by using the relevant conversion factor: 1 year = 3.15 x 10^7 seconds.

By performing these calculations, we find that it takes the star about 2.17 x 10^8 years to complete one revolution around the center of the galaxy. It is important to note that this is a much larger timeframe than a human lifetime, meaning that humans have only seen a small fraction of a galactic year.

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Answer:

10.2 m

Explanation:

A = 70 mm^2 = 70 x 10^-6 m^2

F = 7 N

Pressure sustained by ear drum

P = F / A = 7 / (70 x 10^-6) = 10^5 N/m^2

Let the depth of water is h

Pressure due to the depth of water should be same as the pressure sustained by the ear drum

P = h x d x g

where, d be the density of water.

10^5 = h x 1000 x 9.8

h = 10.2 m

The discomfort and pressure on the eardrums when diving underwater is due to increased pressure caused by depth. The maximum depth a person can dive without rupturing their eardrum can be calculated by dividing the maximum pressure the eardrum can sustain by the pressure increase per meter. In this case, the maximum depth is about 10 meters.

The discomfort and pressure on your eardrums when diving underwater is due to the increased pressure that comes with depth. The pressure exerted on your eardrums underwater is a result of the weight of the water above you and the air above the water's surface. To calculate the maximum depth you can dive without rupturing your eardrum, you need to determine the pressure that the eardrum can sustain and then divide it by the pressure difference caused by the depth.

The maximum force the eardrum can sustain is 7 N, and the area of the eardrum is 7x10-5 m2. The maximum pressure the eardrum can sustain is therefore 7 N / (7x10-5 m2) = 1x105 Pa.

The pressure underwater increases by 1 atmosphere for every 33 feet (10 meters) of saltwater depth. Using this information, you can calculate the maximum depth:

Maximum Depth = Maximum Pressure / Pressure Increase per Meter = 1x105 Pa / (1x105 Pa/m) = 10 meters

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(a) [tex]2.98\cdot 10^{10} J[/tex]

The change in energy of the transferred charge is given by:

[tex]\Delta U = q \Delta V[/tex]

where

q is the charge transferred

[tex]\Delta V[/tex] is the potential difference between the ground and the clouds

Here we have

[tex]q=31 C[/tex]

[tex]\Delta V = 0.96\cdot 10^9 V[/tex]

So the change in energy is

[tex]\Delta U = (31 C)(0.96\cdot 10^9 V)=2.98\cdot 10^{10} J[/tex]

(b) 7921 m/s

If the energy released is used to accelerate the car from rest, than its final kinetic energy would be

[tex]K=\frac{1}{2}mv^2[/tex]

where

m = 950 kg is the mass of the car

v is the final speed of the car

Here the energy given to the car is

[tex]K=2.98\cdot 10^{10} J[/tex]

Therefore by re-arranging the equation, we find the final speed of the car:

[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(2.98\cdot 10^{10})}{950}}=7921 m/s[/tex]

Explanation:

It is given that,

Speed of the plane, v = 110 m/s

Radius of circle, r = 2850 m

(1) Let t is the time taken by a plane. Speed of the plane is given by :

[tex]v=\dfrac{d}{t}=\dfrac{2\pi r}{t}[/tex]

[tex]t=\dfrac{2\pi r}{v}[/tex]

[tex]t=\dfrac{2\pi \times 2850}{110}[/tex]

t = 162.79 s

(2) Mass of a driver, m = 50 kg

Height, h = 6 m

As the driver reaches ground, its potential energy point to zero. So, the change in potential energy is given by :

[tex]\Delta PE=PE_f-PE_i[/tex]

[tex]\Delta PE=0-mgh[/tex]

[tex]\Delta PE=-50\times 9.8\times 6[/tex]

[tex]\Delta PE=-2940\ J[/tex]

So, the change in potential energy is 2940 joules.

Let v is the speed. As the driver reaches water surface, its potential energy is converted to kinetic energy as :

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

[tex]v=\sqrt{\dfrac{2\times PE}{m}}[/tex]

[tex]v=\sqrt{\dfrac{2\times 2940}{50}}[/tex]

v = 10.84 m/s

So, the speed of the reactors the water is 10.84 m/s. Hence, this is the required solution.

It takes approximately 52.3 seconds for a plane, traveling at a constant speed of 110 m/s, to fly once around a circle with a radius of 2850 m. The change in potential energy for a driver of mass 50 kg jumping off a 6 m high cliff is -2940 J. The driver's velocity as they reach the water is approximately 10.8 m/s.

To find the time it takes for a plane to fly once around a circle, we need to calculate the circumference of the circle and divide it by the plane's speed. The formula for the circumference of a circle is C = 2πr, where r is the radius. In this case, the radius is given as 2850 m, so the circumference is 2π(2850) = 5700π m. Dividing the circumference by the plane's speed of 110 m/s gives us the time it takes to complete one lap around the circle: (5700π)/(110) ≈ 52.3 s.

When a driver jumps off a cliff, the force of gravity causes her potential energy to change into kinetic energy as she falls. The formula for potential energy is PE = mgh, where m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s²), and h is the height. In this case, the mass is given as 50 kg and the height is 6 m. Plugging these values into the formula, we get PE = (50)(9.8)(6) = 2940 J. The change in potential energy is simply the negative of the initial potential energy, since it decreases when the driver jumps off the cliff. Therefore, the change in potential energy is -2940 J.

As the driver reaches the water, all of her potential energy has been converted into kinetic energy. The formula for kinetic energy is KE = 0.5mv², where m is the mass and v is the velocity. Since the potential energy at the end of the fall is 0, the entire initial potential energy is now kinetic energy. Plugging in the mass of 50 kg, we can solve for the velocity using the formula 2940 = 0.5(50)v². Simplifying this equation, we get v² = 2940/25 = 117.6. Taking the square root of both sides, we find that v ≈ 10.8 m/s.

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Answer:

6.38 x 10^4 J

Explanation:

d = 0.33 cm = 0.33 x 10^-2 m, Area = 87 x 36 cm^2 = 0.87 x 0.36 m^2

ΔT = 14 degree C, t = 1 min = 60 second

K = 0.8 W / m K

Heat  = K A ΔT t / d

H = 0.8 x 0.87 x 0.36 x 14 x 60 / (0.33 x 10^-2)

H = 6.38 x 10^4 J

The question is related to calculating the rate of heat conduction through a glass window in Physics. For a complete answer, the thermal conductivity of glass is needed, which is not provided in the question. With that value, the formula for heat conduction can be used to find the heat flow per minute.

The subject of this question is Physics, and it applies to a High School grade level. To determine the rate of heat conduction through a window, we can use the formula for heat transfer through conduction:

Q = \(\frac{k \cdot A \cdot \Delta T \cdot t}{d}\)

where:
Q is the heat transferred,
k is the thermal conductivity,
A is the area,
\(\Delta T\) is the temperature difference,
t is the time, and
d is the thickness of the material.

However, the question does not provide the value of the thermal conductivity (k) for glass, which is necessary to calculate the rate of heat conduction. With the thermal conductivity value, we could then insert the given dimensions, temperature difference, and time into this formula to find the heat flow per minute. Without this value, the question cannot be fully answered as the data is incomplete. For a complete solution, thermal conductivity of the material is required.

Explanation:

It is given that,

Mass of concrete pilling, m = 50 kg

Diameter of wire, d = 1 mm

Radius of wire, r = 0.0005 m

Length of wire, L = 11.2

Young modulus of steel, [tex]Y=20\times 10^{10}\ N/m^2[/tex]

The young modulus of a wire is given by :

[tex]Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}[/tex]

[tex]Y=\dfrac{F.L}{A\Delta L}[/tex]

[tex]\Delta L=\dfrac{F.L}{A.Y}[/tex]

[tex]\Delta L=\dfrac{50\ kg\times 9.8\ m/s^2\times 11.2\ m}{\pi (0.0005\ m)^2\times 20\times 10^{10}\ N/m^2}[/tex]

[tex]\Delta L=0.034\ m[/tex]

So, the wire will stretch 0.034 meters. Hence, this is the required solution.

Answer:

The work done on the test charge is 4.5 joules

Explanation:

It is given that,

Electric potential difference, [tex]V=1000\ V[/tex]

Charged particle, [tex]q=4.5\times 10^{-3}\ C[/tex]

Work done on the test charge is given by the product of test charge and potential difference i.e.

W = q × V

[tex]W=4.5\times 10^{-3}\ C\times 1000\ V[/tex]

W = 4.5 Joules

So, the work done on the test charge is 4.5 joules. Hence, this is the required solution.

Answer:

9.4 N

Explanation:

m = mass of the book = 1.83 kg

[tex]\mu _{s}[/tex] = Coefficient of static friction = 0.522

[tex]\mu _{k}[/tex] = Coefficient of kinetic friction = 0.283

[tex]f_{s}[/tex] = Static frictional force

F = force needed to make the book move

force needed to make the book move is same as the magnitude of maximum static frictional force applied by the desk on the book

Static frictional force is given as

[tex]f_{s}[/tex] = [tex]\mu _{s}[/tex] mg

Hence, the force need to move the book is given as

F = [tex]f_{s}[/tex] = [tex]\mu _{s}[/tex] mg

F = [tex]\mu _{s}[/tex] mg

F = (0.522) (1.83 x 9.8)

F = 9.4 N

Answer:

velocity of red ball is either 0 or 1 m/s and then the velocity of green ball is 10 m/s or 8 m/s.

Explanation:

mass of red ball, m1 = 0.05 kg u 1 = 10 m/s

mass of green ball, m2 = 0.1 kg, u2 = 0

Let the velocity of red ball and green ball after the collision is v1 and v2, respectively.

By use of conservation of momentum

m1 u1 + m2 u2 = m1 v1 + m2 v2

0.05 x 10 + 0 = 0.05 x v1 + 0.1 x v2

0.5 = 0.05 (v1 + 2 v2)

v1 + 2v2 = 10 ...... (1)

Now use conservation of kinetic energy

1/2 m1 x u1^2 + 1/2 x m2 x u2^2 = 1/2 m1 v1^2 + 1/2 m2 x v2^2

0.05 x 10 x 10 + 0 = 0.05 x v1^2 + 01. x v2^2

5 = 0.05(v1^2 + 2v2^2)

10 = v1^2 + 2 v2^2 .....(2)

Put teh value of v1 from equation (1) in equation (2)

10 = 10 - 2 v2 + 2 v2^2

0 = v2^2 - v2

v2 =0 , 1 m/s

So, v1 = 10 - 2 x 0 = 10  m/s

or v1 = 10 - 2 x 1 = 8 m/s

Thus, the velocity of red ball is either 0 or 1 m/s and then the velocity of green ball is 10 m/s or 8 m/s.

Answer:

The no of revolutions is 2.032 revolution.

Explanation:

Given that,

Moment of inertia = 0.85 Kgm²

Radius = 170 mm

Force = 32 N

Time =  2s

We need to calculate the angular acceleration

Using formula of torque

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

[tex]\alpha=\dfrac{\tau}{I}[/tex]

[tex]\alpha=\dfrac{F\times r}{I}[/tex]

Where, F = force

r = radius

I = moment of inertia

Put the value into the formula

[tex]\alpha=\dfrac{32\times170\times10^{-3}}{0.85}[/tex]

[tex]\alpha=6.4\ m/s^2[/tex]

We need to calculate the rotational speed

Using equation of angular motion

[tex]\omega_{f}=\omega_{i}+\alpha t[/tex]

[tex]\omega_{f}=6.4\times2[/tex]

[tex]\omega=12.8\ rad/s[/tex]

We need to calculate the angular position

Using equation of angular motion

[tex]\theta=\omega_{i}+\dfrac{1}{2}\alpha t^2[/tex]

[tex]\theta=0+\dfrac{1}{2}\times6.4\times4[/tex]

[tex]\theta=12.8\ radian[/tex]

We need to calculate no of revolutions

[tex]n = \dfrac{\theta}{2\pi}[/tex]

[tex]n=\dfrac{12.8}{2\times3.15}[/tex]

[tex]n=2.032\ revolution[/tex]

Hence, The no of revolutions is 2.032 revolution.

Answer:

a) 17 N

b) 21 N

Explanation:

At the 3 o'clock position, the sum of the forces towards the center is:

∑F = ma

T = m v² / r

19 = m v² / r

At the 12 o'clock position, the sum of the forces towards the center is:

∑F = ma

T + mg = m v² / r

T + (0.18)(9.8) = 19

T = 17.2 N

At the 6 o'clock position, the sum of the forces towards the center is:

∑F = ma

T − mg = m v² / r

T − (0.18)(9.8) = 19

T = 20.8 N

Rounding to two significant figures, the tensions are 17 N and 21 N.

To find the tension in the stick at the twelve o’clock and six o’clock positions, consider the forces acting on the ball at each position and calculate the tension using the equation T = mg + mv^2 / r for twelve o'clock and T = mg - mv^2 / r for six o'clock.

To find the tension in the stick when the ball is at the twelve o’clock and six o’clock positions, we need to consider the forces acting on the ball at each position. At the twelve o'clock position, the tension in the stick is equal to the weight of the ball plus the centripetal force required to keep it moving in a circle. The tension can be calculated using the equation T = mg + mv^2 / r, where T is the tension, m is the mass of the ball, g is the acceleration due to gravity, v is the velocity of the ball, and r is the radius of the circle.

At the six o'clock position, the tension in the stick is equal to the weight of the ball minus the centripetal force. Using the same equation, the tension can be calculated by subtracting mv^2 / r from mg.

Substituting the given values into the equation for each position will give you the tension in the stick.

Answer:

377.78 m/s

Explanation:

Let the speed of bullet is v.

Time taken by the sound is t' and time taken by the bullet is t.

Speed of sound = 340 m/s

Distance = 170 m

Time taken by the sound, t' = distance / speed of sound = 170 / 340 = 0.5 sec

time taken by the bullet, t = 0.95 - 0.5 = 0.45 sec

Speed of bullet = distance / time taken by bullet

Speed of bullet = 170 / 0.45 = 377.78 m/s

Answer:

24 m/s

Explanation:

Let's say that under very icy conditions, there is no friction.

Draw a free body diagram.  There are 2 forces acting on the van.  Gravity straight down, and normal force perpendicular to the surface.

Sum of the forces in the radial (+x) direction:

∑F = ma

N sin θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ − mg = 0

Solve for N in the second equation:

N cos θ = mg

N = mg / cos θ

Substitute into the first equation:

(mg / cos θ) sin θ = m v² / r

mg tan θ = m v² / r

g tan θ = v² / r

v = √(gr tan θ)

Given g = 9.8 m/s², r = 130 m, and θ = 24.8°:

v = √(9.8 m/s² × 130 m × tan 24.8°)

v = 24.3 m/s

Rounded to two significant figures, the maximum velocity is 24 m/s (approximately 54 mph).

Answer:

24 m/s

Explanation:

The maximum speed that a minivan can have and still follow the curve safely under very icy conditions is 24 m/s.

g = 9.8 m/s², r = 130 m, and θ = 24.8°:

v = √(9.8 m/s² × 130 m × tan 24.8°)

v = 24.3 m/s

Answer:

The largest equivalent resistance yu can build using these three resistors is a Serie Resistance with the value of R= 16.74 Ω

Explanation:

Adding Resistances in serie is the way to build de largest equivalent value possible.

Rt= R1+R2+R3

Rt= 6.32 + 8.13 + 2.29

Rt= 16.74Ω

Answer:

2.13 h

Explanation:

Simple harmonic motion is:

x = A sin(2π/T t + φ) + B

where A is the amplitude, T is the period, φ is the phase shift, and B is the midline.

This can also be written in terms of cosine:

x = A cos(2π/T t + φ) + B

Here, A = d/2, T = 12.8, φ = 0, and B = d/2.  I'll use cosine so that the highest level is at t=0.

x = d/2 cos(2π/12.8 t) + d/2

When x = d - 0.250 d = 0.750 d:

0.750 d = d/2 cos(2π/12.8 t) + d/2

0.250 d = d/2 cos(2π/12.8 t)

0.500 = cos(2π/12.8 t)

π/3 = 2π/12.8 t

12.8/6 = t

t = 2.13

It takes 2.13 hours to fall 0.250 d from the highest level.

The time it takes for the water to fall a distance 0.250d from its highest level in the tide cycle can be calculated by determining a quarter of the period, which is 3.2 hours.

The time it takes for the water to fall a distance 0.250d from its highest level can be calculated using the concept of simple harmonic motion. Since the period of the tide is 12.8 hours, to find the time for the water to fall 0.250d, we need to determine the fraction of the period corresponding to this distance.

Given that the complete cycle from highest to lowest level is 12.8 hours, the time to fall 0.250d would be one-quarter of that period, which equals 3.2 hours.

The power delivered to the solenoid is calculated by finding the current through the solenoid and its resistance, then applying the formula P=I²R. This involves applying known laws and formulas of electromagnetism.

The power delivered to the solenoid can be calculated given the field strength inside the solenoid and the physical dimensions and properties of the solenoid and wire. The field strength inside a solenoid is given by the formula B = µon, where n is the number of loops per unit length. We first find n, and then we need to find the current I because the power P delivered to the solenoid is given by P=I²R, where R is the resistance of the wire.

The resistance R of the wire can be found using the formula R = ρL/A, where ρ is the resistivity of copper, L is the length of the wire, and A is the cross-sectional area of the wire. After calculating R and I, we can compute P. Therefore, it's paramount to apply known formulas and laws of electromagnetism to solve this. It's important to note that B is the magnetic field inside the solenoid, µo is the permeability of free space and n is number of turns per unit length.

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To produce a magnetic field of 5.10 mT at the center of a solenoid, a power of 61.4 W must be delivered. This is calculated by determining the number of turns, the required current, the resistance of the copper wire, and then using the power formula. The key parameters are the number of turns (700), current (3.61 A), resistance (4.71 Ω), and power (61.4 W).

To solve this problem, we'll need to follow these steps:

Calculate the number of turns of wire on the solenoid.

Determine the current needed to produce the required magnetic field.

Find the resistance of the wire.

Calculate the power delivered to the solenoid.

1. Calculate the number of turns of wire

The diameter of the wire is 0.100 cm, so the circumference of each turn is approximately equal to the solenoid's length (70.0 cm). The number of turns (N) is:

N = 70.0 cm / 0.100 cm = 700 turns

2. Determine the current needed

The formula for the magnetic field (B) inside a solenoid is given by:

B = μ₀ * (N/L) * I

where:

B = 5.10 mT = 5.10 x 10⁻³ T

μ₀ = 4π * 10⁻⁷ T⋅m/A (permeability of free space)

N = 700 turns

L = 70.0 cm = 0.70 m

Solving for the current (I):

5.10 x 10⁻³ = (4π x 10⁻⁷) * (700 / 0.70) * I

5.10 x 10⁻³ = (4π x 10⁻⁷) * 1000 * I

5.10 x 10⁻³ = (4π x 10⁻⁴) * I

I = 3.61 A

3. Find the resistance of the wire

The resistance (R) of the wire is given by:

R = ρ * (L/A)

where:

ρ = 1.68 x 10⁻⁸ Ω⋅m (resistivity of copper)

L = N * circumference of each turn

A = * (diameter/2)²

Calculating the total length (L) of the wire:

L = 700 * 0.100 * m = 219.91 m (approximately)

Calculating the cross-sectional area (A):

A = * (0.001 m / 2)²

A = 3.14 x (0.0005)²

A = 7.85 x 10⁻⁷ m²

Thus, the resistance (R) is:

R = 1.68 x 10⁻⁸ * (219.91 / 7.85 x 10⁻⁷)

R = 1.68 x 10⁻⁸ * (28.014 x 10⁷)

R  = 4.71 Ω

4. Calculate the power delivered

Power (P) is given by:

P = I² * R

Substituting in the values:

P = (3.61)² * 4.71 = 13.032 * 4.71 =  61.4 W

Therefore, 61.4 W power must be delivered to the solenoid to produce the required magnetic field.

Answer:

Original speed of the bullet = 305.21 m/s

Explanation:

Here momentum is conserved.

Mass of bullet = 9.6 g = 0.0096 kg

Mass of wood = 4.95 kg

Let velocity of bullet be v.

Initial momentum = 0.0096 v

Final mass = 0.0096 + 4.95 = 4.9596 kg

Final velocity = 0.591 m/s

Final momentum = 4.9596 x 0.591 = 2.93 kgm/s

Equating both momentum

           0.0096 v = 2.93

                      v = 305.21 m/s

Original speed of the bullet = 305.21 m/s

Answer:

C) is at the equilibrium point

Explanation:

As we know that the equation of displacement of SHM executing object is given as

[tex]x = A sin(\omega t + \phi)[/tex]

now for velocity of the particle we will have

[tex]v = \frac{dx}{dt}[/tex]

now we will have

[tex]v = A\omega cos(\omega t + \phi)[/tex]

now if velocity is maximum then we will have

[tex]cos(\omega t + \phi) = \pm 1[/tex]

so at this situation we will have

[tex](\omega t + \phi) = N\pi[/tex]

now at this angle the value of

[tex]sin(\omega t + \phi) = 0[/tex]

so the position must be mean position at which the particle is at equilibrium position

In simple harmonic motion, an object oscillating vertically reaches its maximum speed at the equilibrium point. This is the position where the object would rest in the absence of force and is where the maximum velocity is attained during the oscillation.

In the context of simple harmonic motion, an object oscillating in the vertical direction attains its maximum speed at the equilibrium point (C). This equilibrium point represents the position where the object would naturally rest in the absence of force. An object attached to a spring and placed on a frictionless surface serves as a straightforward example of a simple harmonic oscillator. When displaced from its equilibrium, this object executes simple harmonic motion with an amplitude A and a period T. The maximum velocity or speed is achieved as the object moves through the equilibrium point, irrespective of the direction. The maximum displacement from the equilibrium, referred to as amplitude, does not correlate with the object's speed.

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Given an AC source with a known voltage amplitude and frequency, and knowing the current amplitude, we can calculate the capacitance of a capacitor connected to this source using the formula for capacitive reactance and Ohm's law. The calculated value for the capacitance of the capacitor in this scenario is approximately 0.17 microfarads (µF).

The subject of this question is the physics concept of capacitive reactance, which has to do with a capacitor's ability to oppose change in voltage. The question is asking for the capacitance of a capacitor in an arrangement where it is connected across an AC source with a known voltage amplitude and frequency. The current amplitude is also provided.

To calculate the capacitance (C) of the capacitor, we should start with the formula for capacitive reactance (Xc): Xc = 1/(2πfC), where Xc is the capacitive reactance, f is the frequency of the AC voltage source, and C is the capacitance. However, in this case, we need to use Ohm's law (I=V/R or here I=V/Xc), manipulate it for Xc and then substitute it into our reactance formula to solve for C. Doing the math will yield the required capacitance.

Let's do the calculations: First, calculate the reactance Xc from the given amplitude of the voltage (V), of the current (I), Xc=V/I. Hence, Xc= 59.0V/5.05A ≈ 11.68 Ω. Now we substitute Xc into the equation for capacitive reactance Xc =1/(2πfC), manipulate for C to get C = 1/(2πfXc), & replace f & Xc with the given values, we get, C = 1/(2π*77 Hz*11.68 Ω) ≈ 0.176 x 10^-6 F. Thus, the capacitance of the capacitor is approximately 0.17 microfarads (µF).

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The capacitance of the capacitor connected to an alternating current source with a voltage amplitude of 59.0 V, a frequency of 77.0 Hz and a current amplitude of 5.05 A, can be calculated using the formula for capacitive reactance and Ohm's law. The calculated capacitance is approximately 0.217 μF.

The formula for calculating the capacitance (C) of a capacitor in an Alternating Current (AC) circuit is given by the equation Xc = 1/(2πfC), where Xc refers to the capacitive reactance, f is the frequency of the AC source in hertz, and C is the capacitance in farads. In your question, it's stated that the current amplitude is 5.05 A and the voltage amplitude is 59.0 V. The capacitive reactance, Xc, can then be calculated using Ohm's law (Xc = Voltage / Current), which in this case would give us a value of around 11.68 Ohm. Substituting this into the given formula, and equally rearranging for C, gives us a capacitance value of approximately 0.217 x 10⁻⁶ F or 0.217 μF. Hence, the capacitance of the capacitor is approximately 0.217 μF.

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Answer:

The work done is 360 J.

Explanation:

Given that,

Mass = 50 kg

Distance =3 m

We need to calculate the work done

The work done is equal to the product of force and displacement.

Using formula of work done

[tex]W = F\cdot d[/tex]

[tex]W = Fd\cos\theta[/tex]

Where, F = force

D = distance

θ = Angle between force and displacement

Put the value into the formula

[tex]W=120\times3\cos0^{\circ}[/tex]

[tex]W=360\ J[/tex]

Hence, The work done is 360 J.

Answer:

The second child must exert a force of magnitude 23.3N to keep the door from moving.

Explanation:

We have to find the moment that the first child exerts and then match it to that exercised by the second child.

F1= 17.5N

d1= 0.6m

F2= ?

d2= 0.45m

M= F * d

M1= 17.5N * 0.6m

M1= 10.5 N.m

M1=M2

M2= F2 * 0.45m

10.5 N.m= F2 * 0.45m

10.5 N.m/0.45m = F2

F2=23.3 N

The force that the second child must exert to keep the door from moving is 23.33 N.

A balanced force occurs when an object subjected to different forces are at equilibrium.

F1r1 = F2r2

(17.5 x 0.6) = F2(0.45)

F2 = 23.33 N

Thus, the force that the second child must exert to keep the door from moving is 23.33 N.

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The stress in the bungee cord due to the Max's weight is [tex]\fbox{\begin \\ 299511\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\ 2.99\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\3.0\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex].

Further Explanation:

The stress developed in the bungee cord is the amount of tensile force developed inside the bungee cord due to the Max's weight.

Given:  

The diameter d of the bungee cord is [tex]2.5\text{ cm}[/tex].  

The mass m of Max is [tex]15\text{ kg}[/tex].

Concept:

The stress developed in the bungee cord is given by:  

[tex]\fbox{\begin \sigma =\dfrac{F}{A}\end{minispace}}[/tex]                                                                               ... (1)         

Here, [tex]\sigma[/tex] is the stress developed in the bungee cord, [tex]F[/tex]is the force due to Max's weight and [tex]A[/tex] is the area of cross-section of the bungee cord.  

The radius of the bungee cord will be the half of its diameter.  

[tex]r=\dfrac{d}{2}[/tex]

Substitute [tex]2.5\text{ cm}[/tex] for [tex]d[/tex].  

[tex]r= \dfrac{2.5\text{ cm}}{2}\\r=1.25 \text{ cm}[/tex] 

Convert the radius of the bungee cord in meter.  

[tex]r=\dfrac{1.25}{100} \text{ m}\\r=0.0125\text{ m}[/tex] 

The area of cross-section of the bungee cord is given by:  

[tex]A=\pi r^{2}[/tex]  

Substitute [tex]0.0125\text{ m}[/tex] for [tex]r[/tex].  

[tex]A=\pi (0.0125\text{ m})^{2}\\A=4.908\times10^{-4} m^{2}[/tex]  

The force on the bungee cord due to Max's mass is the gravitational force acting on Max's body. The gravitational force acting on Max's body is given by:  

[tex]F=mg[/tex]  

Here, m is the mass of Max's body and g is acceleration due to gravity.  

Consider the value of acceleration due to gravity on Earth be [tex]9.80\text{ m}/\text{s}^{2}[/tex].  

Substitute [tex]15\text{ kg}[/tex] for m and [tex]9.80\text{ m}/\text{s}^{2}[/tex] for g in above expression.  

[tex]F=(15\text{ kg})\times(9.80\text{ m}/\text{s}^{2})\\F=147\text{ N}[/tex]  

Substitute [tex]147\text{ N}[/tex] for F and [tex]4.980\times10^{-4}\text{ m}^{2}[/tex] for A in equation (1).  

[tex]\sigma=\dfrac{147\text{ N}}{4.908\times10^{-4}\text{ m}^{2}}\\\sigma=299511\text{ N}/\text{m}^{2}[/tex]

Thus, the stress developed in the bungee cord is [tex]\fbox{\begin \\ 299511\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\ 2.99\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex] or [tex]\fbox{\begin\\3.0\times10^{5}\text{ N}/\text{m}^{2}\end{minispace}}[/tex].

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Answer Details:

Grade: Senior school

Subject: Physics

Chapter: Stress and Strain

Keywords:

Max, Max's weight, force, stress, area, 299511 N/m2, 299511 N/m^2, 2.99x10^5 N/m2, gravitational, radius, bungee, cord, weight, developed, mass, 3x10^5 N/m^2.

The stress in the bungee cord due to Max’s weight is 294,000 N/m².

The given parameters;

The area of the bungee cord is calculated as follows;

[tex]A = \pi r^2\\\\A = \pi \times (0.0125)^2\\\\ A= 0.0005 \ m^2[/tex]

The weight of Max is calculated as follows;

[tex]F = mg\\\\F = 15 \times 9.8\\\\F = 147 \ N[/tex]

The stress in the bungee cord due to Max’s weight is calculated as;

[tex]\sigma = \frac{F}{A} \\\\\sigma = \frac{147}{0.0005} \\\\\sigma = 294,000 \ N/m^2[/tex]

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Answer:

(a) 1.35 x 10^6 Joule

(b) 321.45 Kcal

Explanation:

Energy consumed in 1 hour = 270 kJ

So, Energy consumed in 5 hour = 270 x 5 = 1350 kJ

(a) Energy in joules = 1350 x 1000 J = 1.35 x 10^6 J

(b) 4.2 Joule = 1 calorie

So, 1.35 x 10^6 Joule = 1.35 x 10^6 / 4.2 = 0.32 x 10^6 Calorie

                                                                =  0.32 x 1000 Kcal = 321.45 Kcal

The electrical energy consumed by the lightbulb is 1..35 x 10⁶J or 322.65 kcal.

The electrical energy consumed by the lightbulb is the product of power and time of energy consumption.

E = Pt

1 h ------------------- 270 kJ

5 h -------------------- ?

= 5 x 270 kJ

= 1,350 kJ

= 1.35 x 10⁶ J.

1 kJ ---------------- 0.239 kcal

1,350 kJ ------------ ?

= 1,350 x 0.239

= 322.65 kcal.

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