A vertical straight wire carrying an upward 28-A current exerts an attractive force per unit length of 7.83 X 10 N/m on a second parallel wire 7.0 cm away. What current (magnitude and direction) flows in the second wire?

Answer:

[tex]u = 0.057 J/m^3[/tex]

Explanation:

Energy density near the surface of the sphere is given by the formula

[tex]u = \frac{1}{2}\epsilon_0 E^2[/tex]

also for sphere surface we know that

[tex]E = \frac{V}{R}[/tex]

R = radius of sphere

V = potential of the surface

now we have

[tex]u = \frac{1}{2}\epsilon_0 (\frac{V^2}{R^2})[/tex]

now from the above formula we have

[tex]u = \frac{1}{2}(8.85 \times 10^{-12})(\frac{8500^2}{0.075^2})[/tex]

[tex]u = 0.057 J/m^3[/tex]

Answer:

[tex]Q = 4.63 \times 10^6 J[/tex]

Explanation:

As we know that melting point of silver is

T = 961.8 degree C

Latent heat of fusion of silver is given as

L = 111 kJ/kg

specific heat capacity of silver is given as

[tex]s = 240 J/kg C[/tex]

now we will have

[tex]Q = ms\Delta T + mL[/tex]

[tex]\Delta T = 961.8 - 20 [/tex]

[tex]\Delta T = 941.8 degree C[/tex]

now from above equation

[tex]Q = (13.74)(240)(941.8) + (13.74)(111 \times 10^3)[/tex]

[tex]Q = 3.1 \times 10^6 + 1.53 \times 10^6[/tex]

[tex]Q = 4.63 \times 10^6 J[/tex]

Given:

Height of tank = 8 ft

and we need to pump fuel weighing 52 lb/ [tex]ft^{3}[/tex] to a height of 13 ft above the tank top

Solution:

Total height = 8+13 =21 ft

pumping dist = 21 - y

Area of cross-section = [tex]\pi r^{2}[/tex] =  [tex]\pi 4^{2}[/tex] =16[tex]\pi[/tex] [tex]ft^{2}[/tex]

Now,

Work done required = [tex]\int_{0}^{8} 52\times 16\pi (21 - y)dy[/tex]

                                  = [tex]832\pi \int_{0}^{8} (21 - y)dy[/tex]

                                  = 832[tex]\pi([/tex][tex][ 21y ]_{0}^{8} - [\frac{y^{2}}{2}]_{0}^{8}\\[/tex])

                                  = 113152[tex]\pi[/tex] = 355477 ft-lb

Therefore work required to pump the fuel is 355477 ft-lb

Answer:

Part a)

v = 16.52 m/s

Part b)

v = 7.47 m/s

Explanation:

Part a)

(a) when the large-mass object is the one moving initially

So here we can use momentum conservation as the net force on the system of two masses will be zero

so here we can say

[tex]m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v[/tex]

since this is a perfect inelastic collision so after collision both balls will move together with same speed

so here we can say

[tex]v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}[/tex]

[tex]v = \frac{(8.4\times 24 + 3.8\times 0)}{3.8 + 8.4}[/tex]

[tex]v = 16.52 m/s[/tex]

Part b)

(b) when the small-mass object is the one moving initially

here also we can use momentum conservation as the net force on the system of two masses will be zero

so here we can say

[tex]m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v[/tex]

Again this is a perfect inelastic collision so after collision both balls will move together with same speed

so here we can say

[tex]v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}[/tex]

[tex]v = \frac{(8.4\times 0 + 3.8\times 24)}{3.8 + 8.4}[/tex]

[tex]v = 7.47 m/s[/tex]

Answer: 3MeV electron

Explanation:

m_e={9.1\times 10^{-31}      m_α=4\times m_e    m_a={9.1\times 10^{-31}

m_p=1.67\times 10^{-27}

(a)  K.E. Energy of electron =[tex]\frac{1}{2}\times{m_e}\times{v_{e} ^{2}}[/tex]=3MeV

[tex]v_{e} ^{2}=\frac{2\times3\times1.6\times10^{-19}\times10^{6}  }{9.1\times 10^{-31} }[/tex]=1.05\times10^{18}

[tex]v_e=\sqrt{1.05\times10^{18} } = 1.025\times10^{9}\frac{m}{s}[/tex]

(b) K.E. Energy of alpha particle =[tex]\frac{1}{2}\times{m_\alpha}\times{v_{\alpha} ^{2}}[/tex]=10MeV

[tex]v_{\alpha} ^{2}= \frac{2\times10\times1.6\times10^{-19}\times10^{6}  }{4\times9.1\times 10^{-31} }[/tex]=0.88\times10^{18}

[tex]v_\alpha=\sqrt{0.88\times10^{18} } =.94\times10^{9}\frac{m}{s}[/tex]

(c) K.E. Energy of auger particle =[tex]\frac{1}{2}\times{m_a}\times{v_{a} ^{2}}[/tex]=0.1MeV

[tex]v_{a} ^{2}=\frac{2\times0.1\times1.6\times10^{-19}\times10^{6}  }{9.1\times 10^{-31} }[/tex]=0.035\times10^{18}

[tex]v_a=\sqrt{0.035\times10^{18} } =.19\times10^{9}\frac{m}{s}[/tex]

(d)  K.E. Energy of proton particle =[tex]\frac{1}{2}\times{m_p}\times{v_{p} ^{2}}[/tex]=400keV

[tex]v_{p} ^{2}=\frac{2\times400\times1.6\times10^{-19}\times10^{3}  }{1.67\times 10^{-27} }[/tex]=0.766\times10^{14}

[tex]v_p=\sqrt{0.766\times10^{14} } =0.875\times10^{7}[tex]\frac{m}{s}[/tex]

from (a),(b),(c),and (d) we can clearly say that the velocity of the electron is more so the penetration of the electron will be deepest.

Answer:

change in momentum of the stone is 1.635 kg.m/s

Explanation:

let m = 0.500kg ball and M  be the mass of the stone, v be the velocity of the ball and V be the velocity of the stone

the  initial kinetic energy of the ball is 1/2(0.500)(20^2) = 100 J

the kinetic energy of the ball after rebounding is 70/100(100) = 70 J

Kb = 1/2mv^2

 v =  \sqrt{2k/m} = \sqrt{2(70)/0.500} = 16.73 m/s  

from the conservation of linear momentum, we know that:

mvi + MVi = mvf + MVf

MVf - MVi = mvi - mvf

MVf - MVi = (0.500)(20) - (0.500)(16.73)

                 = 1.635 kg.m/s  

therefore, the change is momentum of the stone is 1.635 kg.m/s

Answer:

4.89 seconds

Explanation:

The time period of a pendulum is given by

[tex]T = 2\pi \sqrt{\frac{l}{g}}[/tex]

For a second pendulum on earth, T = 2 second

[tex]2 = 2\pi \sqrt{\frac{l}{g}}[/tex]     ...... (1)

Now the time period is T when the pendulum is taken to moon and gravity at moon is 1/6 of gravity of earth

[tex]T = 2\pi \sqrt{\frac{l}{\frac{g}{6}}}[/tex]    ...... (2)

Divide equation (2) by equation (1)

[tex]\frac{T}{2} = \sqrt{6}[/tex]

T = 4.89 seconds

Answer:

Explanation:

There are two types of collision.

(a) Elastic collision: When there is no loss of energy during the collision, then the collision is said to be elastic collision.

In case of elastic collision, the momentum is conserved, the kinetic energy is conserved and all the forces are conservative in nature.

The momentum of the system before collision = the momentum of system after collision

The kinetic energy of the system before collision = the kinetic energy after the collision

(b) Inelastic collision: When there is some loss of energy during the collision, then the collision is said to be inelastic collision.

In case of inelastic collision, the momentum is conserved, the kinetic energy is not conserved, the total mechanical energy is conserved and all the forces or some of the forces are non conservative in nature.

The momentum of the system before collision = the momentum of system after collision

The total mechanical energy of the system before collision = total mechanical of the system after the collision

The magnetic force on the wire has an x-component of 0 N, a y-component of 0.004056224 N in the negative y-direction, and a z-component of 0.000143712 N in the positive z-direction.

The question involves calculating the magnetic force on a current-carrying wire in a magnetic field using the right-hand rule and the formula F = I × (L × B), where F is the force, I is the current, L is the length of the wire, and B is the magnetic field. We know the current I is 0.720 A, the length L is 73.6 cm (which we will convert to meters), and the magnetic field components are By = 0.000270 T and Bz = 0.00770 T.

First, let's convert the length of the wire from centimeters to meters: L = 73.6 cm = 0.736 m.

The force on the wire in the x, y, and z directions (Fx, Fy, Fz) can be calculated using the cross product of the current direction (along the x-axis) and the magnetic field components. Since there is no x component for the magnetic field (Bx = 0), the force in the x-direction (Fx) will be zero.

Using the right-hand rule, the force in the y-direction (Fy) will be:

Fy = I × (L × Bz) = 0.720 A × (0.736 m × 0.00770 T) = 0.004056224 N, pointing in the negative y-direction (since the current is in the positive x-direction and Bz is positive).

Similarly, the force in the z-direction (Fz) is calculated as:

Fz = I × (L × By) = 0.720 A × (0.736 m × 0.000270 T) = 0.000143712 N, pointing in the positive z-direction.

Answer:

v = 10.84 m/s

Explanation:

using the equation of motion:

v^2 = (v0)^2 + 2×a(r - r0)

due to the hammer starting from rest, vo = 0 m/s and a = g , g is the gravitational acceleration.

v^2 = 2×g(r - r0)

v = \sqrt{2×(-9.8)×(4 - 10)}

  = 10.84 m/s

therefore, the velocity at r = 4 meters is 10.84 m/s

Answer:

a) [tex]t=3.199 seconds[/tex]

b) [tex]h = 11.97 m[/tex]

Explanation:

Since this problem belongs to the concept of projectile motion

a) we know,

[tex]Vcos\theta=\frac{R}{t}[/tex]

Where,

V = initial speed

Θ = angle with the horizontal

R = horizontal range

t = Time taken to cover the range 'R'

Given:

V = 27m/s

R = 60m

Θ = 46°

thus,

the equation becomes

[tex]27\times cos46^o=\frac{60}{t}[/tex]

or

[tex]t=\frac{60}{27\times cos46^o}[/tex]

[tex]t=3.199 seconds[/tex]

b)The formula for height is given as:

[tex]h = Vsin\theta \times t-\frac{1}{2}\times gt^2\\[/tex]

where,

g = acceleration due to gravity = 9.8m/s²

substituting the values in the above equation we get

[tex]h = 27\times sin46^o\times 3.199-\frac{1}{2}\times 9.8\times 3.199^2\\[/tex]

or

[tex]h = 62.124-50.14[/tex]

or

[tex]h = 11.97 m[/tex]

It takes 2.22 seconds for the T-shirt to reach the fan. The fan is located at a height of 24.57 meters from the ground.

To find the time it takes for the T-shirt to reach the fan, we need to solve for the time in the horizontal motion. The horizontal distance from the gun to the fan is given as 60 m. Since the horizontal motion is constant velocity, we can use the equation d = vt and solve for t. Plugging in the values, we get t = 60 m / 27 m/s = 2.22 s.

To find the height of the fan from the ground, we need to solve for the vertical motion. The equation for vertical motion is y = yt + (1/2)gt^2, where y is the vertical displacement, yt is the initial vertical velocity, and g is the acceleration due to gravity. In this case, the vertical displacement is unknown, the initial vertical velocity is 0 m/s, and the acceleration due to gravity is 9.8 m/s^2. Plugging in the values and solving for y, we get y = (1/2)(9.8 m/s^2)(2.22 s)^2 = 24.57 m.

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Answer:

0.42 m/s²

Explanation:

r = radius of the flywheel = 0.300 m

w₀ = initial angular speed = 0 rad/s

w = final angular speed = ?

θ = angular displacement = 60 deg = 1.05 rad

α = angular acceleration = 0.6 rad/s²

Using the equation

w² = w₀² + 2 α θ

w² = 0² + 2 (0.6) (1.05)

w = 1.12 rad/s

Tangential acceleration is given as

[tex]a_{t}[/tex] = r α = (0.300) (0.6) = 0.18 m/s²

Radial acceleration is given as

[tex]a_{r}[/tex] = r w² = (0.300) (1.12)² = 0.38 m/s²

Magnitude of resultant acceleration is given as

[tex]a = \sqrt{a_{t}^{2} + a_{r}^{2}}[/tex]

[tex]a = \sqrt{0.18^{2} + 0.38^{2}}[/tex]

[tex]a[/tex] = 0.42 m/s²

The magnitude of the resultant acceleration of the point on the flywheel's rim after it has turned through 60.0° is approximately 0.424 m/s².

To compute the magnitude of the resultant acceleration of a point on the flywheel's rim, we need to use the equation:

θ = ω0t + 0.5αt2

Where θ is the angle turned, ω0 is the initial angular velocity, and α is the angular acceleration.

Plugging in the given values, the equation becomes:

60.0° = 0 + 0.5 * 0.600 rad/s² * t2

Simplifying, we get:

t2 = (60.0° * 2) / 0.600 rad/s²

t2 = 0.200 rad/s²

Therefore, the magnitude of the resultant acceleration is:

a = ωt = 0.600 rad/s² * sqrt(0.200 rad/s²) ≈ 0.424 m/s²

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Answer:

- 0.5 m/s²

Explanation:

m = mass of the clarinet case = 3.230 kg

W = weight of the clarinet case in downward direction

a = vertical acceleration of the case

Weight of the clarinet case is given as

W = mg

W = 3.230 x 9.8

W = 31.654 N

F = Upward force applied = 30.10 N

Force equation for the motion of the case is given as

F - W = ma

30.10 - 31.654 = 3.230 a

a = - 0.5 m/s²

Final answer:

By applying Newton's second law and accounting for the forces acting on the clarinet case, it is found to experience a downward vertical acceleration of 0.489 m/s², due to the net force acting on it being in the downward direction.

Explanation:

To find the vertical acceleration of the clarinet case, first identify the forces acting on it. The force of gravity (weight) pulls it downward, which can be calculated by multiplying the mass of the case by the acceleration due to gravity (9.81 m/s²). The equation for weight is W = mg, where m is mass and g is gravity.

Substituting the given values, W = 3.230 kg * 9.81 m/s² = 31.68 N. The net force [tex]F_{net}[/tex] acting on the case is the upward force by the clarinetist subtracting the weight of the case, [tex]F_{net}[/tex] = 30.10 N - 31.68 N = -1.58 N. Applying Newton's second law, F = ma, and solving for acceleration (a), a = [tex]F_{Net/m}[/tex], we find the case's vertical acceleration as a = -1.58 N / 3.230 kg = -0.489 m/s².

Therefore, the case experiences a downward acceleration of 0.489 m/s², indicating that it is slowing in its ascent or accelerating downward.

Answer:

97.96 [tex]\frac{rad}{s}[/tex]

Explanation:

The initial angular velocity is [tex]w_{0}[/tex] = 126 rad / s.

The constant angular deceleration is 5.00 rad / s2.

The constant angular deceleration is, by definition: dw / dt.

[tex]\frac{dw}{dt}=-5 \frac{rad}{s^{2} }[/tex]

Separating variables

[tex]dw=-5 dt[/tex]

Integration (limits for w: 0 to W0; limits for t: 0 to t)

[tex]w= w_{0}-5t[/tex]

W is by definition [tex]\frac{d\alpha }{dt}[/tex], where [tex]\alpha[/tex] is the angle.

[tex]\frac{d\alpha}{dt}=w_{0} -5t[/tex]

Separating variables

[tex]d\alpha=(w_{0} -5t )dt[/tex]

Integration (limits for [tex]\alpha[/tex]: 0 to 628; limits for t: 0 to t)

[tex]\alpha =w_{0}t-(\frac{5}{2})t^{2}[/tex]

[tex]128=126t-(\frac{5}{2})t^{2}[/tex]

Put in on the typical form of a quadratic equation:

[tex]\frac{5}{2}t^{2}-126t+628=0[/tex]

Solve by using the quadratic equation formula and discard the higher result because it lacks physical sense.

t=5.608 s

Evaluate at this time the angular velocity:

[tex]w(t=5.608)=126-5*5.608[/tex]

[tex]w(5.608)=97.96 \frac{rad}{s}[/tex]

To find the final angular speed of a rotating wheel given the initial speed, angular deceleration and the angle turned, we use an equation for angular motion. We substitute the given values and solve the equation to find the final angular speed.

To solve this question, we will be using the equation for angular motion,

ω² = ω0² + 2αθ

, where ω is the final angular speed we want to find, ω0 = 126 rad/s is the initial angular speed, α = -5.00 rad/s² is the angular deceleration, and θ = 628 rad is the angle turned through. This equation is analogous to the equation for linear motion, v² = u² + 2as, where v is final velocity, u initial velocity, a acceleration and s distance. We substitute the given values into the equation to find the final angular speed. Keep in mind, because we are dealing with deceleration, our α value is a negative.

Solve the equation: ω² - ω0² = 2αθ, which gives: ω² = (ω0² + 2αθ), ω = sqrt((ω0² + 2αθ)), ω = sqrt((126rad/s)² + 2*(-5.00 rad/s²)*628 rad). So, when you calculate the square root of the total, you will find the answer for the final angular speed.

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Answer:

7.55 m/s, 6.56 m/s

Explanation:

v = 10 m/s, theta = 41 degree

Horizontal component of velocity = v x Cos theta = 10 x Cos 41 = 7.55 m/s

Vertical component of velocity = v x Sin theta = 10 x Sin 41 = 6.56 m/s

Final answer:

The ball's horizontal velocity is approximately 7.57 m/s, and its vertical velocity is approximately 6.46 m/s immediately after the serve.

Explanation:

The student's question about the initial horizontal and vertical velocities of a volleyball serve involves breaking down the initial velocity into its component parts using trigonometric functions. Given an initial velocity (v) of 10 m/s and an angle (θ) of 41 degrees from the horizontal, the horizontal component (vx) is calculated using cosine, and the vertical component (vy) is calculated using sine.

Using the formula:

vx = v ∙ cos(θ)

vy = v ∙ sin(θ)

For this problem:

vx = 10 m/s ∙ cos(41°)

vy = 10 m/s ∙ sin(41°)

Plugging in the values yields:

vx ≈ 7.57 m/s

vy ≈ 6.46 m/s

The ball's horizontal velocity is approximately 7.57 m/s, and its vertical velocity is approximately 6.46 m/s immediately after the serve.

Answer:

Part a)

[tex]m_2 = 4.9 \times 10^7 kg[/tex]

Part b)

[tex]q_1 = q_2 = 5312.6 C[/tex]

Explanation:

Part a)

As we know that both charge particles will exert equal and opposite force on each other

so here the force on both the charges will be equal in magnitude

so we will have

[tex]F = m_1a_1 = m_2a_2[/tex]

here we have

[tex]6.3 \times 10^7(7) = m_2(9)[/tex]

now we have

[tex]m_2 = 4.9 \times 10^7 kg[/tex]

Part b)

Now for the force between two charges we can say

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

now we have

[tex](6.3 \times 10^7)(7) = \frac{(9\times 10^9)q^2}{(24\times 10^3)^2}[/tex]

now we have

[tex]q_1 = q_2 = 5312.6 C[/tex]

Answer:

6.93 volts

Explanation:

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

r = radius of the orbit = 2.08 x 10⁻¹⁰ m

V = Electric potential due to proton on electron

Electric potential due to proton on electron is given as

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

Inserting the values

[tex]V = \frac{(9\times 10^{9})(1.6\times 10^{-19})}{2.08\times 10^{-10}}[/tex]

V = 6.93 volts

Final answer:

The electric potential due to the proton on an electron in an orbit with a radius of 2.08 x 10^-10 m in a hydrogen atom model is approximately -13.6 volts.

Explanation:

In the context of the model of a hydrogen atom, the electric potential due to the proton on an electron in an orbit with radius 2.08 x 10-10 m can be calculated using the formula for the electric potential V due to a point charge, which is V = k * Q / r. Here, k is the Coulomb's constant (approximately 8.99 x 109 N m2/C2), Q is the charge of the proton (1.602 x 10-19 C), and r is the distance from the proton to the electron, which is the radius of the orbit.

To find the electric potential, simply plug these values into the formula:

V = (8.99 x 109 N m2/C2) * (1.602 x 10-19 C) / (2.08 x 10-10 m)

After calculating, the electric potential is found to be approximately -13.6 volts, with the negative sign indicating that the potential energy associated with the electron is negative, which is common for bound states such as electrons in an atom.

Answer:

The kinetic energy of the wagon is 967.0 J

Explanation:

Given that,

Force = 120 N

Mass = 55 kg

Height = 8 m

We need to calculate the kinetic energy of the wagon

Using newtons law

[tex]F = ma[/tex]

[tex]\dfrac{120}{55}=a[/tex]

[tex]a =2.2\ m/s^2[/tex]

Using equation of motion

[tex]v^2 =u^2+2as[/tex]

Where,

v = final velocity

u = initial velocity

s = height

Put the value in the equation

[tex]v^2=0+2\times2.2\times8[/tex]

[tex]v=5.93\ m/s[/tex]

Now, The kinetic energy is

[tex]K.E=\dfrac{1}{2}mv^2[/tex]

[tex]K.E=\dfrac{1}{2}\times55\times(5.93)^2[/tex]

[tex]K.E=967.0\ J[/tex]

Hence, The kinetic energy of the wagon is 967.0 J

Answer:

5.44×10⁶ m

Explanation:

For a satellite with period t and orbital radius r, the velocity is:

v = 2πr/t

So the centripetal acceleration is:

a = v² / r

a = (2πr/t)² / r

a = (2π/t)² r

This is equal to the acceleration due to gravity at that elevation:

g = MG / r²

(2π/t)² r = MG / r²

M = (2π/t)² r³ / G

At the surface of the planet, the acceleration due to gravity is:

g = MG / R²

Substituting our expression for the mass of the planet M:

g = [(2π/t)² r³ / G] G / R²

g = (2π/t)² r³ / R²

R² = (2π/t)² r³ / g

R = (2π/t) √(r³ / g)

Given that t = 1.30 h = 4680 s, r = 7.90×10⁶ m, and g = 30.0 m/s²:

R = (2π / 4680 s) √((7.90×10⁶ m)³ / 30.0 m/s²)

R = 5.44×10⁶ m

Notice we didn't need to know the mass of the satellite.

I believe the answer is A: True

Answer:

output work is not possible to have more than 294 kJ value so this is not reasonable claim

Explanation:

As we know that the efficiency of heat engine is given as

[tex]\eta = 1 - \frac{T_2}{T_1}[/tex]

now we will have

[tex]T_2 = 290 K[/tex]

[tex]T_1 = 500 K [tex]

[tex]\eta = 1 - \frac{290}{500}[/tex]

[tex]\eta = 0.42[/tex]

now we know that efficiency is defined as

[tex]\eta = \frac{Work}{Heat}[/tex]

[tex]0.42 = \frac{W}{700}[/tex]

[tex]W = 294 kJ[/tex]

So output work is not possible to have more than 294 kJ value

To determine if the inventor's claim of developing a heat engine is reasonable, we can calculate the maximum theoretical efficiency of the engine and compare it to the claimed net work output and heat input.

A heat engine operates between a hot reservoir and a cold reservoir, absorbing heat from the hot reservoir and converting some of it into work, while rejecting the remaining heat to the cold reservoir. The efficiency of a heat engine is determined by the temperature of the reservoirs. In this case, the inventor claims that the heat engine receives 700 kJ of heat from a source at 500 K and produces 300 kJ of net work while rejecting waste heat to a sink at 290 K. To determine if this claim is reasonable, we can calculate the theoretical maximum efficiency of the heat engine using the Carnot efficiency formula.

The Carnot efficiency formula is given by:

Efficiency = 1 - (Tc / Th)

where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir, both measured in Kelvin. In this case, Tc = 290 K and Th = 500 K.

Plugging in these values into the formula, we have:

Efficiency = 1 - (290 K / 500 K) = 1 - 0.58 = 0.42 or 42%

Therefore, the maximum theoretical efficiency for a heat engine operating between these temperatures is 42%. Since the claimed net work output of the heat engine is 300 kJ, we can calculate the maximum heat input by dividing the net work output by the efficiency:

Maximum heat input = Net work output / Efficiency = 300 kJ / 0.42 = 714.3 kJ

Since the claimed heat input is 700 kJ, which is slightly less than the maximum calculated heat input, it is reasonable to say that the inventor's claim is possible.

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Answer:

655128 ohm

Explanation:

C = Capacitance of the capacitor = 7.8 x 10⁻⁶ F  

V₀ = Voltage of the battery = 9 Volts  

V = Potential difference across the battery after time "t" = 4.20 Volts  

t = time interval = 3.21 sec  

T = Time constant

R = resistance  

Potential difference across the battery after time "t" is given as  

[tex]V = V_{o} (1-e^{\frac{-t}{T}})[/tex]

[tex]4.20 = 9 (1-e^{\frac{-3.21}{T}})[/tex]

T = 5.11 sec  

Time constant is given as  

T = RC  

5.11 = (7.8 x 10⁻⁶) R  

R = 655128 ohm

Answer:

750.25 cm

Explanation:

θ = 13.78°, h = 184 cm

Let the distance between rock and camera is d.

Tan θ = h / d

tan 13.78 = 184 / d

d = 750.25 cm

By utilizing trigonometry and the tangent of an angle in a right triangle, it is determined that the camera (at a height of 184.0 cm from the Mars surface with an angle of depression of 13.78° to a rock) is approximately 761.4 cm away from the rock.

This question involves trigonometry, specifically the tangent of an angle in a right triangle. The robot captures an image of a rock with an angle of depression of 13.78°. The camera is 184.0 cm above the Mars surface. We can create a right triangle where the angle at the camera is 13.78°, the opposite side is 184.0 cm (the height of the camera), and the adjacent side is the distance between the camera and the rock, which we will call x.

By definition, tan(angle) = opposite/adjacent. Here, the angle is 13.78°, the opposite side is 184.0 cm and the adjacent side is x (unknown). To find x, we can use the following formula: x = opposite/tan(angle).

Therefore, x = 184.0 cm / tan(13.78°). Using a calculator, x is approximately 761.4 cm (to the nearest tenth). So the camera is approximately 761.4 cm away from the rock.

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Answer:

[tex]v' = 2.83 m/s[/tex]

Explanation:

Velocity of wave in stretched string is given by the formula

[tex]v = \sqrt{\frac{T}{\mu}}[/tex]

here we know that

T = 4 N

also we know that linear mass density is given as

[tex]\mu = 1 kg/m[/tex]

so we have

[tex]v = \sqrt{\frac{4}{1}} = 2 m/s[/tex]

now the tension in the string is double

so the velocity is given as

[tex]v' = \sqrt{\frac{8}{1}} = 2\sqrt2 m/s[/tex]

[tex]v' = 2.83 m/s[/tex]

Answer:

0.51

Explanation:

m = mass of the book = 3.5 kg

F = force applied by the broom on the book = 21 N

a = acceleration of the book

v₀ = initial speed of the book = 0 m/s

v = final speed of the book = 1.2 m/s

d = distance traveled = 0.74 m

Using the equation

v² = v₀² + 2 a d

1.2² = 0² + 2 a (0.74)

a = 0.973 m/s²

f = kinetic frictional force

Force equation for the motion of the book is given as

F - f = ma

21 - f = (3.5) (0.973)

f = 17.6 N

μ = Coefficient of kinetic friction

Kinetic frictional force is given as

f = μ mg

17.6 = μ (3.5 x 9.8)

μ = 0.51

Answer:

[tex]F_c = 1.4 \times 10^6 N[/tex]

Explanation:

As we know

[tex]f = 24 rpm[/tex]

so we will have

[tex]f = 24 \frac{1}{60} hz = 0.4hz[/tex]

now angular frequency is given as

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

[tex]\omega = 0.8\pi[/tex]

Now the inwards is given as centripetal force

[tex]F_c = m\omega^2 r[/tex]

[tex]F_c = (12000)(0.8\pi)^2(\frac{37}{2})[/tex]

[tex]F_c = 1.4 \times 10^6 N[/tex]

Final answer:

The inward force necessary to provide each blade's centripetal acceleration is approximately 4,755,076 Newtons.

Explanation:

To calculate the inward force necessary to provide each blade's centripetal acceleration, we can use the formula for centripetal force:

[tex]Fc = m \times \omega^2 \times r[/tex]

where Fc is the centripetal force, m is the mass of the blade, ω is the angular velocity, and r is the radius of the blade.

In this case, the mass of the blade is 12,000 kg, the angular velocity is 24 rpm (which can be converted to radians per second by multiplying by 2π/60), and the radius of the blade is 37 m.

Plugging these values into the formula, we get:

Fc = 12000 kg * (24 * 2π/60)^2 * 37 m

Fc ≈ 4,755,076 N

So, the inward force necessary to provide each blade's centripetal acceleration is approximately 4,755,076 Newtons.

Answer:

Acceleration= 1,59 (meters/(second^2))

Direction= NE; 65,22°  above the east direction.

Explanation:

Resulting force= ( ((180N)^2) + ((390N)^2) ) ^ (1/2) = 429,53 N

Angle obove the east direction= ((cos) ^ (-1)) (180N / 429,53 N) = 65,22°

Acceleration= Resulting force / mass = (429,53 N) / (270 kg) =

= (429,53 kg × (meters/(second^2))) / (270kg) = 1,59 (meters/(second^2))

The speed of the electron when it reaches point B, which is 0.395 m east of point A, is approximately 3.86×10⁵ m/s, directed towards the east.

The initial speed of the electron is provided as 4.52×10⁵ m/s. Given the electric field, we can calculate the force on the electron as F = qE, where q is the charge of the electron (-1.60 × 10⁻¹⁹ C) and E is the electric field (1.55 N/C). Hence, the force acting on the electron is F = -1.60 × 10⁻¹⁹ C * 1.55 N/C = -2.48 × 10⁻¹⁹ N.

Using F = ma, we can calculate the acceleration of the electron. Knowing the mass of the electron is 9.11 × 10⁻³¹ kg, the acceleration is a = F/m = -2.48 × 10⁻¹⁹ N / 9.11 × 10⁻³¹ kg = -2.72 × 10¹¹ m/s².

Given the distance of the movement is 0.395 m, we can use the equation v² = u² + 2as to solve for the final velocity 'v', where 'u' is the initial velocity, 'a' is the acceleration and 's' is the distance. Substituting the known values, we get v = sqrt((4.52×10⁵ m/s)² - 2 * 2.72 × 10¹¹ m/s² * 0.395 m) ≈ 3.86 x 10⁵ m/s (approximately).

So, the speed of the electron when it reaches point B is approximately 3.86 x 10⁵ m/s, directed towards the east.

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The rate of cooling provided by a reversible refrigerator, working with a 10-kW compressor and thermal energy reservoirs at 250 K and 310 K, is calculated using the refrigerator's Carnot coefficient and the power of the compressor. With these specific conditions, the cooling rate is calculated to be 50 kW.

The question pertains to the functioning of a reversible refrigerator, specifically the cooling rate provided by the appliance which uses a 10-kW compressor and operates with thermal energy reservoirs at 250 K and 310 K. To determine the cooling rate, we must first understand some basics about the refrigerator's operation.

A reversible refrigerator absorbs heat Qc from a cold reservoir and discards it into a hot reservoir, while work W is done on it. This work is represented by the power exerted by the compressor. The refrigerator functions by moving a coolant through coils, which absorbs heat from the contents of the refrigerator and releases it outside.

The coefficient of performance or Carnot coefficient (KR) of the refrigerator can be computed using the formula KR = Tc / (Th - Tc) where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir, both measured in Kelvin. In this situation, KR = 250 / (310 - 250) = 5.

Since the work done W is the compressor power P, we can use the formula Qc = KR * P to find the cooling rate. Substituting the known values, Qc = 5 * 10 kW = 50 kW. Therefore, the rate of cooling provided by this refrigerator is 50 kW.

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Answer:

(a) 5142.86 m

(b) 317.5 m/s

(c) 49.3 degree C

Explanation:

m = 100 kg, Q = 1200 kcal = 1200 x 1000 x 4.2 = 504 x 10^4 J

(a) Let the altitude be h

Q = m x g x h

504 x 10^4 = 100 x 9.8 x h

h = 5142.86 m

(b) Let v be the speed

Q = 1/2 m v^2

504 x 10^4 =  1/2 x 100 x v^2

v = 317.5 m/s

(c) The temperature of normal human body, T1 = 37 degree C

Let the final temperature is T2.

Q = m x c x (T2 - T1)

504 x 10^4 = 100 x 4.1 x 1000 x (T2 - 37)

T2 = 49.3 degree C