A hunter shoots a bullet horizontally at a rock cliff wall that is 170 m away. She hears the sound of the bullet hitting the rock 0.95 seconds later. Knowing that the speed of sound was 340 m/s, what was the speed of the bullet?

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

To calculate the ratio of rotational energy to translational kinetic energy for a pitched baseball, use the formulas for rotational kinetic energy and translational kinetic energy. The moment of inertia is found by using the radius of the ball, and the rotational and translational kinetic energies can be determined using the mass, radius, and velocity of the ball. Finally, calculate the ratio of the two energies to get the answer.

To calculate the ratio of the rotational energy to the translational kinetic energy, we need to find the rotational kinetic energy and the translational kinetic energy of the pitched baseball.

The rotational kinetic energy can be calculated using the formula:

Rotational Kinetic Energy = (1/2) * I * ω²

Where I is the moment of inertia and ω is the angular speed.

The translational kinetic energy can be calculated using the formula:

Translational Kinetic Energy = (1/2) * m * v²

Where m is the mass of the ball and v is the velocity.

We know that the radius of the ball is 3.87 cm and it moves at a velocity of 36.6 m/s. The angular speed is 112 rad/s. Since the ball is treated as a uniform sphere, its moment of inertia can be calculated using the formula:

Moment of Inertia = (2/5) * m * r²

Substitute the given values into the formulas and calculate the rotational and translational kinetic energies. Then, find the ratio of the rotational energy to the translational energy.

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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:

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:

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:

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:

Radius of cross section, r = 0.24 m

Explanation:

It is given that,

Number of turns, N = 180

Change in magnetic field, [tex]\dfrac{dB}{dt}=3\ T/s[/tex]

Current, I = 6 A

Resistance of the solenoid, R = 17 ohms

We need to find the radius of the solenoid (r). We know that emf is given by :

[tex]E=N\dfrac{d\phi}{dt}[/tex]

[tex]E=N\dfrac{d(BA)}{dt}[/tex]

Since, E = IR

[tex]IR=NA\dfrac{dB}{dt}[/tex]

[tex]A=\dfrac{IR}{N.\dfrac{dB}{dt}}[/tex]

[tex]A=\dfrac{6\ A\times 17\ \Omega}{180\times 3\ T/s}[/tex]

[tex]A=0.188\ m^2[/tex]

or

[tex]A=0.19\ m^2[/tex]

Area of circular cross section is, [tex]A=\pi r^2[/tex]

[tex]r=\sqrt{\dfrac{A}{\pi}}[/tex]

[tex]r=\sqrt{\dfrac{0.19}{\pi}}[/tex]

r = 0.24 m

So, the  radius of a tightly wound solenoid of circular cross-section is 0.24 meters. Hence, this is the required solution.

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:

Fc =  5.41 N

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

Newton's second law for the set of the three blocks

F =  13 N

m=3.0 kg+ 4.0 kg+5.0 kg = 12 kg

F = m*a

13 = 12*a

a = 13 / 12

a = 1.083 m/s² : acceleration of the set of the three blocks

Newton's second law for the  5.0 kg block

m= 5.0 kg

a = 1.083 m/s²

Fc: Contact force of the 4 kg block on the 5 kg block

Fc =  5.0 kg * 1.083 m/s²

Fc =  5.41 N

The 4.0 kg block exerts a force of 5.4 N on the 5.0 kg block when a 13 N force is applied to the entire system of blocks and accelerates the system at 1.08 m/s² on a frictionless surface.

We are dealing with a problem related to Newton's Third Law of Motion where a 13 N force is applied to a system of three blocks in contact with each other on a frictionless surface.

To find the force that the 4.0 kg block exerts on the 5.0 kg block, we need to calculate the acceleration of the system first and then use Newton's second law for just the last two blocks.

The total mass of the system is the sum of the masses of the three blocks: 3.0 kg + 4.0 kg + 5.0 kg = 12.0 kg. The total force provided by the 13 N force gives us an acceleration of the system:

a = F_{total} / m_{total} = 13 N / 12.0 kg = 1.08 m/s² (rounded to two decimal places)

Now, using Newton's second law, F = ma, for the 5.0 kg block, which is being accelerated by the force from the 4.0 kg block (F_{45}), we find:

F_{45} = m_{5.0kg} × a = 5.0 kg × 1.08 m/s²= 5.4 N

Therefore, the 4.0 kg block exerts a force of 5.4 N on the 5.0 kg block.

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Answer:

I = [tex]R^{2}[/tex](K+5)

Explanation:

Given :

J = k+5

Now selecting a thin ring in the wire of radius "r" and thickness dr.

Current through the thin ring is

dI = J X 2πrdr

dI = (K+5) x 2πrdr

Now integrating we get

I = [tex]\int_{0}^{R} = (K+5).2\pi rdr[/tex]

I = (K+5) 2π[tex]\int_{0}^{R} rdr[/tex]

I = (K+5) 2π [tex]\frac{R^{2}}{2}[/tex]

I = [tex]R^{2}[/tex](K+5)

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 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:

[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: 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:

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:

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.

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))

I believe the answer is A: True

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]

Answer:

5.6 mH

Explanation:

i1 = 3.20 A, i2 = 1.90 A, e = 14 mV = 0.014 V,

Let L be the coil's inductance.

[tex]e = -L\times \frac{\Delta i}{\Delta t}[/tex]

[tex]0.014 = -L\times \frac{1.90 - 3.20}{0.52}[/tex]

L = 0.0056 H

L = 5.6 mH

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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-- Water cannot be boled.

-- There is no such thing as a coffe aker.

-- There is no such thing as an immension-type heating element.

Be that as it may, and it very likely still is, as it were . . .

-- The latent heat of vaporization of water at 100°C is about 2250 kilo-joules per kg.

-- To evaporate half of the kg of water in the coffee maker requires 1125 kJ of heat energy.

-- To supply that amount of heat energy over a period of 10 minutes (600 seconds), it must be supplied at a rate of

(1,125,000 Joules / 600 seconds) = 1,875 joules/second.

-- That's 1,875 watts, or 1.875 kilowatts.

-- Choice-a is the choice when the solution is rounded.

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:

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:

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:

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]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:

(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

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