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Ivanshal [37]
3 years ago
5

An ideal heat engine runs with a high temperature reservoir at 750K and a low temperature reservoir at 350K. When the engine is

running, it extracts 400 joules of energy from the hot reservoir and does 250 joules of work each minute. How much energy is expelled to the low temperature reservoir each minute
Physics
1 answer:
Vanyuwa [196]3 years ago
4 0

Answer:

150 joules of energy is expelled to the low temperature reservoir each minute.

Explanation:

Lets assign the terms first.

T_H ,T_L,Q_H,Q_L,W where ...

T_H = 750 K that is, the high temperature of the reservoir.

T_L  = 350 K, low temperature of the reservoir.

Q_H = 400 J,energy extract from the reservoir.

W = 250 J, work done.

Q_L = energy expelled.

And

Formula to be used:

⇒ Q_H=W+Q_L

Re-arranging.

⇒ Q_L=Q_H-W

Now plugging the values of Q_H and W.

⇒ Q_L=400-250

⇒ Q_L=150 Joules

So energy expelled to the low temperature reservoir is of 150 joules.

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nadya68 [22]

Answer:

18.33 Ns

Explanation:

As the pitch back speed has the opposite direction as before, the change in velocity would be

\Delta v = v_2 - v_1 = 29 - (-10) = 39 m/s

So the change in momentum of the ball would be the product of its velocity change and its mass

\Delta p = \Delta v m = 39 * 0.47 = 18.33 kgm/s

This is equals to the impulse acted on the ball by the bat, which is 18.33 Ns

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3 years ago
Which is not a factor that affects the pressure of a gas in a closed container?
maksim [4K]

it is size of of particles because it does not matter about the size in a closed container

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RSB [31]

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Explanation:

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3 years ago
The earth's radius is 6.37×106m; it rotates once every 24 hours. What is the earth's angular speed? What is the speed of a point
Zarrin [17]

Answer:

a) w = 7.27 * 10^-5 rad/s

b) v1 = 463.1 m/s

c) v1 = 440.433 m/s

Explanation:

Given:-

- The radius of the earth,  R = 6.37 * 10 ^6 m

- The time period for 1 revolution T = 24 hrs

Find:

What is the earth's angular speed?

What is the speed of a point on the equator?

What is the speed of a point on the earth's surface located at 1/5 of the length of the arc between the equator and the pole, measured from equator?

Solution:

- The angular speed w of the earth can be related with the Time period T of the earth revolution by:

                                  w = 2π / T

                                  w = 2π / 24*3600

                                  w = 7.27 * 10^-5 rad/s

- The speed of the point on the equator v1 can be determined from the linear and rotational motion kinematic relation.

                                 v1 = R*w

                                 v1 = (6.37 * 10 ^6)*(7.27 * 10^-5)

                                 v1 = 463.1 m/s

- The angle θ subtended by a point on earth's surface 1/5 th between the equator and the pole wrt equator is.

                                 π/2  ........... s

                                 x     ............ 1/5 s

                                 x = π/2*5 = 18°    

- The radius of the earth R' at point where θ = 18° from the equator is:

                                R' = R*cos(18)

                                R' = (6.37 * 10 ^6)*cos(18)

                                R' = 6058230.0088 m

- The speed of the point where θ = 18° from the equator v2 can be determined from the linear and rotational motion kinematic relation.

                              v2 = R'*w

                              v2 = (6058230.0088)*(7.27 * 10^-5)

                              v2 = 440.433 m/s

5 0
3 years ago
Calculate the final velocity right after a 117 kg rugby player who is initially running at 7.45 m/s collides head‑on with a padd
Free_Kalibri [48]

Answer:

v_f = 0.87 m/s

Explanation:

We are given;

F_avg = -17700 N (negative because it's backward)

m = 117 kg

Δt = 5.50 × 10^(−2) s

v_i = 7.45 m/s

Now, formula for impulse is given by;

I = F•Δt = - 17700 x 5.50 × 10^(−2) = - 973.5 kg.m/s

From impulse momentum theory, we know that;

Change in momentum of particle is equal to impulse.

Thus,

Δp = I = m•v_f - m•v_i

Thus,

-973.5= 117(v_f - 7.45)

Thus,

-973.5/117 = (v_f - 7.45)

-8.3205 + 7.45 = v_f

v_f = - 0.87 m/s

We'll take absolute value as;

v_f = 0.87 m/s

5 0
4 years ago
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