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3 years ago

In each cycle, a heat engine performs 710 J of work and exhausts 1480 J of heat. What is the thermal effiency?

Physics

1 answer:

Tomtit [17]3 years ago

6 0

Answer:

32%

Explanation:

For a heat engine, efficiency is work out divided by heat in:

η = Wₒ / Qᵢ

Since energy is balanced, heat in is the sum of work out and heat out:

Qᵢ = Wₒ + Qₒ

Therefore:

η = Wₒ / (Wₒ + Qₒ)

Given Wₒ = 710 J and Qₒ = 1480 J:

η = 710 / (710 + 1480)

η = 0.32

The thermal efficiency is 32%.

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You place a 500 g block of an unknown substance in an insulated container filled 2 kg of water. The block has an initial tempera

Nina [5.8K]

Answer:

3349J/kgC

Explanation:

Questions like these are properly handled having this fact in mind;

Heat loss = Heat gained

Quantity of heat = mcΔ∅

m = mass of subatance

c = specific heat capacity

Δ∅ = change in temperature

m₁c₁(∅₂-∅₁) = m₂c₂(∅₁-∅₃)

m₁ = mass of block = 500g = 0.5kg

c₁ = specific heat capacity of unknown substance

∅₂ = block initial temperature = 50oC

∅₁ = equilibrium temperature of block and water after mix= 25oC

m₂= mass of water = 2kg

c₂ = specific heat capacity of water = 4186J/kg C

∅₃ = intial temperature of water = 20oC

0.5c₁(50-25) = 2 x 4186(25-20)

And we can find c₁ which is the unknown specific heat capacity

c₁ = $\frac{2*4186*5}{0.5*25}$ = 3348.8J/kg C≅ 3349J/kg C

4 0

2 years ago

A bicycle is heading west. It goes 500m in 5 minutes. What is its velocity?

icang [17]

Speed = 500m/5min = 100 m/min. Direction = west. Velocity = 100 m/min west.

4 0

3 years ago

Read 2 more answers

A horizontal force of 300.0 N is used to push a 145-kg mass 30.0 m horizontally in 3.00 s. Calculate the power developed.

Svetradugi [14.3K]

Answer:

3 * 10³J/s

Explanation:

Given :

Force applied, F = 300 N

Distance, d = 30 m

Time, t = 3 seconds

Power, P = Workdone / time

Recall :

Workdone = Force * distance

Workdone = 300 N * 30 m = 9000 Nm

Workdone = 9 * 10³ J

Power = (9 * 10³ J) / 3s

Power = 3 * 10³J/s

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3 years ago

When you look at your speedometer and it reads 60 mph this is your instantaneous speed

11Alexandr11 [23.1K]

true true tire true tire ture

please mark brainlist

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3 years ago

What is the angular displacement of the wheel between t = 5 s and t = 15 s?

Mkey [24]

The question seems incomplete. The complete text is:

a)What is the angular displacement of the wheel between t = 5 s and t = 15 s

b)What is the angular velocity of the wheel at 15 s

And it refers to the attached figure.

a) 25 rad

The graph shown represent the angular position of the wheel at different times.

Therefore, we can simply calculate the angular displacement between two times by calculating the difference between the angular position at t2 and the angular position at t1.

At $t_1 = 5 s$ , the angular position from the graph is $\theta_1 = 100 rad$

At $t_2 = 15 s$ , the angular position from the graph is $\theta_2 = 125 rad$

Therefore, the angular displacement is

$\Delta \theta= \theta_2 - \theta_1 = 125-100 = 25 rad$

2) -5.0 rad/s

For a angular displacement vs time graph, the angular velocity at any time is simply equal to the slope of the curve at that time.

Here we want to calculate the angular velocity at t = 15 s, so we have to calculate the slope at that time.

By noting that the slope is constant in the last part of the motion, we find that the slope between 10 s and 20 s is:

$\frac{\Delta \theta}{\Delta t}=\frac{100 rad - 150 rad}{20 s - 10 s}=-5.0 rad/s$

This slope is constant between 10 s and 20 s, so the angular velocity of the wheel at t = 15 s

$\omega = -5.0 rad/s$

5 0

3 years ago