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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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6 0
3 years ago
The position of a particle is given by ~r(t) = (3.0 t2 ˆi + 5.0 ˆj j 6.0 t kˆ) m
Julli [10]

Answer:

v=(6ti+6k)\ m/s

Explanation:

Given that,

The position of a particle is given by :

r(t) = (3.0 t^2 i + 5.0j+ 6.0 tk) m

Let us assume we need to find its velocity.

We know that,

v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s

So, the velocity of the particle is (6ti+6k)\ m/s.

5 0
3 years ago
Suppose you have two identical sheets of paper and you crumple one of the sheets of paper into a ball. If you drop the crumpled
sammy [17]

Answer:

Because it can easily resist air resistance.

Explanation:

Since air resistance is not negligible, the crumpled paper will reach the ground first because it can easily resist air resistance surrounding it compare to the un-crumpled one that will be influenced by the air thereby causing the un-crumpled paper to spend more time in the air

3 0
3 years ago
You have a pulley 10.4 cm in diameter and with a mass of 2.3 kg. You get to wondering whether the pulley is uniform. That is, is
madreJ [45]

Answer:

Explanation:

Given

Diameter of Pulley=10.4 cm

mass of Pulley(m)=2.3 kg

mass of book(m_0)=1.7 kg

height(h)=1 m

time taken=0.64 s

h=ut+frac{at^2}{2}

1=0+\frac{a(0.64)^2}{2}

a=4.88 m/s^2and [tex]a=\alpha r

where \alphais angular acceleration of pulley

4.88=\alpha \times 5.2\times 10^{-2}

\alpha =93.84 rad/s^2

And Tension in Rope

T=m(g-a)

T=1.7\times (9.8-4.88)

T=8.364 N

and Tension will provide Torque

T\times r=I\cdot \alpha

8.364\times 5.2\times 10^{-2}=I\times 93.84

I=0.463\times 10^{-2} kg-m^2

I_{original}=\frac{mr^2}{2}=0.31\times 10^{-2}kg-m^2

Thus mass is uniformly distributed or some more towards periphery of Pulley

4 0
3 years ago
Which of the following best describes wind?
Afina-wow [57]

Answer:

The answer is D the rising of warm air pushing down cool air.

5 0
3 years ago
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