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nataly862011 [7]
3 years ago
7

A real heat engine operates between temperatures TcTcT_c and ThThT_h. During a certain time, an amount QcQcQ_c of heat is releas

ed to the cold reservoir. During that time, what is the maximum amount of work WmaxWmaxW_max that the engine might have performed
Physics
1 answer:
Nookie1986 [14]3 years ago
8 0

The maximum amount of work performed is

W_{max}=\frac{T_H-T_C}{T_C}Q_C

Explanation:

The efficiency of a real heat engine is given by the equation:

\eta = 1-\frac{T_C}{T_H} (1)

where

T_C is the temperature of the cold reservoir

T_H is the temperature of the hot reservoir

However, the efficiency of a real heat engine can be also written as:

\eta = \frac{W_{max}}{Q_H}

where

W_{max} is the maximum work done

Q_H is the heat absorbed from the hot reservoir

Q_H can be written as

Q_H=W_{max}+Q_C

where

Q_Cis the heat released to the cold reservoir

So the previous equation can be also written as

\eta=\frac{W_{max}}{W_{max}+Q_C} (2)

By combining eq.(1) and (2) we get

1-\frac{T_C}{T_H}=\frac{W_{max}}{W_{max}+Q}

And re-arranging the equation and solving for W_{max}, we find

W_{max}=\frac{T_H-T_C}{T_C}Q_C

Learn more about work and heat:

brainly.com/question/4759369

brainly.com/question/3063912

brainly.com/question/3564634

#LearnwithBrainly

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Dimas [21]
I believe its newtons 3rd law for every action there is an equal but opposite reaction since the squid is moving foward by shooting the water its pushing the squid back as its reaction. Hope this helped !
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4 years ago
A lizard accelerates from 2 m/s to 10 m/s in 4 seconds. what is the lizard average acceleration
VLD [36.1K]

Acceleration = (change in speed) / (time for the change)

change in speed = (ending speed) - (starting speed)

change in speed = (10 m/s) - (2 m/s)  =  8 m/s

Acceleration = (8 m/s) / (4 sec)

Acceleration = (8/4) (m/s²)

<em>Acceleration = 2 m/s²</em>

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3 years ago
If the crate shown here is moving at a constant speed in a straight line and the force applied is 310 N, what is the magnitude o
larisa86 [58]

Answer:

f_k = 310N

the answer is A.

Explanation:

Using the laws of newton:

∑F = ma

where ∑F is the sumatory of forces acting in the system, m the mass and a the acelertion of the system.

Then, if the block is moving with constant velocity, its aceleration is equal to 0, so:

∑F = m(0)

∑F = 0

It means that:

F -f_k = 0

where F is the force applied and f_k is the friction force. Replacing the value of F, we get:

310N -f_k = 0

Finally, solving for f_k:

f_k = 310N

8 0
3 years ago
a car moving at 11 m/s crashes into an obstacle and stops in 0.26s. compute the Force that a seatbelt exerts on a 21-kg child to
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Answer:

890 N

Explanation:

Acceleration is change in velocity over change in time.

a = Δv / Δt

a = (11 m/s − 0 m/s) / 0.26 s

a = 42.3 m/s²

Force is mass times acceleration.

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4 0
4 years ago
At locations A and B, the electric potential has the values VA = 1.83 V and VB = 5.17 V, respectively. A proton released from re
densk [106]

Answer:

a. It starts at point B.

vp = 2.53*10⁴ m/s

a. it starts at point A.

ve= 1.08*10⁶ m/s

Explanation:

a)  As the proton is a positive charge, when released from rest, it will be accelerated due to the potential difference, from the higher potential to the lower one, so it is at the point B when released.

Once released, as the total energy must be conserved, the increase in kinetic energy must be equal (in magnitude) to the change in the electric potential energy, as follows:

ΔK + ΔUe = 0 ⇒ ΔK = -ΔUe =- (e*ΔV)

⇒ -( e* (VA-VB) ) = \frac{1}{2}*mp*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and mp= mass of proton = 1.67*10⁻²⁷ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{1.67e-27kg} } = 2.53e4 m/s

⇒ vp = 2.53*10⁴ m/s

b) If, instead of a proton, the charge realeased from rest, had been an electron, a few things would change:

First, as the electrons carry negative charges, they move from the lower potentials to the higher ones, which means that it would have started at point A.

Second, as its charge is (-e) the change in electric potential energy had been negative also:

ΔUe = -e*ΔV = -e* (VB-VA)

In order to find the speed of the electron when it is just passing point B, we can apply the conservation of energy principle as for the proton, as follows:

-( (-e)* (VB-VA) ) = \frac{1}{2}*me*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and me= mass of electron = 9.1*10⁻³¹ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{9.1e-31kg} } = 1.08e6 m/s

⇒ ve = 1.08*10⁶ m/s

4 0
4 years ago
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