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3 years ago

The diagram shows a heat engine. In which area of the diagram is unusable thermal energy detected?

Physics

2 answers:

Marat540 [252]3 years ago

7 0

Nope, I disagree with the former answer. The answer is definitely Z. <u>W area</u> (boxed with red outline) is represented as the hot reservoir while <u>Z area</u> is the cold reservoir (boxed with blue outline). X area is the heat engine itself and Y area is the work produced from thermal energy from hot reservoir. Typically, all heat engines lose some heat to the environment (based from the second law of thermodynamics) that is symbolically illustrated by the lost energy in the cold reservoir. This lost thermal energy is basically the unusable thermal energy. The higher thermal energy lost, the less efficient your heat engine is.

Minchanka [31]3 years ago

4 0

The answer is Z is unusable

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The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.6 square inches. If a force of 5.6 lb

balu736 [363]

Answer:

16.8 lb is the force on the brake pad of one wheel.

Explanation:

Force applied on the piston = $F_1=5.6 lb$

Area of the piston = $A_1=0.6 inches^2$

Force applied on the brakes = $F_2$

Area of the brakes = $A_2=1.8 inches^2$

Applying Pascal's law: 'For an incompressible fluid pressure at one surface is equal to the pressure at other surface'.

$\frac{F_1}{A_2}=\frac{F_2}{A_2}$

$F_2=\frac{5.6 lb\times 1.8 inhes^2}{0.6 inches^2}=16.8 lb$

16.8 lb is the force on the brake pad of one wheel.

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3 years ago

True or False: For a longitudinal wave, the wavelength is the distance between compressions.

Schach [20]

Answer:

false....

Explanation:

brainliest

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3 years ago

A charged box ( m = 495 g, q = + 2.50 μ C ) is placed on a frictionless incline plane. Another charged box ( Q = + 75.0 μ C ) is

Rudik [331]

Answer:

$a=16.2m/s^{2}$

Explanation:

From the attached file diagram, the total force acting on the charged box is the downward weight and the repulsive force acting in opposite to the weight force . Hence we can write the total force as

$F=masin\alpha -\frac{kQq}{r^{2}} \\\alpha =35^{0}, m=0.495kg, r=0.61m, Q=2.5*10^{-6}, q=75.0*10^{-6}\\$

When fixed,F=o

Hence

$masin\alpha =\frac{kQq}{r^{2}}\\0.495kg*asin35=\frac{9*10^{9}*2.5*10^{-6}*75.0*10^{-6}}{0.61^{2}} \\0.28a=4.5351\\a=\frac{4.5351}{0.28}\\\\ a=16.2m/s^{2}$

The value of the acceleration is 16.2m/s^2

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2 years ago

Two particles P and Q each moves towards ther along Straight Line (M)(N), 51m long. P starts from M with velocity 5m/s and const

bagirrra123 [75]

The time required by the particles are as follows:

a. t = 1.5 seconds

b. t = 3 seconds

c. t = 0.4 seconds

<h3>What is the time required?</h3>

The time required for the particles to be at several distances apart is calculated using the equation of motion given below:

$S = ut + \frac{1}{2}at^{2}$

a) Time required to be 30 m apart:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51 - 30

S1 + S2 = 21

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 21

2t^2 + 11t - 21 = 0

Solving for time, t by factorization, t = 1.5 seconds

b) Time required to meet:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 51

2t^2 + 11t - 51 = 0

Solving for time, t by factorization, t = 3 seconds

c) Time required for velocity of P is ¾ of the velocity of Q:

Using the equation of motion: V = u + at

Vp = 3/4 Vq

4Vp= 3Vq

Substituting the values:

4(5 + t) = 3(6 + 3t)

5t = 2

t = 0.4 seconds

Learn more about distance, velocity and acceleration at: brainly.com/question/14344386

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