Answer:
37.5%
Explanation:
We define first our values, that are,
![Q_H=80MW\\Q_c=50MW](https://tex.z-dn.net/?f=Q_H%3D80MW%5C%5CQ_c%3D50MW)
We can now calculate the net power output, that is
![W=Q_H-Q_c = (80-50) = 30MW](https://tex.z-dn.net/?f=W%3DQ_H-Q_c%20%3D%20%2880-50%29%20%3D%2030MW)
Thermal efficiency is given by
![\eta = \frac{W}{Q_H}\\\eta = \frac{30}{80}\\\eta = 0.375\\\eta = 37.5%](https://tex.z-dn.net/?f=%5Ceta%20%3D%20%5Cfrac%7BW%7D%7BQ_H%7D%5C%5C%5Ceta%20%3D%20%5Cfrac%7B30%7D%7B80%7D%5C%5C%5Ceta%20%3D%200.375%5C%5C%5Ceta%20%3D%2037.5%25)
<span>A transverse wave is a moving wave that consists of oscillations occurring perpendicular (or right angled) to the direction of energy transfer. If a transverse wave is moving in the positive x-direction, its oscillations are in up and down directions that lie in the y–z plane. Light is an example of a transverse wave.</span>
(a) The total momentum of the system before the train cars collide is 1,600 kgm/s.
(b) The total momentum of the system be after the train cars collide is 1,600 kgm/s.
<h3>What is the total momentum of the car system before the collision?</h3>
The total momentum of the car system before the collision is determined by applying the formula for linear momentum.
Pi = m₁u₁ + m₂u₂
where;
- m₁ is the mass of the car on the right
- m₂ is the mass of the car on the left
- u₁ is the initial velocity of the right
- u₂ is the initial velocity of the car on the left
Let the rightward direction = positive
Let the leftward direction = negative
Pi = (600 kg x 4 m/s) + (400 kg) x (-2 m/s)
Pi = 2,400 kgm/s - 800 kgm/s
Pi = 1,600 kgm/s
Based on the law of conservation of linear momentum, the sum of the initial momentum of an isolated system is <u>equal</u> to the sum of the final momentum of the system
Pf = Pi = 1,600 kgm/s.
Learn more about conservation of linear momentum here: brainly.com/question/7538238
#SPJ1
Answer:
![N = 42 rev](https://tex.z-dn.net/?f=N%20%3D%2042%20rev)
Explanation:
As we know that initial angular speed of the tub was zero and then it increases uniformly to 4 rev/s in t = 6.00 s
so we will have
![\theta_1 = \frac{\omega_f + \omega_i}{2} t](https://tex.z-dn.net/?f=%5Ctheta_1%20%3D%20%5Cfrac%7B%5Comega_f%20%2B%20%5Comega_i%7D%7B2%7D%20t)
![\theta_1 = \frac{2\pi f_2 + 2\pi f_1}{2}(t)](https://tex.z-dn.net/?f=%5Ctheta_1%20%3D%20%5Cfrac%7B2%5Cpi%20f_2%20%2B%202%5Cpi%20f_1%7D%7B2%7D%28t%29)
![\theta_1 = \frac{2\pi \times 4 + 0}{2}(6)](https://tex.z-dn.net/?f=%5Ctheta_1%20%3D%20%5Cfrac%7B2%5Cpi%20%5Ctimes%204%20%2B%200%7D%7B2%7D%286%29)
![\theta_1 = 75.4 rad](https://tex.z-dn.net/?f=%5Ctheta_1%20%3D%2075.4%20rad)
Now when the tub will comes to rest uniformly after opening the lid in time interval of t = 15 s
then we have
![\theta_2 = \frac{\omega_f + \omega_i}{2} t](https://tex.z-dn.net/?f=%5Ctheta_2%20%3D%20%5Cfrac%7B%5Comega_f%20%2B%20%5Comega_i%7D%7B2%7D%20t)
![\theta_2 = \frac{2\pi f_2 + 2\pi f_1}{2}(t)](https://tex.z-dn.net/?f=%5Ctheta_2%20%3D%20%5Cfrac%7B2%5Cpi%20f_2%20%2B%202%5Cpi%20f_1%7D%7B2%7D%28t%29)
![\theta_2 = \frac{2\pi \times 4 + 0}{2}(15)](https://tex.z-dn.net/?f=%5Ctheta_2%20%3D%20%5Cfrac%7B2%5Cpi%20%5Ctimes%204%20%2B%200%7D%7B2%7D%2815%29)
![\theta_2 = 188.5 rad](https://tex.z-dn.net/?f=%5Ctheta_2%20%3D%20188.5%20rad)
Now total angular displacement of the tub is given as
![\theta = \theta_1 + \theta_2](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20%5Ctheta_1%20%2B%20%5Ctheta_2)
![\theta = 75.4 + 188.5](https://tex.z-dn.net/?f=%5Ctheta%20%3D%2075.4%20%2B%20188.5)
![\theta = 263.9 rad](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20263.9%20rad)
so number of revolutions is given as
![\N = \frac{\theta}{2\pi}](https://tex.z-dn.net/?f=%5CN%20%3D%20%5Cfrac%7B%5Ctheta%7D%7B2%5Cpi%7D)
![N = 42 rev](https://tex.z-dn.net/?f=N%20%3D%2042%20rev)