Answer:
3/7 ω
Explanation:
Initial momentum = final momentum
I(-ω) + (2I)(3ω) + (4I)(-ω/2) = (I + 2I + 4I) ωnet
-Iω + 6Iω - 2Iω = 7I ωnet
3Iω = 7I ωnet
ωnet = 3/7 ω
The final angular velocity will be 3/7 ω counterclockwise.
Answer:
The temperature of cold reservoir should be 246.818 K for efficiency of 35%
Explanation:
In first case we have given efficiency of Carnot engine = 26 % = 0.26
Temperature of cold reservoir ![T_L=281K](https://tex.z-dn.net/?f=T_L%3D281K)
We know that efficiency of Carnot engine is given by
![\eta =1-\frac{T_L}{T_H}](https://tex.z-dn.net/?f=%5Ceta%20%3D1-%5Cfrac%7BT_L%7D%7BT_H%7D)
![0.26 =1-\frac{281}{T_H}](https://tex.z-dn.net/?f=0.26%20%3D1-%5Cfrac%7B281%7D%7BT_H%7D)
![T_H=379.72K](https://tex.z-dn.net/?f=T_H%3D379.72K)
For second Carnot engine efficiency is given as 35% = 0.35
And temperature of hot reservoir is same so ![T_H=379.72K](https://tex.z-dn.net/?f=T_H%3D379.72K)
So ![0.35=1-\frac{T_L}{379.72}](https://tex.z-dn.net/?f=0.35%3D1-%5Cfrac%7BT_L%7D%7B379.72%7D)
![T_L=246.818K](https://tex.z-dn.net/?f=T_L%3D246.818K)
So the temperature of cold reservoir should be 246.818 K for efficiency of 35%
1.) The object's Velocity
Faster it goes, more kinetic energy it has
Answer:
Explanation:
When 2 gms of steam condenses to water at 100 degree latent heat of vaporization is releases which is calculated as follows
Heat released = mass x latent heat of vaporization
= 2 x 2260 = 4520 J
When 2 gms of water at 100 degree is cooled to ice water at zero degree heat is releases which is calculated as follows
Heat released = mass x specific heat x( 100-0)
= 2 x 4.2 x 100 = 840 J
When 2 gms of water at zero degree condenses to ice at zero degree latent heat of fusion is releases which is calculated as follows
Heat released = mass x latent heat of fusion
= 2 x 334 = 668 J
When 2 grams of steam at 100 degrees Celsius turns to ice at 0 degrees Celsius heat released will be sum of all the heat released as mentioned above ie
4520 + 840 +668 = 6028 J
Answer:
![V_{ft}= 317 cm/s](https://tex.z-dn.net/?f=V_%7Bft%7D%3D%20317%20cm%2Fs)
ΔK = 2.45 J
Explanation:
a) Using the law of the conservation of the linear momentum:
![P_i = P_f](https://tex.z-dn.net/?f=P_i%20%3D%20P_f)
Where:
![P_i=M_cV_{ic} + M_tV_{it}](https://tex.z-dn.net/?f=P_i%3DM_cV_%7Bic%7D%20%2B%20M_tV_%7Bit%7D)
![P_f = M_cV_{fc} + M_tV_{ft}](https://tex.z-dn.net/?f=P_f%20%3D%20M_cV_%7Bfc%7D%20%2B%20M_tV_%7Bft%7D)
Now:
![M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}](https://tex.z-dn.net/?f=M_cV_%7Bic%7D%20%2B%20M_tV_%7Bit%7D%20%3D%20M_cV_%7Bfc%7D%20%2B%20M_tV_%7Bft%7D)
Where
is the mass of the car,
is the initial velocity of the car,
is the mass of train,
is the final velocity of the car and
is the final velocity of the train.
Replacing data:
![(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}](https://tex.z-dn.net/?f=%281.1%20kg%29%284.95%20m%2Fs%29%20%2B%20%283.55%20kg%29%282.2%20m%2Fs%29%20%3D%20%281.1%20kg%29%281.8%20m%2Fs%29%20%2B%20%283.55%20kg%29V_%7Bft%7D)
Solving for
:
![V_{ft}= 3.17 m/s](https://tex.z-dn.net/?f=V_%7Bft%7D%3D%203.17%20m%2Fs)
Changed to cm/s, we get:
![V_{ft}= 3.17*100 = 317 cm/s](https://tex.z-dn.net/?f=V_%7Bft%7D%3D%203.17%2A100%20%3D%20317%20cm%2Fs)
b) The kinetic energy K is calculated as:
K = ![\frac{1}{2}MV^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7DMV%5E2)
where M is the mass and V is the velocity.
So, the initial K is:
![K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2](https://tex.z-dn.net/?f=K_i%20%3D%20%5Cfrac%7B1%7D%7B2%7DM_cV_%7Bic%7D%5E2%2B%5Cfrac%7B1%7D%7B2%7DM_tV_%7Bit%7D%5E2)
![K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2](https://tex.z-dn.net/?f=K_i%20%3D%20%5Cfrac%7B1%7D%7B2%7D%281.1%29%284.95%29%5E2%2B%5Cfrac%7B1%7D%7B2%7D%283.55%29%282.2%29%5E2)
![K_i = 22.06 J](https://tex.z-dn.net/?f=K_i%20%3D%2022.06%20J)
And the final K is:
![K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2](https://tex.z-dn.net/?f=K_f%20%3D%20%5Cfrac%7B1%7D%7B2%7DM_cV_%7Bfc%7D%5E2%2B%5Cfrac%7B1%7D%7B2%7DM_tV_%7Bft%7D%5E2)
![K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2](https://tex.z-dn.net/?f=K_f%20%3D%20%5Cfrac%7B1%7D%7B2%7D%281.1%29%281.8%29%5E2%2B%5Cfrac%7B1%7D%7B2%7D%283.55%29%283.17%29%5E2)
![K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2](https://tex.z-dn.net/?f=K_f%20%3D%20%5Cfrac%7B1%7D%7B2%7D%281.1%29%281.8%29%5E2%2B%5Cfrac%7B1%7D%7B2%7D%283.55%29%283.17%29%5E2)
![K_f = 19.61 J](https://tex.z-dn.net/?f=K_f%20%3D%2019.61%20J)
Finally, the change in the total kinetic energy is:
ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J