Answer:
(a) W=217 J
(b) Tc=378K
(c) e=0.39=39%
Explanation:
For part (a)
We to calculate the mechanical work W the engine does. By knowing QC and QH can obtain the work using equation
W = IQHI — IQcl .....................eq(1)
Put given values for QH and QC into equation (1) to get
the mechanical work of the engine
W = 550 - 335
W=217 J
For part (b)
We want to determine the temperature of low temperature reservoir which means Tc
IQc|/|Qh| =TC/TH
for Tc
Tc=(IQc|/|Qh|)*TH
Now we can put values
Tc= 620K (335/
550.1)
Tc=378K
For part (c)
Here we want to find the thermal efficiency (e) of the cycle
e=1-TC/TH
e=1-(378/620)
e=0.39=39%
It would cost $0.69 to operate the monitor for a day.
Answer:
Yes, Roberto will have twice the kinetic energy
Explanation:
The equation for kinetic energy is Ke = 1/2(<em>m</em>)v^2
The variable <em>m</em> is not squared by anything, which means it is directly proportional to the equation. If you were to multiply <em>m</em> by 2, it would in turn multiply the entire equation by 2, getting you an outcome with twice as much kinetic energy.
Answer:
The total kinetic energy of the system before the collision is 57 Joules.
Explanation:
Given that,
Mass of first block,
Speed of first block,
Mass of second block,
Speed of first block,
We need to find the total kinetic energy of the system before the collision. It is equal to the kinetic energies of both the blocks. It is given by :
So, the total kinetic energy of the system before the collision is 57 Joules.