Answer:
the answer is at the BOTTOM OF THEIR QUESTION
Explanation:
IT IS CORRECT BTW
= Heat released to cold reservoir
= Heat released to hot reservoir
= maximum amount of work
= temperature of cold reservoir
= temperature of hot reservoir
we know that

eq-1
maximum work is given as
=
- 
using eq-1
=
- 
Answer:
U = -3978.8 J
Explanation:
The work of the gravitational force U just depends of the heigth and is calculated as:
U = -mgh
Where m is the mass, g is the gravitational acceleration and h the alture.
for calculate the alture we will use the following equation:
h = L-Lcos(θ)
Where L is the large of the rope and θ is the angle.
Replacing data:
h = 12.2-12.2cos(58.4)
h = 5.8 m
Finally U is equal to:
U = -70(9.8)(5.8)
U = -3,978.8 J
The new natural frequency would be ω/2.
we know that,
= ω. -> equation 1
now, when capacitance is quadrupled,

. -> equation 2
substituting value of equation 1 in equation 2 , we get,

Hence, the new natural frequency of the circuit is ω/2.
what do you mean by frequency ?
The resonant frequency for a particular circuit is the frequency at which this equality stands true. Where L is the inductance in henries and C is the capacitance in farads, this is the LC circuit's resonant frequency.
Learn more about frequency here:-
brainly.com/question/12530980
#SPJ4
Answer:
The instantaneous speed of the object after the first five seconds is 12.5 m/s.
(C) is correct option.
Explanation:
Given that,
An object starts at rest. Its acceleration over 30 seconds.
We need to calculate the instantaneous speed of the object after the first five seconds
We know that,
Area under the acceleration -time graph gives speed.
According to figure,




Hence, The instantaneous speed of the object after the first five seconds is 12.5 m/s.