The work done in the isothermal process is 10 joule.
We need to know about the isotherm process to solve this problem. The isotherm process can be described as a process where the initial temperature system will be the same as the final temperature. Hence, the internal energy change will be zero.
ΔU = 0
Hence,
ΔU = Q - W
0 = Q - W
Q = W
It means that the heat transferred is the same as the work done.
From the question above, we know that the heat transferred is 10 joule. Thus, the work done in the isothermal process is 10 joule.
Find more on isothermal at: brainly.com/question/17097259
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Answer:
the planes wings are lifting at an angle to gravity so the plane isn't lifting as much against gravity when it banks. some of the wing lift is going into turning the plane. :) so it needs more lift to bank and stay up
Explanation:
Answer:
E) momentum and mechanical energy
Explanation:
In the context, an object is attached to the another mass with a spring which is initially at a rest position. Now when the spring is compressed, the two masses moves with the same speed. Now since the both the masses combines with the spring to move together they are considered as one system and in this case the momentum and the kinetic energy will be conserved.
The kinetic energy and momentum of the system after collision and the kinetic energy and momentum of the two masses before collision will be constant.
The given data is incomplete. The complete question is as follows.
At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coefficient of friction was estimated to be 0.36. Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes. (why does the car's mass not matter?)
Explanation:
Let us assume that v is the final velocity and u is the initial velocity of the car. Let s be the skid marks and
be the friction coefficient and m be the mass of car.
Hence, the given data is as follows.
v = 0, s = 84 m,
= 0.36
According to Newton's law of second motion the expression for acceleration is as follows.
F = ma
= ma
= ma
a = 
Also,



= 
= 24.36 m/s
Thus, we can conclude that the speed of the car when the driver slammed on (and locked) the brakes is 24.36 m/s.
A force vector F1 points due
east and has a magnitude of 200 Newtons, A second force F2 is added to F1. The
resultant of the two vectors has a magnitude of 400 newtons and points along
the due east/west line. Find the magnitude and direction of F2. Note that there
are two answers.
<span>The given values are
F1 = 200 N</span>
F2 =?
Total = 400 N
Solution:
F1 + F2 = T
200 N + F2 = 400N
F2 = 400 - 200
F2 = 200
N