A 60 g ball of clay is thrown horizontally at 40 m/s toward a 1.5 kg block sitting at rest on a frictionless surface. the clay h
1 answer:
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Answer:
Explanation:
Given
there is fix Quantity of gas i.e. mass of gas is constant
From ideal gas Equation
PV=nRT
(a) volume always increases is not true as Pressure can also be increased .
(b)If Pressure is constant along with mass then as Temperature increases, Volume also increases.
(c)true , Product of Pressure and volume depends upon temperature thus it also increases with temperature.
(d)Density of gas may or may not increases
As density is
volume may increase or decrease as temperature increase .
(e)false
as it clearly stated that quantity of fixed therefore there is no change in gas
1. -23.2 J
The gravitational potential energy of the ball is given by
where
m = 1.1 kg is the mass of the ball
g = 9.8 m/s^2 is the acceleration of gravity
h is the height of the ball, relative to the reference point chosen
In this part of the problem, the reference point is the ceiling. So, the ball is located 2.16 m below the ceiling: therefore, the heigth is
h = -2.16 m
And the gravitational potential energy is
2. 41.1 J
Again, the gravitational potential energy of the ball is given by
In this part of the problem, the reference point is the floor.
The height of the ball relative to the floor is equal to the height of the floor minus the length of the string:
h = 5.97 m - 2.16 m = 3.81 m
And so the gravitational potential energy of the ball relative to the floor is
3. 0 J
As before, the gravitational potential energy of the ball is given by
Here the reference point is a point at the same elevation of the ball.
This means that the heigth of the ball relative to that point is zero:
h = 0 m
And so the gravitational potential energy is