In this case, volume of the can remains constant. The relationship between pressure and temperature at constant volume is given by:
P/T = Constant
Then
Where
P1 = 40 psi
P2 = ?
T1 = 60°F ≈ 289 K
T2 = 90°F ≈ 305 K (note, 363 K is not right)
Substituting;
Answer:
you could go 12 miles paying $7.80 and $1.75
So in total being $9.55
Answer:
A. The athlete isn’t doing any work because he doesn’t move the weight.
Explanation:
We must remember the definition of work, which says that work is equal to the product of mass by the distance displaced. In this case, the athlete only does work when he lifts the weight from the ground to the point where he holds the weight suspended.
So when he's holding the weight, he doesn't do any work.
Answer:
The work done on the wagon is 37 joules.
Explanation:
Given that,
The force applied by Charlie to the right, F = 37.2 N
The force applied by Sara to the left, F' = 22.4 N
We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :
Work done on the wagon is given by the product of net force and displacement. It is given by :
W = 37 Joules
So, the work done on the wagon is 37 joules. Hence, this is the required solution.
Answer:
Fx1 (6 m) sin 60 = 300 (3 m) cos 60 balancing torques about floor
Fx1 = 900 * 1/2 / 5.20 = 86.6 N this is the horizontal force that must be supplied by the wall to balance torques about the floor
This is also equal to the static force of friction that must be applied at the point of contact with the floor to balance forces in the x-direction.
Fx1 = Fx2 = 86.6 N
