Answer:
The crate was being lifted by a height of 1.48 meters.
Explanation:
In an attempt o move a crate;
Force applied = 2470 N
Work done by the force = 3650 J
We know that the work done is defined as the force used to move an object to a distance.
Given the Force used and the work done by that Force, we need to find out the distance the crate was lifted to.
Work done is defined as:
Work = Force*distance covered in the direction of the force
3650 = 2470*distance
distance = 3650/2470
distance = 1.48 meters
Answer:
No
Explanation:
Unless there are other external forces, this will never be true. Because according to energy conservation, potential energy will be converted to kinetic energy as the ball falls down (so it loses height and gain speed). And vice versa, kinetic to potential when it bounces back. So the potential energy after must be the same (or smaller if losing heat to external environment), so it can only get the the same height or less, but not more.
Answer: The statement first and the fourth statement are true.
Explanation:
According to Newton's gravitational law, every particle in the universe attracts every other particle with the force of attraction between the masses is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.
As we move to higher altitude, the force of gravity on use decreases because the force of gravity is inversely proportional to the distance.
If the masses of the two objects are more then there will be greater force of gravity between them.
Therefore, the statement first and the fourth statement are true.
Note:
The height of a high bar from the floor is h = 2.8 m (or 9.1 ft).
It is not provided in the question, so the standard height is assumed.
g = 9.8 m/s², acceleration due to gravity.
Note that the velocity and distance are measured as positive upward.
Therefore the floor is at a height of h = -2.8 m.
First dismount:
u = 4.0 m/s, initial upward velocity.
Let v = the velocity when the gymnast hits the floor.
Then
v² = u² - 2gh
v² = 16 - 2*9.8*(-2.8) = 70.88
v = 8.42 m/s
Second dismount:
u = -3.0 m/s
v² = (-3.0)² - 2*9.8*(-2.8) = 63.88 m/s
v = 7.99 m/s
The difference in landing velocities is 8.42 - 7.99 = 0.43 m/s.
Answer:
First dismount:
Acceleration = 9.8 m/s² downward
Landing velocity = 8.42 m/s downward
Second dismount:
Acceleration = 9.8 m/s² downward
Landing velocity = 7.99 m/s downward
The landing velocities differ by 0.43 m/s.
Answer:
3311N
Explanation:
r = radius = 600m
V = speed = 150m/s
Mass = weight = 70kg
The weight of pilot when calculated due to circular motion
W = tv
Fv = mv²/r
Fv = 70x150²/600
Fv = 79x22500/600
= 15750000/600
= 2625N
Real Weight of the pilot = m x g
= 70 x 9.8
= 686N
The apparent Weight is calculated by
Mv²/r + mg
= 2625N + 686N
= 3311 N
Therefore the apparent Weight is 3311N
