Answer:
(a) v = 5.42m/s
(b) vo = 4.64m/s
(c) a = 2874.28m/s^2
(d) Δy = 5.11*10^-3m
Explanation:
(a) The velocity of the ball before it hits the floor is given by:
(1)
g: gravitational acceleration = 9.8m/s^2
h: height where the ball falls down = 1.50m

The speed of the ball is 5.42m/s
(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.
You use the following formula:
(2)
vo: velocity of the ball where it starts its motion upward
You solve for vo and replace the values of the parameters:

The velocity of the ball is 4.64m/s
(c) The acceleration is given by:


The acceleration of the ball is 2874.28/s^2
(d) The compression of the ball is:

THe compression of the ball when it strikes the floor is 5.11*10^-3m
In order to describe motion along a straight line, you must state the speed and direction of the motion. Those two quantities, together, comprise what's known as "velocity".
Answer:
8.94*10^22 kg
Explanation:
Given that
Mass of Lo, M = ?
Radius of Lo, r = 1.82*10^6 m
Acceleration on Lo, g = 1.80 m/s²
Gravitational constant, G = 6.67*10^-11
Using the formula
g = GM/r²
Solution is attached below
Answer is 8.94*10^22 kg
5
Explanation:
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Answer:
The tangential speed of the tack is 6.988 meters per second.
Explanation:
The tangential speed experimented by the tack (
), measured in meters per second, is equal to the product of the angular speed of the wheel (
), measured in radians per second, and the distance of the tack respect to the rotation axis (
), measured in meters, length that coincides with the radius of the tire. First, we convert the angular speed of the wheel from revolutions per second to radians per second:


Then, the tangential speed of the tack is: (
,
)


The tangential speed of the tack is 6.988 meters per second.