The speed of the spring when it is released is 3.5 m/s.
The given parameters:
- <em>Mass of the block, m = 2.5 kg</em>
- <em>Spring constant, k = 56 N/m</em>
- <em>Extension of the spring, x = 0.75 m</em>
The speed of the spring when it is released is calculated by applying the principle of conservation of energy as follows;

Thus, the speed of the spring when it is released is 3.5 m/s.
Learn more about conservation of energy here: brainly.com/question/166559
Answer: a) Mr = 2.4×10^-4kg/s
V = 34.42m/a
b) E = 173J
Ø = 2693.1J
c) Er = 0.64J/s
Explanation: Please find the attached file for the solution
Explanation:
It is given that,
Initial velocity of frog, u = 100 m/s
It jumps at an angle of 60 degrees.
(a) Let
are the x and y components of velocity. It can be calculated as :






(b) Let y is the maximum height reached by the frog. It can be calculated using the third equation of motion as :
At maximum height, 
and a = -g


y = 382.63 meters
(c) Let t is the time taken by the frog to reach its maximum height. It can be calculated as :



t = 8.83 seconds
Hence, this is the required solution.
Acceleration and deceleration and momentom
they slow the car down and it loses momentome
Answer:
time of collision is
t = 0.395 s

so they will collide at height of 5.63 m from ground
Explanation:
initial speed of the ball when it is dropped down is

similarly initial speed of the object which is projected by spring is given as

now relative velocity of object with respect to ball

now since we know that both are moving under gravity so their relative acceleration is ZERO and the relative distance between them is 6.4 m



Now the height attained by the object in the same time is given as



so they will collide at height of 5.63 m from ground