The velocity with which the jumper leaves the floor is 5.1 m/s.
<h3>
What is the initial velocity of the jumper?</h3>
The initial velocity of the jumper or the velocity with which the jumper leaves the floor is calculated by applying the principle of conservation of energy as shown below.
Kinetic energy of the jumper at the floor = Potential energy of the jumper at the maximum height
¹/₂mv² = mgh
v² = 2gh
v = √2gh
where;
- v is the initial velocity of the jumper on the floor
- h is the maximum height reached by the jumper
- g is acceleration due to gravity
v = √(2 x 9.8 x 1.3)
v = 5.1 m/s
Learn more about initial velocity here: brainly.com/question/19365526
#SPJ1
False because light microscopes have low resolve and magnification.
Answer:
0.125 m
Explanation:
In this problem, we have:
v = 0.50 m/s is the average velocity of the wave
T = 0.25 s is the period of the wave
We can find the frequency of the wave, which is equal to the reciprocal of the period:

The problem is asking us to find the distance between two crests of the wave: this is equivalent to the wavelength. The wavelength is related to the average velocity and the frequency by the formula:

Substituting the numerical values, we find
