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
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Answer:
see that the correct one is B
Explanation:
To solve this exercise let us use the kinematic relations
v² = v₀² - 2 a x
as they indicate that the car stops, therefore the final speed is yield v = 0
x = v₀² / 2a
let's calculate
x = 2²/(2 0.8)
x = 2.5 m / s²
When reviewing the answers we see that the correct one is B
Answer:
500J
Explanation:
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Answer:
Watts=Volt*Amps
So A=570/120=4.75amps
If voltage drops to 110V We get A=570/110=5.(18...)amps
Answer:
The equation of D = m/V
Where D = density
m = mass
and V = volume
We are solving for V, so with the manipulation of variables we multiply V on both sides giving us
V(D) = m
now we divide D on both sides giving us
V = m/D
We know our mass which is 600g and our density is 3.00 g/cm^3
so
V = 600g/3.00g/cm^3 = 200cm^3 or 200mL
a cubic centimeter (cm^3) is one of the units for volume. It's exactly like mL. 1 cm^3 = 1 mL
If you wish to change it to L, you'd have to convert
Explanation:
