Answer:
The speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.
Explanation:
Given;
mass of block, m = 4 kg
coefficient of kinetic friction, μk = 0.25
angle of inclination, θ = 30°
initial speed of the block, u = 5 m/s
From Newton's second law of motion;
F = ma
a = F/m
Net horizontal force;
∑F = mgsinθ + μkmgcosθ

At the top of the ramp, energy is conserved;
Kinetic energy = potential energy
¹/₂mv² = mgh
¹/₂ v² = gh
¹/₂ x 5² = 9.8h
12.5 = 9.8h
h = 12.5/9.8
h = 1.28 m
Height of the ramp is 1.28 m
Now, calculate the speed of the block (in m/s) when it has returned to the bottom of the ramp;
v² = u² + 2ah
v² = 5² + 2 x 7.022 x 1.28
v² = 25 + 17.976
v² = 42.976
v = √42.976
v = 6.56 m/s
Therefore, the speed of the block when it has returned to the bottom of the ramp is 6.56 m/s.
Explanation:
Ultraviolet is a form of electromagnetic radiation with wavelength from 10 nm to 400 nm, shorter than that of visible light, but longer than X-rays. UV radiation is present in sunlight, and constitutes about 10% of the total electromagnetic radiation output from the Sun.
The internal energy of the gas is 49,200 J
Explanation:
The internal energy of a diatomic gas, such as
, is given by

where
n is the number of moles
R is the gas constant
T is the absolute temperature of the gas
For the gas in this problem, we have:
n = 4.50 (number of moles)
R = 8.31 J/(mol·K) (gas constant)
(absolute temperature)
Substituting, we find:

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Answer:
0 N
Explanation:
because he is taller and more mass he never go back