Answer:
F₁> F₂
Explanation:
For this exercise Newton's second law is used, in the adjoint we can see the unapplied forces in this exercise.
Y axis y
N- W = 0
in this axis there is no movement
X axis
F -fr = m a
as they indicate that the velocity is consonant the acceleration is worth zero
F - fr = 0
friction force has the expression
fr = μ N
fr = μ mg
we substitute
F = μ m g
by the time the block is stopped the deferred force is
F₁ = μ_s m g
when it begins to move the force should decrease to
F₂ = μ_k k m g
as the static coefficient is greater than the dynamic coefficient
F₁> F₂
Answer:
0.888 ohm
Explanation:
We have given that the battery has an emf of 12 volt that E =12 volt
Terminal voltage V =18.4 volt
Current through the battery is =7.20 A
We have top find the internal resistance of the battery
Now according to Kirchhoff's law V=E+IR
So 
So the value of internal resistance of battery is 0.888 ohm
Answer:
Hi myself Shrushtee.
Explanation:
Artificial gravity is a must for any space station if humans are to live there for any extended length of time. Without artificial gravity, human growth is stunted and biological functions break down. An effective way to create artificial gravity is through the use of a rotating enclosed cylinder, as shown in the figure. Humans walk on the inside edge of the cylinder, which is sufficiently large (diameter of 2235 meters) that its curvature is not readably noticeable to the inhabitants. (The space station in the figure is not drawn to the scale of the human.) Once the space station is rotating at the necessary speed, how many minutes would it take the space station to make one revolution?
The distance traveled by the man in one revolution is simply the circumference of the space station, C = 2p R. From this result, you should be able to deduce the time it takes for the space station to sweep out a complete revolution.
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Answer:
(a) the work done by the student is 110.1 J
(b) The gravitational force that acts on the amplifier is 102.9 N
Explanation:
Given;
mass of the amplifier, m = 10.5 kg
initial position of the amplifier, x₀ = 1.82 m
final position of the amplifier, x₁ =0.75 m
The dispalcement of the amplifier Δx = x₁ - x₀ = 1.82 m - 0.75 m = 1.07 m
(b) The gravitational force that acts on the amplifier;
F = mg
F = 10.5 x 9.8
F = 102.9 N
(a) the work done by the student is calculated as;
W = FΔx
W = 102.9 x 1.07
W = 110.1 J
