Question: 18 kilogram Mass Block rest on level surface if the coefficient of static friction between the Block and the surface is 0.6 what horizontal force is required to just move the blcok ( take gravity as 10m/s2
)
Answer:
108 N
Explanation:
From the question,
Applying
F' = mgμ................ Equation 1
Where F' = Frictional force = horizontal force required to just move the block, m = mass of the block, g = acceleration due to gravity, μ = coefficient of static friction.
From the question,
Given: m = 18 kg, μ = 0.6, g = 10 m/s²
Substitute these values into equation 1
F' = 18×0.6×10
F' = 108 N
Answer:
4.24m/s
Explanation:
Potential energy at the top= kinetic energy at the button
But kinetic energy= sum of linear and rotational kinetic energy of the hoop
PE= mgh
KE= 1/2 mv^2
RE= 1/2 I ω^2
Where
m= mass of the hoop
v= linear velocity
g= acceleration due to gravity
h= height
I= moment of inertia
ω= angular velocity of the hoop.
But
I = m r^2 for hoop and ω = v/r
giving
m g h = 1/2 m v^2 + 1/2 (m r^2) (v^2/r^2) = 1/2 m v^2 + 1/2 m v^2 = m v^2
and m's cancel
g h = v^2
Hence
v= √gh
v= √10×1.8
v= 4.24m/s
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The correct option will be
D. Time, initial velocity and final velocity
The Formula can be written as,
Acceleration=Final velocity-Initial Velocity/Time
Answer:I believe it is D I might be wrong
Explanation:
