Answer:
Explanation:
<u><em>Finding the net force:</em></u>
<u><em>Firstly , we'll find force of Friction:</em></u>
Where 
is the coefficient of friction and m = 13.6 kg
<u><em>Now, Finding the net force:</em></u>
<u><em>Finding Acceleration:</em></u>
Answer:
q = 2.65 10⁻⁶ C
Explanation:
For this exercise we use Coulomb's law
F =
In this case they indicate that the load is of equal magnitude
q₁ = q₂ = q
the force is attractive because the signs of the charges are opposite
F = 
q = 
we calculate
q = 
q = 
Ra 7 10-12
q = 2.65 10⁻⁶ C
Answer:
Explanation:
To find out the angular velocity of merry-go-round after person jumps on it , we shall apply law of conservation of ANGULAR momentum
I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω
I₁ is moment of inertia of disk , I₂ moment of inertia of running person , I is the moment of inertia of disk -man system , ω₁ and ω₂ are angular velocity of disc and man .
I₁ = 1/2 mr²
= .5 x 175 x 2.13²
= 396.97 kgm²
I₂ = m r²
= 55.4 x 2.13²
= 251.34 mgm²
ω₁ = .651 rev /s
= .651 x 2π rad /s
ω₂ = tangential velocity of man / radius of disc
= 3.51 / 2.13
= 1.65 rad/s
I₁ ω₁ + I₂ ω₂ = ( I₁ + I₂ ) ω
396.97 x .651 x 2π + 251.34 x 1.65 = ( 396.97 + 251.34 ) ω
ω = 3.14 rad /s
kinetic energy = 1/2 I ω²
= 3196 J
