The 50N group of force would be greater
So whichever object is being pulled will be pulled towards the 50N force
Answer:
t = 3.82 s
Ax = 147 N (←)
Ay = 294 N (+↑)
Explanation:
Given
m = 30.0 Kg
ωinitial = 125 rad/s
ωfinal = 0 rad/s
μC = 0.5
R = 0.3 m
t = ?
Ax = ?
Ay = ?
For the disk, we can apply
∑ τ = I*α
where I = m*R²/2
then
⇒ R*Ffriction = (m*R²/2)*α
⇒ R*(-μC*N) = R*(-μC*m*g) = (m*R²/2)*α
⇒ α = -2*μC*g / R
⇒ α = -2*(0.5)*(9.8) / 0.3 = -32.666 rad/s²
we can use the equation to get t:
α = Δω / t ⇒ t = Δω / α = (0 - 125) / (-32.666)
⇒ t = 3.82 s
The horizontal and vertical components of force which the member AB exerts on the pin at A during this time are
∑ Fx = 0 (+→)
Ax - Ffriction = 0
⇒ Ax = Ffriction = μC*m*g = (0.5)*(30)*(9.8) = 147 N
⇒ Ax = 147 N (←)
∑ Fy = 0 (+↑)
⇒ Ay - m*g = 0
⇒ Ay = m*g
⇒ Ay = 30*9.8 = 294 N
⇒ Ay = 294 N (+↑)
Answer:
Can two like charges attract each other : True ( Yes)
Answer:
× 
Explanation:
Knowing that, the volume of the sphere is given by, 
Thus, the fractional change in volume is given by,
× 
Pressure at the bottom of the sea is,
Δp =pgh =
×
×
pa.
Knowing that,
Bulk modulus: 


Answer =
× 
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