Answer:
(a) 4.21 m/s
(b) 24.9 N
Explanation:
(a) Draw a free body diagram of the object when it is at the bottom of the circle. There are two forces on the object: tension force T pulling up and weight force mg pulling down.
Sum the forces in the radial (+y) direction:
∑F = ma
T − mg = m v² / r
v = √(r (T − mg) / m)
v = √(0.676 m (54.7 N − 1.52 kg × 9.8 m/s²) / 1.52 kg)
v = 4.21 m/s
(b) Draw a free body diagram of the object when it is at the top of the circle. There are two forces on the object: tension force T pulling down and weight force mg pulling down.
Sum the forces in the radial (-y) direction:
∑F = ma
T + mg = m v² / r
T = m v² / r − mg
T = (1.52 kg) (4.21 m/s)² / (0.676 m) − (1.52 kg) (9.8 m/s²)
T = 24.9 N
D), increases. The object absorbs light energy which in turn (energy is energy) usually involves absorbing heat as well.
Answer:
For any collision occurring in an isolated system, momentum is conserved. The total amount of momentum of the collection of objects in the system is the same before the collision as after the collision.
Explanation:
Hope this helps
Answer:D
Explanation:
Given
mass of object 
Distance traveled 
velocity acquired 
conserving Energy at the moment when object start falling and when it gains 12 m/s velocity
Initial Energy
Final Energy
where 
is friction work if any
Since Friction is Present therefore it is a case of Open system and net external Force is zero
An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .
Answer:
Q = c M ΔT where c is the heat capacity and M the mass present
Q2 / Q1 = M2 / M1 since the other factors are the same
M = ρ V where ρ is the density
M = ρ Π (d / 2)^2 where d is the diameter of the sphere
M2 / M1 = (2 D/2)^2 / (D/2)^2 = 4
It will take 4Q heat to heat the second sphere
