Answer:
(D) friction from the ground changes the ball's
kinetic energy into heat
Explanation:
When a ball is roll on the ground, the electrons in the atoms on the surface of the ground push against the electrons in the atoms on the surface of your ball that is touching the ground. A rolling ball stops because the surface on which it rolls resists its motion,that is when two surfaces come in contact with each other, the surface of one tends to oppose the motion of the other. A rolling ball stops because of friction.
To answer this question, we would need the formula for the kinetic energy which is Kinetic Energy = ½ x m x v^2
Where the following means: m is the mass of the object and v is the velocity
So At 7 m/s: Kinetic Energy = ½ x 5 x 7^2 = 122.5 J is the answer
Answer:
the initial velocity of the car is 12.04 m/s
Explanation:
Given;
force applied by the break, f = 1,398 N
distance moved by the car before stopping, d = 25 m
weight of the car, W = 4,729 N
The mass of the car is calculated as;
W = mg
m = W/g
m = (4,729) / (9.81)
m = 482.06 kg
The deceleration of the car when the force was applied;
-F = ma
a = -F/m
a = -1,398 / 482.06
a = -2.9 m/s²
The initial velocity of the car is calculated as;
v² = u² + 2ad
where;
v is the final velocity of the car at the point it stops = 0
u is the initial velocity of the car before the break was applied
0 = u² + 2(-a)d
0 = u² - 2ad
u² = 2ad
u = √2ad
u = √(2 x 2.9 x 25)
u =√(145)
u = 12.04 m/s
Therefore, the initial velocity of the car is 12.04 m/s
Newton's first law states that an object at rest remains so until acted upon by external forces or an object in motion does so without accelerating (that is, at constant speed).
Therefore, x can only describe an object coming to rest. That, the last option is more correct.
Explanation:
(a) You can solve this using kinematics or energy.
Using kinematics:
a = F/m = 90 N / 4 kg = 22.5 m/s²
v₀ = 0 m/s
Δx = 5 m
Find: v
v² = v₀² + 2aΔx
v² = (0 m/s)² + 2 (22.5 m/s²) (5 m)
v = 15 m/s
Using energy:
W = ΔKE
Fd = ½ mv²
(90 N) (5 m) = ½ (4 kg) v²
v = 15 m/s
(b) ΔU = mg Δh
ΔU = (4 kg) (9.8 m/s) (12 m sin 40° − 15 m)
ΔU = -290 J
W = ΔU = -290 J
