Answer:
0.3 m
Explanation:
Initially, the package has both gravitational potential energy and kinetic energy. The spring has elastic energy. After the package is brought to rest, all the energy is stored in the spring.
Initial energy = final energy
mgh + ½ mv² + ½ kx₁² = ½ kx₂²
Given:
m = 50 kg
g = 9.8 m/s²
h = 8 sin 20º m
v = 2 m/s
k = 30000 N/m
x₁ = 0.05 m
(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²
x₂ ≈ 0.314 m
So the spring is compressed 0.314 m from it's natural length. However, we're asked to find the additional deformation from the original 50mm.
x₂ − x₁
0.314 m − 0.05 m
0.264 m
Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.
Answer: Charge = -2.4x10^-9 Coulombs
Explanation:
The charge of one electron is e = -1.6x10^-19 C
Then, the charge of 1.5 x 10^10 electrons is equal to 1.5 x 10^10 times the charge of one electron:
Here i will use the relation (a^b)*(a^c) = a^(b + c)
Charge = ( 1.5 x 10^10)*( -1.6x10^-19 C) = -2.4x10^(10 - 19) C
Charge = -2.4x10^-9 C
A 300-kg bear grasping a vertical tree slides down at constant velocity. The friction force between the
tree and the bear is
Answer:
t₁ > t₂
Explanation:
A coin is dropped in a lift. It takes time t₁ to reach the floor when lift is stationary. It takes time t₂ when lift is moving up with constant acceleration. Then t₁ > t₂, t₁ = t₂, t₁ >> t₂ , t₂ > t₁
Solution:
Newton's law of motion is given by:
s = ut + (1/2)gt²;
where s is the the distance covered, u is initial velocity, g is the acceleration due to gravity and t is the time taken.
u = 0 m/s, t₁ is the time to reach ground when the light is stationary and t₂ is the time to reach ground when the lift is moving with a constant acceleration a.
hence:
When stationary:
Hence t₂ < t₁, this means that t₁ > t₂.
