Answer:
16.4287
Explanation:
The force and displacement are related by Hooke's law:
F = kΔx
The period of oscillation of a spring/mass system is:
T = 2π√(m/k)
First, find the value of k:
F = kΔx
78 N = k (98 m)
k = 0.796 N/m
Next, find the mass of the unknown weight.
F = kΔx
m (9.8 m/s²) = (0.796 N/m) (67 m)
m = 5.44 kg
Finally, find the period.
T = 2π√(m/k)
T = 2π√(5.44 kg / 0.796 N/m)
T = 16.4287 s
Answer:
the velocity of the bullet-wood system after the collision is 2.48 m/s
Explanation:
Given;
mass of the bullet, m₀ = 20 g = 0.02 kg
velocity of the bullet, v₀ = 250 m/s
mass of the wood, m₁ = 2 kg
velocity of the wood, v₁ = 0
Let the velocity of the bullet-wood system after collision = v
Apply the principle of conservation of linear momentum to calculate the final velocity of the system;
Initial momentum = final momentum
m₀v₀ + m₁v₁ = v(m₀ + m₁)
0.02 x 250 + 2 x 0 = v(2 + 0.02)
5 + 0 = v(2.02)
5 = 2.02v
v = 5/2.02
v = 2.48 m/s
Therefore, the velocity of the bullet-wood system after the collision is 2.48 m/s
Answer:
the velocity of the boats after the collision is 4.36 m/s.
Explanation:
Given;
mass of fish, m₁ = 800 kg
mass of boat, m₂ = 1400 kg
initial velocity of the fish, u₁ = 12 m/s
initial velocity of the boat, u₂ = 0
let the final velocity of the fish-boat after collision = v
Apply the principle of conservation of linear momentum for inelastic collision;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
800 x 12 + 1400 x 0 = v(800 + 1400)
9600 = 2200v
v = 9600/2200
v = 4.36 m/s
Therefore, the velocity of the boats after the collision is 4.36 m/s.
Answer:
The distance covered by the balloon is 47.52 meters.
Explanation:
Given that,
Initial speed of the balloon, u = 1.14 m/
Let us assumed we need to find the distance covered by the balloon after t = 3 second. Let d is the distance covered by the balloon. It can be given by :
Here, a = g
d = 47.52 meters
So, the distance covered by the balloon is 47.52 meters. Hence, this is the required solution.
Answer:
The y-component of the normal force is 45.74 N.
Explanation:
Given that,
Mass of the crate, m = 5 kg
Angle with hill, 
We need to find the y component of the normal force. We know that the y component of the normal force is given by :
So, the y-component of the normal force is 45.74 N. Hence, this is the required solution.
