Answer:
v= 1.71 m/s
Explanation:
Given that
Distance between two successive crests = 4.0 m
λ = 4 m
T= 7 sec
T is the time between 3 waves.
3 waves = 7 sec
1 wave = 7 /3 sec
So t= 7/3 s
We know that frequency f
f= 1/t= 3/7 Hz
Lets take speed of the wave is v
v= f λ
f=frequency
λ=wavelength
v= 3/7 x 4 = 12 /7
v= 1.71 m/s
Answer:
v₁f = 0.5714 m/s (→)
v₂f = 2.5714 m/s (→)
e = 1
It was a perfectly elastic collision.
Explanation:
m₁ = m
m₂ = 6m₁ = 6m
v₁i = 4 m/s
v₂i = 2 m/s
v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i + ((2m₂) / (m₁ + m₂)) v₂i
v₁f = ((m – 6m) / (m + 6m)) * (4) + ((2*6m) / (m + 6m)) * (2)
v₁f = 0.5714 m/s (→)
v₂f = ((2m₁) / (m₁ + m₂)) v₁i + ((m₂ – m₁) / (m₁ + m₂)) v₂i
v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)
v₂f = 2.5714 m/s (→)
e = - (v₁f - v₂f) / (v₁i - v₂i) ⇒ e = - (0.5714 - 2.5714) / (4 - 2) = 1
It was a perfectly elastic collision.
If a man pushes on a wall with some force then according to Newton's third law, wall will also apply force on man with same magnitude but opposite in direction.
Answer:
The size of the force that pushes the wall is 12,250 N.
Explanation:
Given;
mass of the wrecking ball, m = 1500 kg
speed of the wrecking ball, v = 3.5 m/s
distance the ball moved the wall, d = 75 cm = 0.75 m
Apply the principle of work-energy theorem;
Kinetic energy of the wrecking ball = work done by the ball on the wall
¹/₂mv² = F x d
where;
F is the size of the force that pushes the wall
¹/₂mv² = F x d
¹/₂ x 1500 x 3.5² = F x 0.75
9187.5 = 0.75F
F = 9187.5 / 0.75
F = 12,250 N
Therefore, the size of the force that pushes the wall is 12,250 N.
