Answer:
u₂ = 32.29 m/s
Explanation:
m₁ = 12 kg
u₁ = 0 m/s
m₂ = 417 g = 0.417 kg
d = 15 cm = 0.15 m
μ = 0.40
u₂ = ?
It is an inelastic collision. After the collision
We can apply ∑ F = m*a for the system (m₁ + m₂)
where m = m₁ + m₂ = 12 kg + 0.417 kg = 12.417 kg
then
∑ F = - Ff = m*a
if
Ff = μ*N = μ*W = μ*(m*g) = μ*m*g
⇒ - μ*m*g = m*a ⇒ a = - μ*g = - 0.40*9.8 m/s² = - 3.92 m/s²
⇒ a = - 3.92 m/s²
Since vf = 0 m/s (for the system)
we use the equation
vf² = vi²+2*a*d ⇒ 0 = vi²+2*a*d ⇒ vi = √(-2*a*d)
⇒ vi = √(-2*(- 3.92 m/s²)*0.15 m)
⇒ vi = 1.0844 m/s
we can use the equation for an inelastic collision
m₁*u₁ + m₂*u₂ = (m₁ + m₂)*vs
since m = m₁ + m₂; u₁ = 0 m/s and vs = vi
we have
m₁*0 + m₂*u₂ = m*vi
⇒ u₂ = m*vi / m₂
⇒ u₂ = 12.417 kg*1.0844 m/s / 0.417 kg
⇒ u₂ = 32.29 m/s (→)
Answer:
Explanation:
If you are asking for how much a cubic meter of iron would weight, you need to find its density.
The density of iron taken from the periodic table is , therefore a single cubic meter would weight 7870 kg, which written in scientific notation would be
The Doppler effect, is one effect on light which shows that it's source is moving
Answer:
1120 N
Explanation:
The velocity with which he hits the water can be found with kinematics:
v² = v₀² + 2aΔy
v² = (0 m/s)² + 2 (-9.8 m/s²) (-9.00 m)
v = -13.3 m/s
Or it can be found with conservation of energy.
PE = KE
mgh = ½ mv²
v = √(2gh)
v = √(2 × -9.8 m/s² × -9.00 m)
v = -13.3 m/s
Sum of forces on the diver after he hits the water:
∑F = ma
F − mg = m Δv/Δt
F − (74.0 kg) (9.8 m/s²) = (74.0 kg) (0 m/s − (-13.3 m/s)) / (2.50 s)
F = 1120 N
Answer:
<em>The force applied to the car was 30,600 N</em>
Explanation:
According to the second Newton's law, the net force applied to an object of mass m is:
Where a is the acceleration at which the object moves. The net force can be also calculated as the sum of all forces acting on the body.
We have a car of m=2000 Kg, being accelerated at 5.5 m/s^2 by a force F (unknown) directed upwards.
Considering the force is upwards and the weight of the car (W) is directed downwards, the net force is:
Being W=m.g
Equating [1] and [2]:
Adding W:
Substituting:
F=30,600 N
The force applied to the car was 30,600 N