Using Newton's second law of motion:
F=ma ; [ F = force (N: kgm/s^2);m= mass (kg); a = acceleration (m/s^2)
Given: Find: Formula: Solve for m:
F: 2500N mass:? F=ma Eq.1 m=F/a Eq. 2
a= 200m/s^2
Solution:
Using Eq.2
m= (2500 kgm/s^2)/ (200m/s^2) = 12.5 kg
Answer: Velocity terminal = 0.093m/s
Explanation:
1. We start by evaluating the gap distance between the two cylinders as h = R(sleeve) - R(cylinder)
= (0.0604/2 - 0.06/2)m
= 2×10^-4
Surface are of the cylinder in the drop, which is required in order to evaluate the shearing stress can be expressed as A(cylinder) = π.d.L
= (π×0.06×0.4)m²
= 0.075m²
Since the force of the cylinder's weight is going to balance the shearing force on the walls, we can express the next equation and derive terminal velocity from it.
Shearing stress = u×V.terminal/h = 0.86×V/0.0002
= 4300Vterminal
Therefore, Fw = shearing stress × A
30N = 4300Vterminal × 0.075
V. terminal = 30/4300 m.s
V. terminal = 0.093m/s
Explanation:
Given that,
Mass of the car, m₁ = 1250 kg
Initial speed of the car, u₁ = 7.39 m/s
Mass of the truck, m₂ = 5380 kg
It is stationary, u₂ = 0
Final speed of the truck, v₂ = 2.3 m/s
Let v₁ is the final velocity of the car. Using the conservation of momentum as :



So, the final velocity of the car is 2.5 m/s but in opposite direction. Hence, this is the required solution.
I believe the answer would be mass. Low mass stars and medium mass stars often become white dwarfs when they die while high mass stars explode in violent explosions called supernovas and usually leave behind a black hole or a neutron star.
This would be typical of an elastic collision.