Using kinematic equation, v^2 - u^2 = 2as. 5^2 - 3^2 = 2a x 16. a = 0.5m/s^2. So particle will deaccelerate at 0.5m/s^2. ( v = final velocity, u= initial velocity, a= acceleration, s= displacement.)
We want to find how much momentum the dumbbell has at the moment it strikes the floor. Let's use this kinematics equation:
Vf² = Vi² + 2ad
Vf is the final velocity of the dumbbell, Vi is its initial velocity, a is its acceleration, and d is the height of its fall.
Given values:
Vi = 0m/s (dumbbell starts falling from rest)
a = 10m/s² (we'll treat downward motion as positive, this doesn't affect the result as long as we keep this in mind)
d = 80×10⁻²m
Plug in the values and solve for Vf:
Vf² = 2(10)(80×10⁻²)
Vf = ±4m/s
Reject the negative root.
Vf = 4m/s
The momentum of the dumbbell is given by:
p = mv
p is its momentum, m is its mass, and v is its velocity.
Given values:
m = 10kg
v = 4m/s (from previous calculation)
Plug in the values and solve for p:
p = 10(4)
p = 40kg×m/s
Answer:
The extension of a material or a spring is its increase in length when pulled. Hooke’s Law says that the extension of an elastic object is directly proportional to the force applied to it. In other words:
Explanation:
Answer:
F = 4399 KN
Explanation:
given,
mass of automobile = 890 kg
initial speed = 48 km/h
= 48 × 0.278 = 13.34 m/s
using equation of motion
v² = u² + 2 a × s
0 = 13.34² - 2 a ×0.018
a = 4943.21 m/s²
F = m × a
F = 890 × 4943.21
F = 4399456.9 N
F = 4399 KN
hence, the Net force is F = 4399 KN
