Answer:
change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum
Explanation:
Let's look for the speed of the car
F = m a
a = F / m
We use kinematics to find lips
v = v₀ + a t
v = v₀ + (F / m) t
The moment is defined by
p = m v
The moment change
Δp = m v - m v₀
Let's replace the speeds in this equation
Δp = m (v₀
+ F / m t) - m v₀
Δp = m v₀ + F t - m v₀
Δp = F t
We see that the change of momentum does not depend on the mass of the cars, as the force and time are the same all vehicles have the same change of momentum
Answer: 37.5 g
Explanation: you multiply 0.0375 by 1000 which equals 37.5
Answer:
the magnitude of the average contact force exerted on the leg is 3466.98 N
Explanation:
Given the data in the question;
Initial velocity of hand v₀ = 5.25 m/s
final velocity of hand v = 0 m/s
time interval t = 2.65 ms = 0.00265 s
mass of hand m = 1.75 kg
We calculate force on the hand F
using equation for impulse in momentum
F
× t = m( v - v₀ )
we substitute
F
× 0.00265 = 1.75( 0 - 5.25 )
F
× 0.00265 = 1.75( - 5.25 )
F
× 0.00265 = -9.1875
F
= -9.1875 / 0.00265
F
= -3466.98 N
Next we determine force on the leg F
Using Newton's third law of motion
for every action, there is an equal opposite reaction;
so, F
= - F
we substitute
F
= - ( -3466.98 N )
F
= 3466.98 N
Therefore, the magnitude of the average contact force exerted on the leg is 3466.98 N
Answer:
k = 
b = 
t = 
Solution:
As per the question:
Mass of the block, m = 1000 kg
Height, h = 10 m
Equilibrium position, x = 0.2 m
Now,
The velocity when the mass falls from a height of 10 m is given by the third eqn of motion:

where
u = initial velocity = 0
g = 10
Thus

Force on the mass is given by:
F = mg = 
Also, we know that the spring force is given by:
F = - kx
Thus

Now, to find the damping constant b, we know that:
F = - bv

Now,
Time required for the platform to get settled to 1 mm or 0.001 m is given by:
