Answer:
Force(F) = -80,955.01 N
Explanation:
We need to first determine the impulse that the truck driver received from the car during the collision
So; m₁v₁ - m₂v₂ = (m₁m₂)v₀
where;
m₁ = mass of the truck = 4280 kg
v₁ = v₂ = speed of the each vehicle = 7.69 m/s
m₂ = mass of the car = 810 kg
Substituting our data; we have:
(4280×7.69) - (810×7.69) = (4280+810)v₀
32913.2 - 6228.9 = (5090)v₀
26684.1 = (5090)v₀
v₀ = 
v₀ = 5.25 m/s
NOW, Impulse on the truck = m (v₀ - v)
= 4280 × (5.25 - 7.69)
= 4280 × (-2.44)
= -10,443.2 kg. m/s
Force that the seat belt exert on the truck driver can be calculated as:
Impulse = Force × Time
-10,443.2 kg. m/s = F (0.129)
F = 
Force(F) = -80,955.01 N
Thus, the Force that the seat belt exert on the truck driver = -80,955.01 N
Answer:
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If an automobile moving at high speed suddenly comes to a stop, you would have a large change in momentum. This relates to Newton's second law in the form F = delta p / delta t, where p is momentum (mv).
You could lessen the effect of the sudden stop on the passengers by changing the average force exerted on them. If you look at Newton's second law again, you can see that given some delta p, you can decrease F by increasing delta t. What this means is that if you increase the length of time over which the change in momentum occurs, you can decrease the average force exerted to obtain that change in momentum. This is the reason why landing on a soft cushion is preferable to landing on a concrete surface. The cushion gives way to any object falling on it while still providing some resistance (you don't stop as abruptly), so while your change in momentum is the same in both cases, you have a larger delta t in the case of the cushion.
Answer:
The shortest distance is 
Explanation:
The free body diagram of this question is shown on the first uploaded image
From the question we are told that
The speed of the bicycle is 
The distance between the axial is 
The mass center of the cyclist and the bicycle is 
behind the front axle
The mass center of the cyclist and the bicycle is 
above the ground
For the bicycle not to be thrown over the
Momentum about the back wheel must be zero so
=> 
=> 
Here 
So 
Apply the equation of motion to this motion we have
Where 
and 
since the bicycle is coming to a stop
=>
Hi there!
On a level road:
∑F = Ff (Force due to friction)
The net force is the centripetal force, so:
mv²/r = Ff
Rewrite the force due to friction:
mv²/r = μmg
Cancel out the mass:
v²/r = μg
Solve for v:
v = √rμg
v = √(25)(9.81)(0.8) = 14.01 m/s
