The crumple zones in the second car will improve the chance of survival of the driver because it will act as shock absorber, reducing the impact of the force on the driver.
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Newton's third law of motion</h3>
According to Newton's third law of motion, action and reaction are equal and opposite.
The car with rigid body will exert maximum force to the driver while the car with crumple zone will exert lesser force to the driver since the crumple zone will act as shock absorber, reducing the impact of the force on the driver.
Thus, the crumple zones in the second car will improve the chance of survival of the driver because it will act as shock absorber, reducing the impact of the force on the driver.
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Answer:
here given is a weight
then force becomes mg
that is F=Mg
=4*9.8
then by using the formula
F=Ma
a=F/M
=4*9.8/9.8
=4
Explanation:
You have to solve this by using the equations of motion:
u=3
v=0
s=2.5
a=?
v^2=u^2+2as
0=9+5s
Giving a=-1.8m/s^2
Then using the equation:
F=ma
F is the frictional force as there is no other force acting and its negative as its in the opposite direction to the direction of motion.
-F=25(-1.8)
F=45N
Then use the formula:
F=uR
Where u is the coefficient of friction, R is the normal force and F is the frictional force.
45=u(25g)
45=u(25*10)
Therefore, the coefficient of friction is 0.18
Hope that helps
Actually Welcome to the Concept of the Projectile Motion.
Since, here given that, vertical velocity= 50m/s
we know that u*sin(theta) = vertical velocity
so the time taken to reach the maximum height or the time of Ascent is equal to
T = Usin(theta) ÷ g, here g = 9.8 m/s^2
so we get as,
T = 50/9.8
T = 5.10 seconds
thus the time taken to reach max height is 5.10 seconds.
Solution:
initial sphere mvr = final sphere mvr + Iω
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m²
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω
where: ω = 2.87 rad/s
So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)²
E = 12.64 J becomes PE = mgh, so
12.64 J = 2.3 kg * 9.8m/s² * h
h = 0.29 m
h = L(1 - cosΘ) → where here L is the distance to the CM
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ
Θ = arccos((1-0.29)/1) = 44.77 º
