Answer:
v = 14 m/s
= 31.3 mph
The answer would be the same if the mass of the car were 2000 kg
Explanation:
Let V be the final velocity of the car after skidding, and v be the initial velocity of the car. Let a be the acceleration of the car and Δx be the distance the car travels after applying brakes (length of the skid marks). Let Fk be the force of friction between the tyres and the road. Let N be the normal force exerted on the car and μ be the co efficient of kinetic friction.
V^2 = v^2 + 2×a×Δx
Now V, the final velocity is zero as the car stops
0 = v^2 + 2×a×Δx
v^2 = -2×a×Δx
v =√-2×a×Δx .....*
Now applying Newton's Second Law
Fnet = m×a
-Fk = m×a
-μ×N = m×a
-μ×m×g = m×a (The mass cancels out)
a = -μ×g
Substituting the value of a back to equation *
v = √-2×(-μ×g)×Δx
v = √-2×(-0.5×9.8)×20
v = 14 m/s
Therefore the speed the car was travelling with v = 14 m/s
which is equal to 31.3 mph
Now if you were to change the mass of the car to 2000 kg the value for v would still be the same. As it is seen above mass cancels out so it does not influence or affect the value of the velocity obtained.
Hello, I see you are in a jam. Lemme help.
1.) True
2.) True
3.) True
4.) True
5.) True
LOL these are all true ;)
I am not sure about this one
Answer:
We know that for a pendulum of length L, the period (time for a complete swing) is defined as:
T = 2*pi*√(L/g)
where:
pi = 3.14
L = length of the pendulum
g = gravitational acceleration = 9.8 m/s^2
Now, we can think on the swing as a pendulum, where the child is the mass of the pendulum.
Then the period is independent of:
The mass of the child
The initial angle
Where the restriction of not swing to high is because this model works for small angles, and when the swing is to high the problem becomes more complex.
Answer:
12294.31 m/s
Explanation:
Momentum = (mass)(velocity)
Momentum before = Momentum after
(momentum of bullet)+(momentum of block)=(momentum of bullet and block)
0.035v+50(0)=(0.035+50)(8.6)
0.035v=430.301
v=12294.3142857m/s
