initial velocity of the car given as

final velocity is given as

as we know that

now we can convert final speed into m/s

now acceleration is rate of change in velocity



so the acceleration of the car is 3 m/s^2
Objects dropped straight or thrown horizontally from the same height
change their vertical velocity at the same rate, and fall through equal
vertical distances in equal time intervals.
The statement is false.
Answer:

Explanation:
As we know by radioactivity law

so here we will have


now we will have


now we also know that



According to the law of conservation of momentum:

m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision
Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know.
m1 = 125 kg v1 = 12m/s v'1 = -12.5m/s
m2 = 235kg v2 = -13m/s v'2 = ?




Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation.




The velocity of the 2nd car after the collision is
0.03m/s.
Answer:
α(0) = 0 rad/s²
α(5) = 15 rad/s²
Explanation:
The angular velocity of the flywheel is given as follows:
w(t) = A + B t²
where, A and B are constants.
Now, for the angular acceleration, we must take derivative of angular velocity with respect to time:
Angular Acceleration = α (t) = dw/dt
α(t) = (d/dt)(A + B t²)
α(t) = 2 B t
where,
B = 1.5
<u>AT t = 0 s</u>
α(0) = 2(1.5)(0)
<u>α(0) = 0 rad/s²</u>
<u></u>
<u>AT t = 5 s</u>
α(5) = 2(1.5)(5)
<u>α(5) = 15 rad/s²</u>