Force = (mass) x (acceleration)
= (0.025 kg) x (5 m/s²)
= 0.125 Newton
1.5 m/s is the velocity.
9.3 m is the length of aisle, over which Distance will be covered.
Time is demanded in which the child will move the cart over the aisle with 1.5 m/s.
v=S/t
and,
t=S/v
Put values,
t=9.3/1.5=6.2 s
M = mass of the first sphere = 10 kg
m = mass of the second sphere = 8 kg
V = initial velocity of the first sphere before collision = 10 m/s
v = initial velocity of the second sphere before collision = 0 m/s
V' = final velocity of the first sphere after collision = ?
v' = final velocity of the second sphere after collision = 4 m/s
using conservation of momentum
M V + m v = M V' + m v'
(10) (10) + (8) (0) = (10) V' + (8) (4)
100 = (10) V' + 32
(10) V' = 68
V' = 6.8 m/s
Answer:
The displacement of the volleyball is 2.62 m
Explanation:
Given;
initial velocity of the volleyball, u = 7.5 m/s
final velocity of the volleyball, v = 2.2 m/s
displacement of the volleyball, d = ?
Apply the following kinematic equation;
v² = u² - 2gd
2gd = u² - v²
Therefore, the displacement of the volleyball is 2.62 m
Consider a long train moving at speed v. Now consider a passenger throwing a ball inside this train, towards the back of the train, with same velocity v (but in the opposite direction of the train movement).
- A passenger inside the train will see the ball moving with speed v
- For an observer outside the train, however, the ball will appear as still. In fact, for him the ball will have a speed v (given by the movement of the train) -v (velocity of the ball but moving in the opposite direction), so the net velocity will be v+(-v)=0.
