Answer:
Momentum of block B after collision =
Explanation:
Given
Before collision:
Momentum of block A = =
Momentum of block B = =
After collision:
Momentum of block A = =
Applying law of conservation of momentum to find momentum of block B after collision .
Plugging in the given values and simplifying.
Adding 200 to both sides.
∴
Momentum of block B after collision =
Answer:
f = q
Explanation:
In the attachment we can see a diagram of the parallel rays.
The dotted line represents the normal to the mirror surface
These rays when reflected using the constructor equation
where p and q are the distance to the object and the image respectively.
Since the rays are parallel P = inf
1 / f = 1 / inf + 1 / q
f = q
this means that all the rays focus on one focal point.
Impulse = Change in momentum.
The ball was moving with a momentum of 0.45 * 22 = 9.9
The ball comes to rest in the receivers arm; this means the ball's final velocity = 0. So mv2 = 0.45 * 0
The magnitude of the impact is just the change in momentum. 9.9 - (0.45 * 0) = 9.9
About half of the people mostly MEN..?
Answer:
t = 0.657 s
Explanation:
given,
initial vertical velocity = 7.5 m/s
initial horizontal velocity = 0 m/s
angle = 49◦
using kinetic equation
final velocity in vertical direction
v sinθ = u_y - gt ........................(1)
final velocity in horizontal direction
v cosθ = u_x + a_x × t
here u_x = 0.0 m/s
v cosθ = a_x×t ......................(2)
Dividing equation (1) / (2)
solving for time t
u_y = initial velocity along x direction
acceleration along a_x = 1.4 m/s²
g = acceleration due to gravity = 9.8 m/s²
θ = 43° , u_y = 7.5 m/s
t = 0.657 s
time taken by the particle is t = 0.657 s