Explanation:
Momentum is conserved.
a) In the first scenario, Olaf and the ball have the same final velocity.
mu = (M + m) v
(0.400 kg) (10.9 m/s) = (70.2 kg + 0.400 kg) v
v = 0.0618 m/s
b) In the second scenario, the ball has a final velocity of 8.10 m/s in the opposite direction.
mu = mv + MV
(0.400 kg) (10.9 m/s) = (0.400 kg) (-8.10 m/s) + (70.2 kg) v
v = 0.108 m/s
Answer:
suvat is an acronym of five variables that describe a system in motion. a, s, u, v and t
Answer:
t = 23.255 s, x = 2298.98 m, v_y = - 227.90 m / s
Explanation:
After reading your extensive writing, we are going to solve the approach.
The initial speed of the plane is 250 miles / h and it is at an altitude of 2650 m; In general, planes fly horizontally for launch, therefore this is the initial horizontal speed.
As there is a mixture of units in different systems we are going to reduce everything to the SI system.
v₀ₓ = 250 miles h (1609.34 m / 1 mile) (1 h / 3600 s) = 111.76 m / s
y₀ = 2650 m
Let's set a reference system with the x-axis parallel to the ground, the y-axis is vertical. As time is a scalar it is the same for vertical and horizontal movement
Y axis
y = y₀ + v₀ t - ½ g t²
the initial vertical velocity when the cargo is dropped is zero and when it reaches the floor the height is zero
0 = y₀ + 0 - ½ g t²
t =
t = √(2 2650/ 9.8)
t = 23.255 s
Therefore, for the cargo to reach the desired point, it must be launched from a distance of
x = v₀ₓ t
x = 111.76 23.255
x = 2298.98 m
at the point and arrival the speed is
vₓ = v₀ₓ = 111.76
vertical speed is
v_y = v_{oy} - gt
v_y = 0 - gt
v_y = - 9.8 23.25 555
v_y = - 227.90 m / s
the negative sign indicates that the speed is down
in the attachment we have a diagram of the movement
Answer:231.16 N/C
Explanation:
Given
Electric Flux
Area(A)
Given Field point above 
Therefore angle between Area vector Electric Field =90-31.6=
We know that Flux is given by



E=231.16 N/C