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
That's 1/2 of what it is on the surface.
The distance between the center of the Earth and any object
on the surface is 1 Earth radius ... about 3960 miles.
Gravitational force is inversely proportional to the square of
the distance between the centers of the objects, so in order
to reduce the acceleration of gravity by 1/2, you increase the
distance by √2 .
(3960 miles) x (√2) = 5,600 miles from the center
= 1,640 miles above the surface.
Answer:
force and leverage distance
Explanation:
the formula for torque if = force x distance
(the distance above is the leverage distance on the crow bar)
therefore if there is an increase in either the torque or the leverage distance, or both, the torque exerted by the crow bar also increases.
for example
- lets assume a force of 5 n is applied on the crow bar with a leverage distance of 2 m.
the torque = 5 x 2 = 10 N.m
- but if the force was increased to 7 N
torque = 7 x 2 = 12 N.m
from the illustration above, we can see that the torque increased with an increase in force. There would also be an increase in torque if the distance were to be increased.
Answer:
the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is 
Explanation:
The free-body diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;
- the peripheral velocity that is directed downward
along the y-axis
- the linear velocity
that is directed along the x-axis
Now;
Also,
where 
(angular velocity) = 
∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is
