Answer:
2081.65 m
Explanation:
We'll begin by calculating the time taken for the load to get to the target. This can be obtained as follow:
Height (h) = 3000 m
Acceleration due to gravity (g) = 10 m/s²
Time (t) =?
h = ½gt²
3000 = ½ × 10 × t²
3000 = 5 × t²
Divide both side by 5
t² = 3000 / 5
t² = 600
Take the square root of both side
t = √600
t = 24.49 s
Finally, we shall determine the distance from the target at which the load should be released. This can be obtained as follow:
Horizontal velocity (u) = 85 m/s
Time (t) = 24.49 s
Horizontal distance (s) =?
s = ut
s = 85 × 24.49
s = 2081.65 m
Thus, the load should be released from 2081.65 m.
Answer:
v = 60 m/s
Explanation:
It is given that,
A wave is represented by the equation :
We need to find the velocity of the wave
The general equation of a wave is given by :
....(1)
Equation (1) can be written as :
...(2)
If we compare equation (1) and (2) we get :
The velocity of a wave is given by :
So, the velocity of the wave is 60 m/s.
Answer:
By increasing the perpendicular distance between the force and the pivot point
Explanation:
Due to the formula ,
An increase in distance would cause the turning force to increase
<em>Feel free to mark it as brainliest :D</em>
Answer:
The magnitude is 63 kg m²/s, and the direction is -k.
Explanation:
Plug the values into the equation:
L = m v×r
L = (3.0 kg) (<5 i + 3j> m/s × <2i − 3j> m)
Take the cross product. The cross product of two dimensional vectors is:
<v₁ i + v₂ j> × <r₁ i + r₂ j> = <(v₁ r₂ − v₂ r₁) k>
Therefore:
L = (3.0 kg) <((5)(-3) − (3)(2)) k m²/s>
L = (3.0 kg) <-21 k m²/s>
Multiply:
L = <-63 k kg m²/s>
The magnitude is 63 kg m²/s, and the direction is -k.
The force of gravity starts to decrease since your traveling away from the atmosphere.