Answer:
The bomb will remain in air for <u>17.5 s</u> before hitting the ground.
Explanation:
Given:
Initial vertical height is, 
Initial horizontal velocity is, 
Initial vertical velocity is, 
Let the time taken by the bomb to reach the ground be 't'.
So, consider the equation of motion of the bomb in the vertical direction.
The displacement of the bomb vertically is 
Acceleration in the vertical direction is due to gravity, 
Therefore, the displacement of the bomb is given as:

So, the bomb will remain in air for 17.5 s before hitting the ground.
Answer:
The terminal speed of this object is 12.6 m/s
Explanation:
It is given that,
Mass of the object, m = 80 kg
The magnitude of drag force is,

The terminal speed of an object is attained when the gravitational force is balanced by the gravitational force.



On solving the above quadratic equation, we get two values of v as :
v = 12.58 m/s
v = -15.58 m/s (not possible)
So, the terminal speed of this object is 12.6 m/s. Hence, this is the required solution.
Answer:
no becaus force is mass multiplied by acceleration. the mass of the elephant does not change
Answer: -2.5
Explanation:
1/2(-5)= -2.5
-2.5(1)= -2.5
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The crate only moves horizontally, so its net vertical force is 0. The only forces acting in the vertical direction are the crate's weight (pointing downward) and the normal force of the surface on the crate (pointing upward). By Newton's second law, we have
∑ <em>F</em> (vertical) = <em>n</em> - <em>mg</em> = 0 → <em>n</em> = <em>mg</em> = 1876 N
where <em>n</em> is the magnitude of the normal force.
In the horizontal direction, the crate is moving at a constant speed and thus with no acceleration, so it's completely in equilibrium and the net horizontal force is also 0. The only forces acting on it in this direction are the 747 N push (pointing in the direction of the crate's motion) and the kinetic friction opposing it (pointing in the opposite direction). By Newton's second law,
∑ <em>F</em> (horizontal) = 747 N - <em>f</em> = 0 → <em>f</em> = 747 N
The frictional force is proportional to the normal force by a factor of the coefficient of kinetic friction, <em>µ</em>, such that
<em>f</em> = <em>µn</em> → <em>µ</em> = <em>f</em> / <em>n</em> = (747 N) / (1876 N) ≈ 0.398188 ≈ 0.40