Answer:
yes; yes
Explanation:
Phases of the moon refers to the shapes of the moon due to the lit part of it visible from the Earth. On a new moon day, the moon comes between the sun and the earth such that the lit portion is not visible from the Earth. On a full moon day, the earth comes between the sun and the moon and the whole lit part is visible.
When one would view the earth from the moon, the earth would also be visible as going through the phases. The order would be reversed. Understand this with the following example, On a new moon day, the Earth would be visible completely lit from the moon. So it will be full Earth day on the moon. On a full moon day, the lit side of the Earth would be completely away and hence, from the moon, new earth would be there.
Sylvester James Gates' biggest discovery was the string theory which involves computer coding. I hope this answer helped.
As we know that here gun + bullet system is isolated from all other external force system
so here the momentum of system is always conserved
so we will say

here we know that




now we will have

by solving above equation we will have

Answer:
497.6 N
Explanation:
From the question,
The net force on the skydiver = weight of the skydiver- the resistive force of air
F' = W-F...................... Equation 1
Where W = weight of the skydiver, F = resistive force of air.
But,
W = mg................ Equation 2
Where m = mass of the skydiver, g = acceleration due to gravity.
Substitute equation 2 into equation 1
F' = mg-F............ Equation 3
Given: m = 87 kg, F = 355 N, g = 9.8 m/s²
Substitute these values into equation 3
F' = 87(9.8)-355
F' = 852.6-355
F' = 487.6 N
Answer:
v₁
Explanation:
A crew member is walking on a tugboat that is pulling a barge. The tugboat is moving at a constant speed upstream in a river that has a constant downstream current. The velocity of the crew member with respect to the tugboat is v₁. What is the velocity of the crew member with respect to the barge?
velocity is the change in displacement per time.
the velocity of the crew member with respect to the barge is v₁ since the barge is moving in the velocity of tugboat