Answer:

Explanation:
Time can be found by dividing the distance by the speed.

The distance is 45 meters and the speed is 12.5 meters per second.


Divide. Note that the meters, or "m" will cancel each other out.


It will take the dolphin 3.6 seconds to swim a distance of 45 meters are 12.5 meters per second.
The temperature on the Earth would increase.
The temperature of the stratosphere would decrease.
Answer:
what's your question I can't understand
Answer with Explanation:
We are given that
Diameter of pipe,


Volume flow rate of the petroleum along the pipe=

By equation of continuity







1 m=100 cm
Sure !
Start with Newton's second law of motion:
Net Force = (mass) x (acceleration) .
This formula is so useful, and so easy, that you really
should memorize it.
Now, watch:
The mass of the box is 5.25 kilograms, and the box is
accelerating at the rate of 2.5 m/s² .
What's the net force on the box ?
Net Force = (mass) x (acceleration)
= (5.25 kilograms) x (2.5 m/s²)
Net force = 13.125 newtons .
But hold up, hee haw, whoa ! Wait a second !
Bella is pushing with a force of 15.75 newtons, but the box
is accelerating as if the force on it is only 13.125 newtons.
What happened to the rest of Bella's force ? ?
==> Friction is pushing the box in the opposite direction,
and cancelling some of Bella's force.
How much ?
(Bella's 15.75 newtons) minus (13.125 that the box feels)
= 2.625 newtons backwards, applied by friction.