Near the shore, winds tend to blow out toward the water at night. what drives this process?

a. warm air above the water rises and is replaced by cooler air from the land.
b. warm air above the land conducts heat to the cooler air above the water.
c. cool air above the land conducts heat to the warmer air above the water.
d. cool air above the water rises and is replaced by warmer air from the land.

Winds at the shore process - Option C

Cool air above the land conducts heat to the warmer air above the water drives this process. Near the shore, winds tend to blow out toward the water at night. It is understood that at night air is cool and due to sun air is warm in the day.

Therefore, cool air on the land close to the coast, this air leds heat to the hot air which is present on the water (H2O). This indicates that air on land is cool and air on water is warm. Hence, it only conducts heat.

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j

the speed of the cart after 3 seconds of low fan speed is ⇒ 54.0 cm/s.

the speed of the cart after 5 seconds of medium fan speed is ⇒ 120.0 cm/s.

the speed of the cart after 2 seconds of high fan speed is ⇒ 64.0 cm/s. 

just did the lab.