Answer:(a) With our choice for the zero level for potential energy of the car-Earth system when the car is at point B ,
U
B
=0
When the car is at point A, the potential energy of the car-Earth system is given by
U
A
=mgy
where y is the vertical height above zero level. With 135ft=41.1m, this height is found as:
y=(41.1m)sin40.0
0
=26.4m
Thus,
U
A
=(1000kg)(9.80m/s
2
)(26.4m)=2.59∗10
5
J
The change in potential energy of the car-Earth system as the car moves from A to B is
U
B
−U
A
=0−2.59∗10
5
J=−2.59∗10
5
J
(b) With our choice of the zero configuration for the potential energy of the car-Earth system when the car is at point A, we have U
A
=0. The potential energy of the system when the car is at point B is given by U
B
=mgy, where y is the vertical distance of point B below point A. In part (a), we found the magnitude of this distance to be 26.5m. Because this distance is now below the zero reference level, it is a negative number.
Thus,
The boat's initial velocity is:
While the boat's acceleration is
with a negative sign, since the boat is slowing down, so it is a deceleration. The distance traveled by the boat until it comes to a stop can be found by using the following equation:
where vf=0 is the final velocity of the boat and S is the distance covered. Re-arranging the formula, we can find S:
This shows the sunlight shining on the moon is creating a shadow. The part of the shadow that we don’t see from Earth is called the eclipse. Since the Earth gets a shadow from the moon, that causes the rest of the Earth to light up which causes a lunar eclipse.
Explanation:
10 is the number that is subtracted from it is the MINUEND
4 is the number being subtracted its then the SUBTRAHEND
6 is the result it is then the DIFFERENCE
so the correct term for 4 is SUBTRAHEND
