A.) LUNAR MARIA..............
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,
Answer:
(a) 98.548 mPa
(b) N = 942346
9.42 X 10∧5 cycles
Explanation:
The solved solution is in the attach document.
Answer:
Explanation:
= Permittivity of free space = 
A = Area = 
d = Thickness = 
k = Dielectric constant = 5.4
V = Voltage = 86.2 mV
Charge is given by
The charge on the outer surface is
Answer:
The electrical loads in parallel circuits each have the same voltage drop, with equals the total applied voltage of the circuit.
Explanation:
I did some research and the voltage drop across any branch of a parallel circuit is the same as the applied voltage.
