Answer:opposite
Explanation:for a capacitor to discharge (after charging) the polarities of the current and voltage have to be reversed
Let the data is as following
mass of payload = "m"
mass of Moon = "M"
now we know that we place the payload from the position on the surface of moon to the position of 5r from the surface
So in this case we can say that change in the gravitational potential energy is equal to the work done to move the mass from one position to other
so it is given by

we know that


now from above formula


so above is the work done to move the mass from surface to given altitude
Answer:
Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.
Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.
Explanation:
Part a
When Road is Level
The stopping sight distance is given as

Here
- SSD is the stopping sight distance which is to be calculated.
- u is the speed which is given as 60 mi/hr
- t is the perception-reaction time given as 2.5 sec.
- a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
- G is the grade of the road, which is this case is 0 as the road is level
Substituting values

So the minimum stopping sight distance is 563.36 ft.
Part b
When Road has a maximum grade of 4%
The stopping sight distance is given as

Here
- SSD is the stopping sight distance which is to be calculated.
- u is the speed which is given as 60 mi/hr
- t is the perception-reaction time given as 2.5 sec.
- a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
- G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade
For upgrade of 4%, Substituting values

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>
For downgrade of 4%, Substituting values

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>
As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.
The gap between the two flasks is partially evacuated of air creating a near vacuum which significantly reduces heat transfer by conduction or convection
Answer:
the magnitude of the charge Q on each plate is 
Explanation:
Given that :
mass (m) = 
charge (q) = +0.155 µC = 
angle 
Area A on each plate = 0.0135 m²
From the diagram below;
----- equation (1)
Also by using Gauss Law ;

----- equation (2)
Combination equation 1 and 2 together ; we have


