Answer:
FORCE - rate of change of momentum, ie its changing velocity [change in velocity is of more concern] NEWTON
WORK - product of force and displacement, ie [velocity may be constant or variable but change in position with certain force is of more concern] JOULES
I hope you understood from this..
To solve this problem we will begin by finding the necessary and effective distances that act as components of the centripetal and gravity Forces. Later using the same relationships we will find the speed of the body. The second part of the problem will use the equations previously found to find the tension.
PART A) We will begin by finding the two net distances.
And the distance 'd' is
Through the free-body diagram the tension components are given by
Here we can watch that,
Dividing both expression we have that,
Replacing the values,
PART B) Using the vertical component we can find the tension,
Answer:
44100 N
Explanation:
Each wall will have dimension of 4 m x 1.5 m
Whole force will act on central point of wall situated at a depth of 1.5 /2 = .75m
pressure at CM = h d g , h = .75 , d ( density of water = 10³ )
pressure at CM = .75 x 10³ x 9.8
= 7350 N / m²
Total force on each wall
= pressure x area
= 7350 x 4 x 1.5
= 44100 N Ans
b ) If h = 1.5 x 2 = 3
Pressure = hdg
1.5 x 10³ x 9.8
= 14700 N / m²
Force
= pressure x area
14700 x 3 x 4
= 176400 N
Which is 4 times 44100 N
So force will quadruple.
It is so because both area and height have become twice.
A. The English system uses one unit for each category of measurement.
The electric potential at point A in the electric field= 0.099 x 10 ⁻¹v
<u>Explanation</u>:
Given data,
charge = 5.5 x 10¹² C
k =9.00 x 10⁹
The electric potential V of a point charge can found by,
V= kQ / r
Assuming, r=5.00×10⁻² m
V= 5.5 x 10⁻¹²C x 9.00 x 10⁹ / 5.00×10⁻² m
V= 49.5 x 10⁻³/ 5.00×10⁻²
Electric potential V= 0.099 x 10⁻¹v
