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Gnesinka [82]
3 years ago
15

A point charge q1 is held stationary at the origin. A second charge q2 is placed at point a, and the electric potential energy o

f the pair of charges is 5.4 x 10^-8J. When the second charge is moved to point b, the electric force on the charge does -1.9x10^-8J of work. What is the electric potential energy of the pair of charges when the second charge is at point b?
Physics
1 answer:
MAVERICK [17]3 years ago
7 0

Answer:

U_{b}=+7.3*10^{-8}J

Explanation:

When a particle moves from a point where the potential energy is Ua to point where it is Ub,the change in potential energy is is equal to work done

So

W_{a-b}=U_{a}-U_{b}\\ U_{b}=U_{a}-W_{a-b}\\

Where Wa-b here is negative this means Ub is greater Ua. Therefore potential energy increases

So

U_{b}=(+5.4*10^{-8}J )-(-1.9*10^{-8}J)\\U_{b}=+7.3*10^{-8}J  

You might be interested in
An object’s motion remains constant when acted upon by what?
igomit [66]

Answer:

An outside force

Explanation:

Newton's law an object in motion stays in motion an object at rest stays at rest unless acted on by an outside force.

6 0
2 years ago
You are sitting on a deck of your house surrounded by oak trees. You hear the sound of an acorn hitting the deck. You wonder if
Black_prince [1.1K]

Answer: 96N

Explanation:

To calculate the velocity of the impact On the persons head, we have

h = gt²/2

14 = 9.81t²/2

t² = 28/9.8

t² = 2.86

t = 1.69s

V = u + at

V = 0 + 9.81*1.69

V = 16.58m/s

a(average) = (v1² + v2²) /2Δy

a(average) = 16.58² + 0)/2 * 0.005

a(average) = 274.8964/0.01

a(average) = 27489.64m/s²

Using newton's second law of motion,

F(average) = m * a(average)

F(average) = 0.0035 * 27489.64

F(average) = 96.21N

Therefore the force needed by the acorn to do much damage starts from 96N

8 0
3 years ago
In the diagram, the trough of the wave is shown by:
Kisachek [45]

Answer:

the correct representation of the trough is b

3 0
3 years ago
Of the following, which is the best reason for why a woman's risk for cardiovascular disease increases after menopause?
MatroZZZ [7]

Answer:

The best reason for the increased risk of cardiovascular diseases in women after menopause is the significant decrease in the levels of Estrogen in the body.

Detailed answer:

  • Due to the approach of menopause the estrogen levels in the blood decrease drastically.
  • Estrogen helps in relaxing the vessels and keeping them in their natural diameter.
  • But due to a drastic decrease in its level due to menopause, there is a high chance of cholesterol building up on the walls of the artery in the heart, which can cause many cardiovascular diseases.

6 0
2 years ago
Determine the normal boiling point of a substance whose vapor pressure is 55.1 mm hg at 35°c and has a δhvap of 32.1 kj/mol.
Novosadov [1.4K]

Answer:

389.78681 K

Explanation:

P_1 = Initial pressure = 55.1 mmHg

P_2 = Final pressure = 1 atm = 760 mmHg

T_2 = Boiling point

T_1 = Initial temperature = 35°C

\Delta H_{vap} = Heat of vaporization = 32.1 kJ/mol

From the Clausius-Claperyon equation

ln\dfrac{P_2}{P_1}=(-\dfrac{\Delta H_{vap}}{R})(\dfrac{1}{T_2}-\dfrac{1}{T_1})\\\Rightarrow \dfrac{1}{T_2}=-ln\dfrac{P_2}{P_1}\dfrac{R}{\Delta H_{vap}}+\dfrac{1}{T_1}\\\Rightarrow \dfrac{1}{T_2}=-ln\dfrac{760}{55.1}\dfrac{8.314}{32.1\times 10^{3}}+\dfrac{1}{273.15+35}\\\Rightarrow T_2=\left(-ln\left(\frac{760}{55.1}\right)\frac{8.314}{32.1\times \:10^3}+\frac{1}{273.15+35}\right)^{-1}\\\Rightarrow T_2=389.78681\ K

The normal boiling point of the substance is 389.78681 K

3 0
3 years ago
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