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2 years ago

8

For a point charge, how does the potential vary with distance from the point charge, r?

1 answer:

mr_godi [17]2 years ago

3 0

For a point charge, how does the potential vary with distance from the point charge, r?

a constant

b. r.

c. 1/r.

d. $1/r^2$ .

e. $r^2$ .

Answer:

The correct option is C

Explanation:

Generally for a point charge the electric potential is mathematically represented as

$V = \frac{k Q }{r }$

Here we can deduce that the electric potential varies inversely with the distance i.e

$V \ \alpha \ \frac{1}{r}$

So

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Klio2033 [76]

Answer:

Based off the word "conserved" I would say

A. Conservation of Momentum.

Explanation:

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2 years ago

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jonny [76]

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[Your explanation is correct]

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3 years ago

Please choose the answer that describes the scientific notation 5098000

bonufazy [111]

So there is a decimal after the last zero and it looks like this 5098000. You have to move the decimal point six back to get in between the five and the zero which looks like this 5.098000

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3 years ago

A powerful motorcycle can produce an acceleration of 3.00 m/s2 while traveling at 106.0 km/h. At that speed, the forces resistin

otez555 [7]

Answer:

"1155 N" is the appropriate solution.

Explanation:

Given:

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Forces resisting motion,

$F_f=432 \ N$

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By using Newton's second law, we get

⇒ $F-F_f=ma$

Or,

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By putting the values, we get

⇒ $=(3\times 241)+432$

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7 0

2 years ago

A(n) 12500 lb railroad car traveling at 7.8 ft/s couples with a stationary car of 7430 lb. The acceleration of gravity is 32 ft/

navik [9.2K]

To solve this problem we will apply the concepts related to the conservation of momentum. That is, the final momentum must be the same final momentum. And in each state, the momentum will be the sum of the product between the mass and the velocity of each object, then

$\text{Initial Momentum} = \text{Final Momentum}$

$m_1u_1 +m_2u_2 = m_1v_1+m_2v_2$

Here,

$m_{1,2}$ = Mass of each object

$u_{1,2}$ = Initial velocity of each object

$v_{1,2}$ = Final velocity of each object

When they position the final velocities of the bodies it is the same and the car is stationary then,

$m_2u_2 = (m_1+m_2)v_f$

Rearranging to find the final velocity

$v_f = \frac{m_2u_2}{ (m_1+m_2)}$

$v_f = \frac{ 12500*7.8}{ 12500+7430}$

$v_f = 4.8921ft/s$

The expression for the impulse received by the first car is

$I = m_1 (v-u)$

$I = \frac{W}{g} (v-u)$

Replacing,

$I = \frac{12500}{32.2}(4.89-7.8)$

$I = -1129.65lb\cdot s$

The negative sign show the opposite direction.

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2 years ago