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2 years ago

The electric potential a distance r from a small charge is proportional to what power of the distance from the charge?

Physics

1 answer:

gizmo_the_mogwai [7]2 years ago

3 0

Answer: r^-1

Explanation:

V = (k × Q) / r, where V is electric potential, k is constant and r is distance from charge.

Since Q will not change, it can be considered a constant too.

So, we can write V ∝ 1/r

which is same as V ∝ r^-1

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Hi! I have a word bank I need help with please! I have the questions as an attachment

Katena32 [7]

Scott needs to determine the density of a metallic rod. First, he should determine the mass of his sample on the laboratory balance. Second, he should measure the volume of his sample by water displacement. Finally, he can calculate the density by dividing mass/volume.
Hope this helped ;)

4 0

2 years ago

If the magnitude of the magnetic force on a proton is F when it is moving at 18.0 ∘ with respect to the field, what is the magni

Lelu [443]

Answer:

The force when θ = 33° is 1.7625 times of the force when θ = 18°

Explanation:

The force on a moving charge through a magnetic field is given by

F = qvB sin θ

q = charge of the moving particle

v = Velocity of the moving charge

B = Magnetic field strength

θ = angle between the magnetic field and the velocity (direction of the motion) of the moving charge

Because qvB are all constant, we can call the expression K.

F = K sinθ

when θ = 18°,

F = K sin 18° = 0.309K

when θ = 33°, let the force be F₁

F₁ = K sin 33° = 0.5446K

(F₁/F) = (0.5446K/0.309K) = 1.7625

F₁ = 1.7625 F

Hope this Helps!!!

5 0

2 years ago

A spring has a force constant k, and an object of mass m is suspended from it. The spring is cut in half and the same object is

kenny6666 [7]

Answer:

$f2/f1 = \sqrt{2}$

Explanation:

From frequency of oscillation

$f = 1/2pi *\sqrt{k/m}$

Initially with the suspended string, the above equation is correct for the relation, hence

$f1 = 1/2pi *\sqrt{k/m}$

where k is force constant and m is the mass

When the spring is cut into half, by physics, the force constant will be doubled as they are inversely proportional

$f2 = 1/2pi *\sqrt{2k/m}$

Employing f2/ f1, we have

$f2/f1 = \sqrt{2}$

3 0

2 years ago

Two taut strings of identical mass and length are stretched with their ends fixed, but the tension in one string is 1.10 times g

ollegr [7]

Answer:

The beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

Explanation:

Given;

velocity of wave on the string with lower tension, v₁ = 35.2 m/s

the fundamental frequency of the string, F₁ = 258 Hz

velocity of wave on the string with greater tension;

$v_1 = \sqrt{\frac{T_1}{\mu }$

where;

v₁ is the velocity of wave on the string with lower tension

T₁ is tension on the string

μ is mass per unit length

$v_1 = \sqrt{\frac{T_1}{\mu} } \\\\v_1^2 = \frac{T_1}{\mu} \\\\\mu = \frac{T_1}{v_1^2} \\\\ \frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}\\\\v_2^2 = \frac{T_2v_1^2}{T_1}$

Where;

T₁ lower tension

T₂ greater tension

v₁ velocity of wave in string with lower tension

v₂ velocity of wave in string with greater tension

From the given question;

T₂ = 1.1 T₁

$v_2^2 = \frac{T_2v_1^2}{T_1} \\\\v_2 = \sqrt{\frac{T_2v_1^2}{T_1}} \\\\v_2 = \sqrt{\frac{1.1T_1*(35.2)^2}{T_1}}\\\\v_2 = \sqrt{1.1(35.2)^2} = 36.92 \ m/s$

Fundamental frequency of wave on the string with greater tension;

$f = \frac{v}{2l} \\\\2l = \frac{v}{f} \\\\thus, \frac{v_1}{f_1} =\frac{v_2}{f_2} \\\\f_2 = \frac{f_1v_2}{v_1} \\\\f_2 =\frac{258*36.92}{35.2} \\\\f_2 = 270.6 \ Hz$

Beat frequency = F₂ - F₁

= 270.6 - 258

= 12.6 Hz

Therefore, the beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

6 0

2 years ago

if the mass of the objects stays the same and their distance from eachother decreases, then the force of gravity between the two

dedylja [7]

Then the force of gravity between them would be quadrupled and so on but the gravitational force is inversely proportional to the square of the separation distance between the two interacting objects which makes more separation distance will result in weaker gravitational forces

5 0

2 months ago