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Sveta_85 [38]
3 years ago
11

? What does the number, or density, of field lines on a source charge indicate?

Physics
2 answers:
Oxana [17]3 years ago
6 0

A.  

direction of charge on the source charge

B.  

amount of charge on the test charge

C.  

amount of charge on the source charge

D.  

direction of charge on the test charge



Those are the anwser choices

Alenkinab [10]3 years ago
3 0
The density of field lines on a source charge indicates the strength of the charge there is. The field lines all represent the movements of field charges. Naturally, if there are a large number of them compressed in an area, then that would mean the charge at that certain point is strong 
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One charge is decreased to one-third of its original value, and a second charge is decreased to one-half of its original value.
Nadya [2.5K]
The original Coulomb force between the charges is:

Fc=(k*Q₁*Q₂)/r², where k is the Coulomb constant and k=9*10⁹ N m² C⁻², Q₁ is the first charge, Q₂ is the second charge and r is the distance between the charges.

The magnitude of the force is independent of the sign of the charge so I can simply say they are both positive. 

Q₁ is decreased to Q₁₁=(1/3)*Q₁=Q₁/3 and
Q₂ is decreased to Q₂₂=(1/2)*Q₂=Q₂/2. 

New force:

Fc₁=(k*Q₁₁*Q₂₂)r², now we input the decreased values of the charge

Fc₁=(k*{Q₁/3}*{Q₂/2})/r², that is equal to:

Fc₁=(k*(1/3)*(1/2)*Q₁*Q₂)/r²,

Fc₁=(k*(1/6)*Q₁*Q₂)/r²

Fc₁=(1/6)*(k*Q₁*Q₂)/r², and since the original force is: Fc=(k*Q₁*Q₂)/r² we get:

Fc₁=(1/6)*Fc

So the magnitude of the new force Fc₁ with decreased charges is 6 times smaller than the original force Fc. 
4 0
3 years ago
Read 2 more answers
An object that starts at a position of 5 m and travels for 3 seconds at a velocity of -9 m/s ends up at what position?
Rina8888 [55]

Answer:

Sorry don't know the answer

5 0
3 years ago
Calculate the kinetic energy of a 2kg ball moving at 5m/s
shepuryov [24]

Answer:

25

Explanation:

The kinetic energy is 25

8 0
3 years ago
O
lbvjy [14]

Hi there!

1.

The period of a pendulum can be calculated using the following equation:
\large\boxed{T = 2\pi \sqrt{\frac{L}{g}}}

T = period (s)

L = length of string (m)
g = acceleration due to gravity (m/s²)

Plug in the values:

T = 2\pi \sqrt{\frac{4}{9.8}} = \boxed{4.014 s}

2.

Calculate the period:


T = \frac{\text{Time}}{\# of oscillations} = \frac{5}{20} = \boxed{0.4 s }

Frequency is the reciprocal of the period, so:


f = \frac{1}{T} = \frac{1}{0.4} = \boxed{2.5 Hz}

6 0
3 years ago
A bus changes its velocity from 5 m/s to 15 m/s in 5 seconds. What is its acceleration?
Masja [62]
A = V-Vo/change in T

a = 15-5/5

a = 2
8 0
4 years ago
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