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2 years ago

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four equal charges of +3microcolumb are placed at the four corners of a square that is 40cm on a side. Find the force on any one

of the charge.

1 answer:

Reil [10]2 years ago

6 0

Answer:

i can see your question

pls fast

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Please please help me :)

mrs_skeptik [129]

Answer:

Explanation:

2) From F=ma

Force =15×40=600N or kgm/s2

3)From the same equation making acceleration the subject of the formula will give

a=f÷m

=24÷4=6m/s2

4)m=f÷a

=45÷15=3kg

4 0

2 years ago

Licemer1 [7]

By V=IR
A: 24=I*20
I = 1.2A

B: 220 = I*250
I = 0.88A

C: 6= I*3
I = 2 A

C,A,B

3 0

3 years ago

Read 2 more answers

Which is a key criterion for the design of an automobile bumper?

Mrac [35]

Answer: The answer is A

Explanation: The bumper is the first part of an automobile to be impacted when in a head-on accident

3 0

2 years ago

A miner develops cancer of the esophagus. Ten years before, he had been exposed to radiation when he worked for a year in a mine

slamgirl [31]

D.) It is unlikely that a specific cause can be determined, but the treatment would likely be the same in either case.

3 0

2 years ago

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The distance between two particles is 2 centimeters. If the distance is increased to 4 centimeters, the force will be ?

postnew [5]

Answer:

The new force is 1/4 of the previous force.

Explanation:

Given

$Initial\ Distance = 2cm$ ---- $r_1$

$New\ Distance = 4cm$ --- $r_2$

Required

Determine the new force

Let the two particles be q1 and q2.

The initial force F1 is:

$F_1 = \frac{kq_1q_2}{r_1^2}$ --- Coulomb's law

Substitute 2 for r1

$F_1 = \frac{kq_1q_2}{2^2}$

$F_1 = \frac{kq_1q_2}{4}$

The new force (F2) is

$F_2 = \frac{kq_1q_2}{r_2^2}$

Substitute 4 for r2

$F_2 = \frac{kq_1q_2}{4^2}$

$F_2 = \frac{kq_1q_2}{4*4}$

$F_2 = \frac{1}{4}*\frac{kq_1q_2}{4}$

Substitute $F_1 = \frac{kq_1q_2}{4}$

$F_2 = \frac{1}{4}*F_1$

$F_2 = \frac{F_1}{4}$

The new force is 1/4 of the previous force.

3 0

2 years ago