V A block of 12 kg mass is at rest

Three +3.0-μC point charges are at the three corners of a square of side 0.50 m. The last corner is occupied by a −3.0-μC charge

Naddik [55]

Answer:

4.30 x 10⁵ N/C

Explanation:

Two positive +3.0-μC point charges at opposite corners of the square creates equal and opposite electric field at the center. hence the electric field by these two positive charges at opposite corners becomes zero.

$a$ = length of the square of side = 0.50 m

$r$ = distance of the center from each corner = $\frac{a}{sqrt(2)} = \frac{0.50}{sqrt(2)} = 0.354 m$

Magnitude of net electric field at the center is given as

$E = \frac{2kq}{r^{2}} \\E = \frac{2(8.99\times10^{9})(3\times10^{-6})}{(0.354)^{2}} \\E = 4.30\times10^{5} NC^{-1}$

6 0

3 years ago

Can someone help me PLEASE? I need to finish this because I have more homework and I need to go to bed

Dafna1 [17]

When the computer is on the circuit has positive and negative charges it will move fluenlty to get the power when it is in on position.

8 0

3 years ago

You're driving down the highway late one night at 22 m/s when a deer steps onto the road 38 m in front of you. Your reaction tim

sleet_krkn [62]

speed of the car is given on the road

$v = 22 m/s$

reaction time is given as

$t = 0.50 s$

now we can find the distance that it will move during this reaction time

$d_1 = 22* 0.50 = 11 m$

now the deceleration of the car is 10 m/s^2

so the distance that it will move before stop is given by

$v_f^2 - v_i^2 = 2 a d$

$0^2 - 22^2 = 2*(-10)*d$

$d = 24.2 m$

so the total distance that it require to stop is given as

$d = 24.2 + 11 = 35.2 m$

while the deer is standing at distance 38 m

so the car will stop 2.8 m before the position of deer

3 0

3 years ago

If the distance between two charged particles is increased 2.69 times, the ratio of new to old electric force is _______ times t

Tasya [4]

Answer:

F = K Q1 Q2 / R^2 force between 2 charged partices

F2 / F1 = (R1 / R2)^2 = (1 / 2.69)^2 = .139

F2 = .139 F1

8 0

1 year ago

Alex is standing still and throws a football with a speed of 10 m/s to his friend, who is also standing still. The two friends a

Phantasy [73]

The question is incomplete. It comes with a set of answer choices.

These are the answer choices:

Alex observes it as 10 m/s, and his friend observes it as less than 10 m/s.

Alex observes it as less than 10 m/s, and his friend observes it as 10 m/s.

Both Alex and his friend observe it as 10 m/s.

Both Alex and his friend observe it as less than 10 m/s.

Answer: Both Alex and his friend observe it as 10 m/s.

Justification:

1) The speed is relative to the frame of reference.

2) It is said that the both Alex and his friend are standing still.

3) Then, the speed they both see is the same, 10 m/s, respect the Earth (where they are standing still).

Of course, Alex is watching the ball moving away and his friend is seing it approaching, but it is not relevant for the question, as it deals with the speed which is only about magnitude, not direction.

7 0

3 years ago

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