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Anit [1.1K]
3 years ago
7

A small glass bead has been charged to + 30.0 nC . A small metal ball bearing 2.60 cm above the bead feels a 1.80×10−2 N downwar

d electric force. Part A What is the charge on the ball bearing? Express your answer with the appropriate units.
Physics
1 answer:
BlackZzzverrR [31]3 years ago
8 0

Answer:

The charge on the ball bearing 4.507 × 10^-8 C

Explanation:

From Coulomb's law

F = kq1q2/r²

make q2 the subject

q2 = Fr²/kq1

q2 = (1.8×10^-2 × 0.026²) ÷ (9×10^9 × 30×10^-9)

q2 = 4.507 × 10^-8 C

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An object is launched with an initial speed of 30 m/s at an angle of 60° above the horizontal . What is the maximum height reach
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H = 34.43 m

Explanation:

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H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(30)^2\times \sin^2(60)}{2\times 9.8}\\\\H=34.43\ m

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Two trains A and B of length 400 m each are moving on two parallel tracks with
Vinvika [58]

<u>Answer:</u>

<em>The initial distance between the trains is 1450 m. </em>

<u>Explanation:</u>

In the question two trains are of equal length 400 m and moves at a uniform speed of 72 km/h. train A is moving ahead of train B. If the train B has to overtake train A it should accelerate.

Train B’s acceleration  is 1m/s^2   and it accelerated for 50 seconds.

<em>a=1 m/s^2</em>

<em>t=50 s </em>

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<em>we have to convert this speed into m/s  </em>

<em>u=72 \times 5/18=20 m/s</em>

<em>Distance covered in accelerating phase  S=ut+1/2  at^2  </em>

<em>=20 \times 50+1/2 \times 1 \times 50^2</em>

<em>=1000+1250=2250 m </em>

If  a train is just behind another, the distance covered by the train located behind during overtaking phase will be equal to the sum of the lengths of the trains.

<em>Here length of train A+length of train B=400+400=800 m</em>

<em>Hence the initial distance between the trains = 2250-800=1450 m</em>

6 0
3 years ago
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