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Vesnalui [34]
2 years ago
11

If a body p with a positive charge is placed in contact with a body q (initially uncharged), what will be the nature of the char

ge left on q?
Physics
1 answer:
uysha [10]2 years ago
5 0

If a body p with a positive charge is placed in contact with a body q (initially uncharged), then the nature of charge gained by q must be positive, because rubbing an uncharged body with a charged body or placed in contact with a positive charged body, helps gain a charge to the uncharged body.

There are a variety of methods to charge an object. One method is known as induction. In the induction process, a charged object is brought near but not touched to a neutral conducting object.

Let's know, how a element gain positive charge?

A positive charge occurs when the number of protons exceeds the number of electrons. A positive charge may be created by adding protons to an atom or object with a neutral charge. A positive charge also can be created by removing electrons from a neutrally charged object.

To learn more about Positive charge here

brainly.com/question/2903220

#SPJ4

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You drop a batitin a stationary elevator and the ball hits the floor in o 50 s. How long does it take for the ball to hit the fl
solmaris [256]

Answer:

option (a) 0.61 s

Explanation:

Given;

Time taken by the ball to reach the ground  = 0.50 s

Let us first calculate the distance through which the ball falls on the ground

from the Newton's equation of motion, we have

s=ut+\frac{1}{2}at^2

where,  

s is the distance

a is the acceleration

t is the time

here it is the case of free fall

thus, a = g = acceleration due to gravity

u =  initial speed of the ball = 0

on substituting the values, we get

s=0\times 0.50+\frac{1}{2}\times9.8\times0.50^2

or

s = 1.225 m

Now,

when the elevator is moving up with speed of 1.0 m/s

the initial speed of the ball = -1.0 m/s   (as the elevator is moving in upward direction)

thus , we have

s=ut+\frac{1}{2}at^2

or

1.225=-1.0\times\ t+\frac{1}{2}\times9.8t^2

or

4.9t^2 - t  - 1.225 = 0

or

t = 0.612 s

hence, the correct answer is option (a) 0.61 s

4 0
3 years ago
Eating 2500 Cal every day a friend of mine maintains a stable weight of 70 kg. One day, after eating 3500 Cal, he decided to do
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Answer:

Explanation:

Calories to be burnt = 3500 - 2500 = 1000 Cals .

Efficiency of conversion to mechanical work  is 25 % .

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4000 Cals = 4.2 x 4000 = 16800 J  .

Work done in one jump = kinetic energy while jumping

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= .5 x 70 x 3.3²

= 381.15 J .

Number of jumps required = 16800 / 381.15

= 44 .

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