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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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A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal
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Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

v=\sqrt{v_x^2+v_y^2}           (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

v_y^2=v_{oy}^2+2gh   (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s  (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}

vx is calculated by using the information about the horizontal range of the ball:

R=v_o\sqrt{\frac{2h}{g}}    (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}

The velocity of the ball just before it touches the ground is 46.99 m/s

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Does anyone know 08fortnitebeast?
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Answer:

nah bruh

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i just dont know

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