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SCORPION-xisa [38]
3 years ago
13

8. two +1 C charges are separated by 3000m. What is the magnitude of the electric force between them?

Physics
1 answer:
Sidana [21]3 years ago
8 0

Answer:

1000 N

Explanation:

The magnitude of the electrostatic force between two charged object is given by

F=k\frac{q_1 q_2}{r^2}

where

k is the Coulomb constant

q1, q2 is the magnitude of the two charges

r is the distance between the two objects

Moreover, the force is:

- Attractive if the two forces have opposite sign

- Repulsive if the two forces have same sign

In this problem:

q_1=q_2=+1C are the two charges

r = 3000 m is their separation

Therefore, the electric force between the charges is:

F=(9\cdot 10^9)\frac{(1)(1)}{3000^2}=1000 N

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Kelly is riding a bicycle, moves with an initial velocity of 5 m/s. Ten seconds later, she is moving at 15 m/s. What is her acce
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Answer:

her acceleration is 1 m/sec

Explanation:

The following information is given in the question

The initial velocity is 5 m/s

After 10 seconds, she would be moved at 15 m/s

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As we know that

Acceleration = Change in speed ÷ time

Acceleration = (15 - 5) ÷ (10)

= 1 m/sec

Hence, her acceleration is 1 m/sec

The same would be considered  

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2 years ago
A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (Figur
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Answer with Explanation:

We are given that mass of block=0.0600 kg

Initial speed of block=0.63 m/s

Distance of block  from the hole when the block is revolved=0.47 m

Final speed=3.29 m/s

Distance of block  from the hole when the block is revolved=9\times 10^{-2}m

a.We have to find the tension in the cord in the original situation when the block has speed =v_0=0.63 m/s

T=\frac{mv^2}{r}

Because tension is equal to centripetal force

Substitute the values

T=\frac{0.06\times (0.63)^2}{0.47}=0.05 N

b.v=3.29 m/s

T=\frac{mv^2}{r}=\frac{0.06\times (3.29)^2}{0.09}=7.2 N

c.Work don=Final K.E-Initial K.E

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W=\frac{1}{2}(0.06)((3.29)^2-(0.63)^2)

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According to the law of reflection, the angle of incidence is equal to the angle of
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Before going to answer this question first we have to understand reflection and laws of reflection.

Reflection is the optical phenomenon in which light will bounce back to the same medium from which it had originated .

Whenever a light ray will incident on a mirror or any reflecting surface, it will be reflected. The ray which falls on the reflecting surface is called incident ray and the ray which is reflected is called reflected ray.

Let us consider a normal to the point of incidence.The angle made by incident ray with the normal is called angle of incidence.Let it be denoted as[ i ]

The angle made by the reflected ray with the normal is called angle of incidence.Let it be denoted as [r]

There are two types of reflection.One is called regular and other one is called as irregular.The laws of reflection is valid for both the types of reflection.

There are two laws of reflection.

FIRST LAW -It states that the incident ray,reflected ray and the normal to the point of incidence,all lie in one plane.

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Hence the correct answer will be angle of reflection.

                                         

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