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ValentinkaMS [17]
3 years ago
7

Can someone please help? Thank u!

Physics
1 answer:
Snowcat [4.5K]3 years ago
7 0

Answer:

B. 7.5 m/s^2

Explanation:

To find acceleration you need to subtract the final velocity by the starting velocity then divide that by the time

a= v-v/t

a= 60-0/8

a= 60/8

a=7.5 m/s^2

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Which statement accurately describes the chemical symbol of an element?
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The chemical symbol always begins with a capital letter.

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Answer:

7.82 s

Explanation:

Given:

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a = 9.8 m/s²

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t = 7.82 s

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The Coulomb force between two charges q1 and q2 at separation r in air is F. If half of the separation is filled with medium of
vodomira [7]

Answer:

The new Coulomb force is q₁q₂/9πε₀r²

Explanation

The coulomb force between the two charges q₁ and q₂ at a distance r in air is given by F = q₁q₂/4πε₀r².

Now, let us assume the material of dielectric constant κ = 9 is placed between them on the side of the q₁ charge. The value of its effective charge is now q₃ = q₁/κ at a distance of d = r/2 from the q₂ charge.

Since we have air between q₂ and q₃, the coulomb force between them is

F' = q₂q₃/4πε₀d²

= q₂(q₁/κ)/4πε₀(r/2)²

=  4q₂q₁/κ4πε₀r²

= 4/κ(q₂q₁/4πε₀r²)

= 4/9 × (q₂q₁/4πε₀r²)

= q₁q₂/9πε₀r²

So, the new Coulomb force is q₁q₂/9πε₀r²

3 0
4 years ago
Newton’s Second law of Motion (F = ma) is one of the basis of forces and motion. How is it very similar to his First law and why
koban [17]

Answer:

First law can be deduced from second law.

Acceleration may determine the position and velocity of the system.

Explanation:

When net force is zero, the second law is 0 = ma, or the motion is at constant speed. Thus first law establishes that when there are no forces, the object moves at constant speed, so first law is explained by using the second.

If you determine the acceleration of a system, you may use calculus or kinematic equations to determine velocity and position of the particle and determine how it moves. This is very important in mechanics and engineering, for example, for spacecrafts, forensic situations, etc.

4 0
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