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3 years ago

How does the electric force between two charged objects change when the

Physics

1 answer:

castortr0y [4]3 years ago

3 0

Answer:

When the electric force between two charged objects change, the distance between the objects increases in Iicreasing the separation distance between objects and decreases the force of attraction or repulsion between objects.

Hope it helps.

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The phases of the moon depend on how much of the lighted side of the moon can be seen from Earth.

Ostrovityanka [42]

Answer:

true

Explanation:

if there is no light it's different from when there is

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3 years ago

Why is understanding the concept of forces, friction and gravity important?

nydimaria [60]

Because it's the basis of how everything around you works

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3 years ago

A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block

MariettaO [177]

Answer:

v₀ = 0.5058 m/s

Explanation:

From the question, for the block to hit the bottle, the elastic potential energy of the spring at the bottle (x = 0.08 m) should be equal to the sum of the elastic potential energy of the spring at x = 0.05 m and the kinetic energy of block at x = 0.05 m

Now, the potential energy of the block at x = 0.08 m is ½kx²

where;

k is the spring constant given by; k = ω²m

ω is the angular velocity of the oscillation

m is the mass of the block.

Thus, potential energy of the spring at the bottle(x = 0.08 m) is;

U = ½ω²m(0.08m)²

Also, potential energy of the spring at the bottle(x = 0.05 m) is;

U = ½ω²m(0.05m)²

and the kinetic energy of the block at x = 0.05 m is;

K = ½mv₀²

Thus;

½ω²m(0.08)² = ½ω²m(0.05)² + ½mv₀²

Inspecting this, ½m will cancel out to give;

ω²(0.08)² = ω²(0.05)² + v₀²

Making v₀ the subject, we have;

v₀ = ω√((0.08)² - (0.05)²)

So,

v₀ = 8.1√((0.08)² - (0.05)²)

v₀ = 0.5058 m/s

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3 years ago

Lloyd is standing on a scaffolding 12 meters above the ground to clean the windows of a tall building. His bucket, which has a m

Sever21 [200]

Answer:

U₂ = 20 J

KE₂ = 40 J

v= 12.64 m/s

Explanation:

Given that

H= 12 m

m = 0.5 kg

h= 4 m

The potential energy at position 1

U₁ = m g H

U₁ = 0.5 x 10 x 12 ( take g= 10 m/s²)

U₁ = 60 J

The potential energy at position 2

U₂ = m g h

U ₂= 0.5 x 10 x 4 ( take g= 10 m/s²)

U₂ = 20 J

The kinetic energy at position 1

KE= 0

The kinetic energy at position 2

KE= 1/2 m V²

From energy conservation

U₁+KE₁=U₂+KE₂

By putting the values

60 - 20 = KE₂

KE₂ = 40 J

lets take final velocity is v m/s

KE₂= 1/2 m v²

By putting the values

40 = 1/2 x 0.5 x v²

160 = v²

v= 12.64 m/s

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3 years ago

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Anika [276]

Answer:

Explanation:

If Ami is saying she likes it then it it personal. If you are speaking from statistics and studies it is impersonal and technically not from there perspective. All of these do this except C.

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3 years ago

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