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GREYUIT [131]
2 years ago
10

A box is at rest on a table. What can you say about the forces acting on the box?

Physics
2 answers:
Nikitich [7]2 years ago
6 0
You can tell a lot about an object that's not moving,
and also a lot about the forces acting on it:

==> If the box is at rest on the table, then it is not accelerating.

==> Since it is not accelerating, I can say that the forces on it are balanced.

==> That means that the sum of all forces acting on the box is zero,
and the effect of all the forces acting on it is the same as if there were
no forces acting on it at all.

==> This in turn means that all of the horizontal forces are balanced,
AND all of the vertical forces are balanced.

Horizontal forces:
sliding friction, somebody pushing the box

All of the forces on this list must add up to zero. So ...

(sliding friction force) = (pushing force), in the opposite direction.

If nobody pushing the box, then sliding friction force = zero.

Vertical forces:
gravitational force (weight of the box, pulling it down)
normal force (table pushing the box up)

All of the forces on this list must add up to zero, so ...

(Gravitational force down) + (normal force up) = zero

(Gravitational force down) = -(normal force up) .
Troyanec [42]2 years ago
4 0

1) normal force 2) gravitational force

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Calculate the net force on the right charge due to the other two. Enter a positive value if the force is directed to the right a
lbvjy [14]

Answer:

Answer:

A. - 0.017N. It acts to the left.

B. - 0.043N. It acts to the left.

C. 0.060N. It acts to the right.

Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

Where K = Coloumbs constant =

Q(1) = charge on the leftmost side.

Q(2) = charge in the middle.

Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

It has a negative sign, hence, it acts to the left.

B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

Read more on Brainly.com - brainly.com/question/14592748#readmore

Explanation:

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Answer:

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Answer:

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