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Ksivusya [100]
3 years ago
6

The coefficients of static and kinetic friction between a 3.0-kg box and a horizontal desktop are 0.40 and 0.30, respectively. w

hat is the force of friction on the box when a 15-n horizontal push is applied to the box
Physics
1 answer:
luda_lava [24]3 years ago
6 0

The force of friction on the box is 8.82 N.

The force of friction is given by the formula

F=μN

Here N is the normal force of the box=mg=3*9.8=29.4 N

Now the static friction force on the box is=0.4*29.4=11.76 N

Since the box is applied with the horizontal pull of 15 N, which is greater than the static friction, therefore the box must be in the moving condition, in that case there is dynamic friction force is applied on the box=0.3*29.4=8.8 N

Therefore the force of friction on the box is 8.8 N.

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Nesterboy [21]

Answer;

The temperature change for the second pan will be lower compared to the temperature change of the first pan

Explanation;

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That is; Q = mcΔT

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3 0
3 years ago
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sdas [7]

Answer:

The new distance is     d = 0.447 d₀

Explanation:

The electric out is given by Coulomb's Law

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We reduce the charge of sphere B to 1/5 of its initial value (q_{B}=q₂ = q₂ / 5) than new distance (d = n d₀)

dat

     q₁ = q_{A}

     q₂ = q_{B}

     r = d₀

In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points

         F = k q₁ q₂ / d₀²

         F = k q₁ (q₂ / 5) / (n d₀)²

         .k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)

          5 n² = 1

          n = √ 1/5

          n = 0.447

The new distance is

         d = 0.447 d₀

6 0
3 years ago
Steve does 80 J of work to a rope threaded through a pulley and attached to a crate of anvils. The tension of the rope is 10 N a
Ivahew [28]

Answer:

The Answer is (B)

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Answer:

αβ = Ma

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By Newton's 2nd Law, the equation governing the motion of the rocket while the rocket is burning fuel is

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