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charle [14.2K]
2 years ago
6

A box of weight 74 N is sliding on a horizontal surface at a constant velocity due to an external force F-> of magnitude 4.8

N. A normal force (n->) and a friction force (f->) also act on the box. What is the magnitude of the friction force?
74 N


2.4 N


69 N


4.8 N


9.6 N

Physics
1 answer:
n200080 [17]2 years ago
6 0

Answer

4.8 N

If the box is moving with a constant velocity, then we can say that the system is in equilibrium. This is because if the external force (F->) was greater than other forces the box would be accelerating. This tells us that this force (F->) is just enough to overcome friction and so it must be equal to 4.8 N.

The normal force has no effect to the horizontal velocities or forces. It is equal to -Weight. That is -74 N. The negative sign shows that the force is in opposite direction.

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What happens to the object if the line crosses the x-axis from the positive portion of a velocity versus time diagram?
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7 0
3 years ago
A crate is sliding on the floor. If there is a total force acting on the crate in the same direction as it is sliding, the crate
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7 0
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A
True [87]

Answer:

\Delta T=3.615^{\circ}C is the drop in the water temperature.

Explanation:

Given:

  • mass of ice, m_i=14.7\ g=0.0147\ kg
  • mass of water, m_w=324\ g=0.324\ kg

Assuming the initial temperature of the ice to be 0° C.

<u>Apply the conservation of energy:</u>

  • Heat absorbed by the ice for melting is equal to the heat lost from water to melt ice.

<u>Now from the heat equation:</u>

Q_i=Q_w

m_i.L=m_w.c_w.\Delta T ......................(1)

where:

L= latent heat of fusion of ice =333.55\ J.g^{-1}

c_w= specific heat of water =4.186\ J.g^{-1}.^{\circ}C^{-1}

\Delta T= change in temperature

Putting values in eq. (1):

14.7 \times 333.55=324\times 4.186\times \Delta T

\Delta T=3.615^{\circ}C is the drop in the water temperature.

8 0
2 years ago
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