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ch4aika [34]
3 years ago
12

A 12 kg box is pulled across the floor with a 48 N horizontal force. If the force of friction is 12 N, what is the acceleration

of the box?
Physics
1 answer:
Oksi-84 [34.3K]3 years ago
7 0

Answer:

The acceleration of the box is 3 m/s²

Explanation:

Given;

mass of the box, m = 12 kg

horizontal force pulling the box forward, Fx = 48 N

frictional force acting against the box in opposite direction, Fk = 12 N

The net horizontal force on the box, F = 48 N - 12 N

The net horizontal force on the box, F = 36 N

Apply Newton's second law of motion to determine the acceleration of the box;

F = ma

where;

F is the net horizontal force on the box

a is the acceleration of the box

a = F / m

a = 36 / 12

a = 3 m/s²

Therefore, the acceleration of the box is 3 m/s²

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The filament of a certain lamp has a resistance that increases linearly with temperature. When a constant voltage is switched on
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Answer:

The change in temperature is \Delta T  = 1795 K

Explanation:

From the question we  are told that

   The temperature coefficient is  \alpha  =  4 * 10^{-3 }\  k^{-1 }

The resistance of the filament is mathematically represented as

           R  =  R_o [1 + \alpha  \Delta T]

Where R_o is the initial resistance

Making the change in temperature the subject of the formula

     \Delta T = \frac{1}{\alpha } [\frac{R}{R_o} - 1 ]

Now from ohm law

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This implies that current varies inversely with current so

           \frac{R}{R_o} = \frac{I_o}{I}

Substituting this we have

       \Delta T  = \frac{1}{\alpha } [\frac{I_o}{I} - 1 ]

From the question we are told that

    I  = \frac{I_o}{8}

Substituting this we have

   \Delta T  = \frac{1}{\alpha } [\frac{I_o}{\frac{I_o}{8} } - 1 ]

=>     \Delta T  = \frac{1}{3.9 * 10^{-3}} (8 -1 )

        \Delta T  = 1795 K

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A light ray hits a plane surface at 20 degrees. What is the angle between the incident and reflected rays.
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140°

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When light rays hit surface at 20°, they also leave the surface at the same angle

Since the whole surface has 180° then subtracting these two angles from total angle gives the the angle between the incident and reflected rays.

180°-20°-20°=140°

The angle of incidence and reflection are equal hence 140/2=70°

The question needed the angle between the incident and reflected rays which is already calculated as 140°

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