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3 years ago

How does the mass of an object affect the outcome when the forces acting upon it are unbalanced? I need six sentences...

Physics

1 answer:

PilotLPTM [1.2K]3 years ago

6 0

<h3><u>Mass of an object affect the outcome of unbalanced forces:</u></h3>

Newton’s second law of motion deals with the result of motion of an object when unbalanced forces are applied. The second law of motion establish the relationship between mass, acceleration and unbalanced forces of the object. The below points explains the relation between unbalanced forces acting on mass of the object. The acceleration of object would be higher when the unbalanced force is higher.

The equation for the unbalanced force is,

$Unbalanced\ force = Mass \times Acceleration.$

For Example: Two masses of 500 kg and 1 kg is applied with same unbalanced forces. The change in motion of 500 kg would be very much less than the change in motion of 1 kg.

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3 years ago

3. A model rocket takes 0.05 seconds to speed up from rest to its maximum velocity of 80 m/s.

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Answer:

$1600 \frac{m}{s^2}$

Explanation:

Acceleration is defined as the change in velocity divided by the time it took to produce such change. The formula then reads:

$a = \frac{change-in-velocity}{time} = \frac{Vf-Vi}{t}$

Where Vf is the final velocity of the object, (in our case 80 m/s)

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and t is the time it took to change from the Vi to the Vf (in our case 0.05 seconds.

Therefore we have:

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2 years ago

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A transformer is intended to decrease the value of the alternating current from 500 amperes to 25 amperes. The primary coil cont

EastWind [94]

Answer:

The number of turns in secondary coil is 4000

Explanation:

Given:

Current in primary coil $I_{P} = 500$ A

Current in secondary coil $I_{S} = 25$ A

Number of turns in primary coil $N_{P} = 200$

In case of transformer the relation between current and number of turns is given by,

$\frac{N_{S} }{N_{P} } = \frac{I_{P} }{I_{S} }$

For finding number of turns in secondary coil,

$N_{S} = \frac{I_{P} }{I_{S} } N_{P}$

$N_{S} = \frac{500}{25} \times 200$

$N_{S} = 4000$

Therefore, the number of turns in secondary coil is 4000

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2 years ago

A spring with a mass on the end of it hangs in equilibrium a distance of 0.4200 m above the floor. The mass is pulled down a dis

den301095 [7]

Answer:

0.48 m

Explanation:

I'm assuming that this takes place in an ideal situation, where we neglect a host of factors such as friction, weight of the spring and others

If the mass is hanging from equilibrium at 0.42 m above the floor, from the question, and it is then pulled 0.06 m below that particular position. This pulling is a means of adding more energy into the spring, when it is released, the weight compresses the spring and equals its distance (i.e, 0.06 m) above the height.

0.42 m + 0.06 m = 0.48 m

At the highest point thus, the height is 0.48 m above the ground.

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3 years ago