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icang [17]
2 years ago
5

if an object is being acted on by two forces a push and a pull of 6N what is the net force of the object

Physics
1 answer:
Paul [167]2 years ago
6 0
The concepts of force<span>, mass, and weight play critical roles. Newton's Laws of. Motion ... the person stops </span>pushing<span>? ... F </span>net<span> =10 N </span>2<span> N. = 8 N (to the right) a = F </span>net<span> m. = 8 N. 5 kg. =1.6 m s. </span>2<span> ... </span>Two equal forces<span> act on an </span>object<span> in the directions shown. </span>If<span> these ... </span>Two<span> connected carts </span>being accelerated by a force<span> F applied by.</span>
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The amount of work done by two boys who apply 200 N of force in an
Aleksandr [31]

The amount of work done by two boys who apply 200 N of force in an unsuccessful attempt to move a stalled car is 0.

Answer: Option B

<u>Explanation: </u>

Work done is the measure of work done by someone to push an object from its present position. We can also define work done as the amount of forces needed to move an object from its present position to another position. So the amount of work done is directly proportionate to the product of forces acting on the object and the displacement of the object.  

           \text { Work done }=\text { Force } \times \text { displacement }

So in this present case, as the two boys have done an unsuccessful attempts to push a stalled car so that means the displacement of the car is zero as there is no change in the position of the car. But they have applied a force of 200 N each. So the amount of work done will be

           \text { Work done }=200 \mathrm{N} \times 0=0

Thus, the amount of work done by two boys will be zero due to their unsuccessful attempt to move a stalled car.

8 0
2 years ago
2. How long must a 400 W electrical engine work in order to produce 300 kJ of work?
yanalaym [24]

Answer:

Explanation:

400 W = 400 J/s

300000 J / 400 J/s = 750 s or 12.5 minutes

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3 years ago
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Answer:

Explanation:

As the dielectric is inserted between the plates of a capacitor, the capacitance becomes K times and the electric field between the plates becomes 1 / K times the original value. Where, K be the dielectric constant.

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