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Vsevolod [243]
3 years ago
6

A 300-n force acts on a 25-kg object. the acceleration of the object is?

Physics
2 answers:
Over [174]3 years ago
8 0
To solve the answer use the equation: a = fnet / m

a = 300 N / 25 kg

300 N / 25 kg = 12m/s

The acceleration of the object is 12m/s

Evgesh-ka [11]3 years ago
4 0
This can be obtained by following the Newton's Second Law of motion. It states that the acceleration of an object which is produced by a net force is directly related to the net force and inversely related to the mass of the object. We use the equation:

a = F / m
a = 300 / 25
a = 12 m/s²
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The magnitude of the average induced emf in the wire during this time is 9.533 V.

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Radius r= 0.63 m

Magnetic field B= 0.219 T

Time t= 0.0572 s

We need to calculate the average induce emf in the wire during this time

Using formula of induce emf

E=-\dfrac{d\phi}{dt}

E=-B\dfrac{dA}{dt}

E=-B\dfrac{A_{2}-A_{1}}{dt}

E=B\dfrac{A_{1}-A_{2}}{dt}.....(I)

In reshaping of wire, circumstance must remain same.

We calculate the length when wire is in two loops

l=2\times 2\pi\times r_{1}

l=2\times 2\pi\times 0.63

l=7.916\ m

The length when wire is in one loop

l=2\pi\times r_{2}

7.916=2\times \pi\times r_{2}

r_{2}=\dfrac{7.916}{2\times \pi}

r_{2}=1.259\ m

We need to calculate the initial area

A_{1}=N\times\pi\times r_{1}^2

Put the value into the formula

A_{1}=2\times3.14\times(0.63)^2

A_{1}=2.49\ m^2

The final area is

A_{2}=N\times\pi\times r_{2}^2

A_{2}=1\times\pi\times(1.259)^2

A_{2}=4.98\ m^2

Put the value of initial area and final area in the equation (I)

E=0.219\dfrac{2.49-4.98}{0.0572}

E=-9.533\ V

Negative sign shows the direction of induced emf.

Hence, The magnitude of the average induced emf in the wire during this time is 9.533 V.

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