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faltersainse [42]
3 years ago
14

A rifle of mass 2 kg is horizontally suspended by a pair of strings so that recoil can be measured. The rifle fires a bullet of

mass 1/100 kg at a speed of 200 m/s. The recoil velocity of the rifle is about ___________.
Physics
1 answer:
aliya0001 [1]3 years ago
3 0

Answer: 1m/s

Explanation: according to the law of conservation of linear momentum in an isolated system, the momentum of the gun equals that of the bullet.

Mathematically

Mb×Vb = Mg×Vg

Where Mb = mass of bullet = 1/100 = 0.01 kg

Vb = velocity of bullet = 200 m/s

Mg = mass of gun = 2kg

Vg = recoil velocity of gun =?

0.01×200 = 2×Vg

Vg = 0.01×200/2

Vg = 0.01×100

Vg = 1m/s

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Answer:

303

Explanation:

We are given that

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Emf induced in coil 2,E_2=4.16 V

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We have to find the number of loops in coil 2.

Rat of change of magnetic flux in a single loop is same.

Let \phi_1 and \phi_2 be the magnetic flux in coil 1 and coil 2.

\frac{d\phi_1}{dt}=\frac{d\phi_2}{dt}

\frac{V_2}{V_1}=\frac{N_2}{N_1}

Using the formula

\frac{E_2}{E_1}=\frac{N_2}{N_1}

\frac{4.16}{3.13}=\frac{N_2}{228}

N_2=\frac{4.16}{3.13}\times 228

N_2=303

Hence, the number of loops in coil 2=303

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Explanation:

3 0
3 years ago
1.2 Define the following terms and in each case give the symbol and the unit: 1.2.1 wavelength (4) ·​
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U = 102.8 J (100 J to two significant digits)

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