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nikdorinn [45]
3 years ago
13

The muzzle velocity of a 50.0g shell leaving a 3.00 kg rifle is 400m/s what is the recoil velocity of the rifle

Physics
1 answer:
serg [7]3 years ago
6 0

Here if we consider bullet and gun as a system then we can say that momentum of the system will remain conserved

so we will have

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

now we know that

m_1 = 50 g = 0.050 kg

m_2 = 3 kg

v_{1i} = v_{2i} = 0

v_{1f} = 400 m/s

now we will have

0.050(0) + 3(0) = 0.050(400) + 3(v)

20 + 3v = 0

v = - \frac{20}{3} = - 6.67 m/s

so gun will recoil with speed 6.67 m/s

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8 0
3 years ago
A 0.260 m radius, 525 turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a
Sophie [7]

Answer:

B = 0.37T

Explanation:

In order to calculate the needed magnitude of the magnetic force you use the following formula, which calculate the induced emf of the solenoid when there is a change in the magnetic flux:

emf=-N\frac{\Delta \Phi_B}{\Delta t}=-N\frac{\Delta (BAcos\theta)}{\Delta t}       (1)

emf: induced voltage in the solenoid = 10,000V

N: turns of the solenoid = 525

ФB: magnetic flux

B: magnitude of the magnetic field = ?

A: cross-sectional area of the solenoid = π*r^2

r: radius of the cross-sectional area = 0.260m

Δt: interval time of the change of the magnetic flux = 4.17ms = 4.17*10^-3s

First, you have the magnetic field direction perpendicular to the plane of the solenoid, after, the angle between them is 90°  (quarter of a revolution)

In the equation (1) the only parameter that changes on time is the angle, then, you can solve for B from the equation (1):

emf=-NBA\frac{cos(90\°)-cos(0\°)}{\Delta t}=\frac{NBA}{\Delta t}\\\\B=\frac{(\Delta t)(emf)}{NA}=\frac{(\Delta t)(emf)}{N(\pi r^2)}\\\\

Finally, you replace the values of the parameters to calculate B:

B=\frac{(4.17*10^{-3}s)(10000V)}{(525)(\pi(0.260m)^2)}=0.37T

The strength of the magnetic field is 0.37T

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3 years ago
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