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LenaWriter [7]
3 years ago
10

A bullet in a gun is accelerated from the firing chamber to the end of the barrel at an average rate of 6.20×105m/s2 for 8.10×10

-4s. What is its muzzle velocity (that is, its final velocity)?
Physics
1 answer:
kati45 [8]3 years ago
7 0

Answer:

502.2 m/s

Explanation:

initial velocity, u = 0 m/s

Let the final speed is v.

acceleration, a = 6.2 x 10^5 m/s^2

time taken, t = 8.10 x 10^-4 s

The muzzle speed is the final speed.

Use first equation of motion

v = u + a t

v = 0 + 6.2 x 10^5 x 8.10 x 10^-4

v = 502.2 m/s

Thus, the muzzle speed of the bullet is 502.2 m/s.

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How to find acceleration in non uniform motion. Please give two ways
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6 0
3 years ago
A 500-eV electron and a 300-eV electron trapped in a uniform magnetic field move in circular paths in a plane perpendicular to t
Alina [70]

Answer:ratio of the radii of their orbits = 1.3 --- C

Explanation:

1- eV = to the kinetic energy of the electrons

and kinetic energy is given as

K.E= 1/2mv2

v = √(2E/m)----- equation 1

The force on the particles  relating to the magnetic and circular motion ( centripetal force is given as

F = magnetic force = centripetal force

F= qvB = mv2/r

qvB = mv2/r

  r = mv/qB ------ equation 2

We know from equation 1 that v = √(2E/m)

Therefore,

r = √(2mE)/qB------ equation 3

We can now say that the  ratio of the two radii of their orbits can be calculated as

r1/r2 =(√(2mE1)/qB) /(√(2mE2)/qB

Where E1 = 500-eV  and E2 = 300-eV (1- eV = to the kinetic energy of the electrons)

r1/r2 = (√(2m x500)/qB) /(√(2mx 300)/qB

Cancelling out common variables, we are left with

r1/r2 =\sqrt{500/300}

r1/r2= 1.29 ≈ 1.3

   

8 0
2 years ago
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