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MAXImum [283]
2 years ago
7

A velocity-time graph is shown below: (2 points)

Physics
1 answer:
BabaBlast [244]2 years ago
5 0

Answer:

2m/s²

Explanation:

  • V=Vstart+at
  • rewrite that to find a so a=(V-Vstart)/t

part A of graph

  • a=(20m/s–0m/s)/5s
  • a=4m/s²

part B of graph

  • a=(0m/s–0m/s)/5s
  • a=0m/s²

the average between the two is both answers added divided by the number of answers

  • (4m/s²–0m/s²)/2
  • 4m/s²/2
  • 2m/s²
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A beam of electrons moving in the x-direction enters a region where a uniform 208-G magnetic field points in the y-direction. Th
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Answer:

1.26\cdot 10^7 m/s

Explanation:

When a charged particle moves perpendicularly to a magnetic field, the force it experiences is:

F=qvB

where

q is the charge

v is its velocity

B is the strength of the magnetic field

Moreover, the force acts in a direction perpendicular to the motion of the charge, so it acts as a centripetal force; therefore we can write:

qvB=m\frac{v^2}{r}

where

m is the mass of the particle

r is the radius of the orbit of the particle

The equation can be re-arranges as

v=\frac{qBr}{m}

where in this problem we have:

q=1.6\cdot 10^{-19}C is the magnitude of the charge of the electron

B=208 G=208\cdot 10^{-4}T is the strength of the magnetic field

The beam penetrates 3.45 mm into the field region: therefore, this is the radius of the orbit,

r=3.45 mm = 3.45\cdot 10^{-3} m

m=9.11\cdot 10^{-31} kg is the mass of the electron

So, the electron's speed is

v=\frac{(1.6\cdot 10^{-19})(208\cdot 10^{-4})(3.45\cdot 10^{-3})}{9.11\cdot 10^{-31}}=1.26\cdot 10^7 m/s

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