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alexandroperez13
06.09.2019 •
Physics
An electron is released from rest in a uniform electric field. the electron accelerates vertically upward, traveling 4.5 meters in the first 2.84 μs after it is released. what is magnitude of the electric field? what is the direction of the magnetic field? are we justified if ignoring gravity on the electron in this situation?
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Ответ:
The electric and magnetic field are 6.34 N/C and
.
Explanation:
Given that,
Distance = 4.5 m
Time = 2.84 μs
We need to calculate the acceleration
Using equation of motion
The distance covered by the electron is
When, the electron at rest
Where, s = distance
a = acceleration
t = time
Put the value into the formula
We need to calculate the electric field
Using formula of the electric field
Put the value into the formula
We need to calculate the magnetic field
Using formula of magnetic field
Put the value into the formula
According to the right hand rule,
The direction of magnetic field is outward because the direction of force is upward.
Hence, The electric and magnetic field are 6.34 N/C and
.
Ответ:
31.68 meters per second.
Explanation:
Speed is equals to Distance divided by Time, so, this means that speed is inversely proportional to time and directly proportional to distance.
Here, Distance is given which is 274 meters.
Time taken to cover the given distance is 8.65 seconds.
So, by putting the values of distance and time in the formula of speed we will get the top speed of Cheetah which is 31.68 meters per second.