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matrenka [14]
3 years ago
15

An electron moves through a region with a uniform magnetic field of unknown magnitude directed in the +y direction. At the insta

nt that the electron is moving in the +z direction with a velocity of 2.83e+6 m/s​​, it experiences a magnetic force of 2.87 pN in the +x direction. The magnetic field is 6.34 T.If the electron were replaced with a proton moving with the same velocity, how would the resulting acceleration of the particle change?a) The resulting acceleration of the proton would be weaker in magnitude and in the same direction as the acceleration of the electron.b) The resulting acceleration of the proton would be weaker in magnitude and oppositely directed compared to the acceleration of the electron.c)The resulting acceleration of the proton would be greater in magnitude and in the same direction as the acceleration of the electron.d)The resulting acceleration of the proton would be equal in magnitude and oppositely directed compared to the acceleration of the electron.e) The resulting acceleration of the proton would be greater in magnitude and oppositely directed compared to the acceleration of the electron.f)The resulting acceleration of the proton would be equal in magnitude and in the same direction as the acceleration of the electron.
Physics
1 answer:
suter [353]3 years ago
6 0

Answer:

Since proton is more massive than electron, thus its acceleration will be lesser in magnitude since acceleration is inversely proportional to acceleration.

However, its charge is opposite to that of electron. Thus it will experience the acceleration in the opposite direction.

Thus, the correct option is:

b) The resulting acceleration of the proton would be weaker in magnitude and oppositely directed compared to the acceleration of the electron.

Explanation:

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aliya0001 [1]

Answer:

15.3 s and 332 m

Explanation:

With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon

    gm = 1/6 ge

    gm = 1/6  9.8 m/s² = 1.63 m/s²

We calculate the range

    R = Vo² sin 2θ  / g

    R = 25² sin (2 30) / 1.63

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We will calculate the time of flight,

   Y = Voy t – ½ g t2  

   Voy = Vo sin θ

When the ball reaches the end point has the same initial  height Y=0

0 = Vo sin  t – ½  g t2

0 = 25 sin (30)  t – ½ 1.63 t2

0= 12.5 t –  0.815 t2

We solve the equation

0= t ( 12.5 -0.815 t)

 t=0 s

t= 15.3 s

The value of zero corresponds to the departure point and the flight time is 15.3 s

Let's calculate the reach on earth

R2 = 25² sin (2 30) / 9.8

R2 = 55.2 m

R/R2 = 332/55.2

R/R2 = 6

Therefore the ball travels a distance six times greater on the moon than on Earth

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