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2 years ago

A group of students conduct an experiment to study Newton's second law of motion. They applied a force to a toy car and

Physics

1 answer:

Anika [276]2 years ago

4 0

Answer:

Explanation:

If N increases then the Acceleration will also increase

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3 years ago

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3 years ago

What is the frequency for a beam of electrons orbiting in a field of 4.62 x 10^-3 T? Let the mass of an electrons m = 9.31 x 10^

melisa1 [442]

Frequency: $1.27\cdot 10^8 Hz$

Explanation:

The force experienced by an electron in a magnetic field is

$F=qvB$

where

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

v is the speed of the electron

B is the strength of the magnetic field

Since the force is perpendicular to the direction of motion of the electron, the force acts as centripetal force, so we can write:

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

where

r is the radius of the orbit

$m=9.31\cdot 10^{-31}kg$ is the mass of the electron

Re-arranging the equation,

$\frac{v}{r}=\frac{qB}{m}$ (1)

We also know that in a circular motion, the speed is equal to the ratio between circumference of the orbit and orbital period (T):

$v=\frac{2\pi r}{T}$

Substituting into (1),

$\frac{2\pi}{T}=\frac{qB}{m}$

We also know that 1/T is equal to the frequency f, so

$f=\frac{qB}{2\pi m}$

In this problem,

$B=4.62\cdot 10^{-3}T$

Therefore, the frequency of the electrons is

$f=\frac{(1.6\cdot 10^{-19})(4.62\cdot 10^{-3})}{2\pi(9.31\cdot 10^{-31})}=1.27\cdot 10^8 Hz$

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3 years ago

How much net force is required to accelerate a 0.5 kg toy car, initially at rest to a velocity of 2.4 m/s in 6s?

Jet001 [13]

Force = 0.20N .F = m ×a .& a = v/t then the f = m×v/t

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4 years ago