3. If the radius of Earth decreased, with no change in mass, your weight would

Physics

1 answer:

m_a_m_a [10]3 years ago

6 0

Answer:

The weight of the person would increase

Explanation:

The Universal Law of Gravitation gives the magnitude of the force between the masses of two objects (m1 and m2) separated a given distance "d" as:

$F_g=G\,\frac{m_1*m_2}{d^2}$

where G is the universal gravitational constant.

Our weight on Earth is this force between the Earth (of mass M) and ourselves (our mass m) at a distance that is the Earth's radius R:

$Weight=G\frac{M*m}{R^2}$

Now, if we keep all the values equal (mass of the Earth M and our mass m) except for the distance between the center of the Earth and our center of gravity (the radius of the Earth), we are going to have now a smaller radius (r) in the formula above:

$Weight=G\frac{M*m}{r^2}$

and dividing by a smaller number (r is smaller than R), will render a larger quotient. This means that the actual force (weight) will become larger, so the weight would clearly increase.

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A charged particle enters a uniform magnetic field B with a velocity v at right angles to the field. It moves in a circle with p

alukav5142 [94]

A) d. 10T

When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

$F=qvB$

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.

This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

$qvB = \frac{mv^2}{r}$

which can be rewritten as

$v=\frac{qB}{mr}$

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

$\frac{2\pi r}{T}=\frac{qB}{mr}$

So, we get:

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

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.

B) $a. f/10$

The frequency of revolution of a particle in uniform circular motion is

$f=\frac{1}{T}$

where

f is the frequency

T is the period

We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:

T' = 10 T

Then its frequency of revolution will be:

$f'=\frac{1}{T'}=\frac{1}{(10T)}=\frac{f}{10}$