Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The forces along the line joining the centre of the two objects.
❍ Let us consider two masses m1 and m2 line at a separation distance d. Let the force of attraction between the two objects be F.
According to universal law of gravitation,
Also,
Combining both, We will get
Or, We can write it as,
Where, G is the constant of proportionality and it is called 'Universal Gravitational constant'.
☯️ Hence, derived !!
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<u>Answer</u>
Mean
<u>Explanation</u>
For any set of date there is a typical value for the probability distribution called central tendency.
There three types of central tendency<em> mean, median and mode.</em>
Mean is the average value of all data set in a distribution.
median is the the central value when the data is arranged in ascending or descending order.
Mode is the value that occurs most in a set of data.
Using the formula v=f times lambada
then v=the speed of light.
and f=what’s we’re looking for
and lambada=the wavelength.
so then you sub what you have (v and lambada) in the formula.
then multiply the frequency(f) by the given wavelength and then solve for f
Answer:
<em>The period of the motion will still be equal to T.</em>
<em></em>
Explanation:
for a system with mass = M
attached to a massless spring.
If the system is set in motion with an amplitude (distance from equilibrium position) A
and has period T
The equation for the period T is given as
where k is the spring constant
If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.
Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, <em>increasing the amplitude has no effect on the period of the mass and spring system.</em>
