Answer:
f.The period is independent of the suspended mass.
Explanation:
The period of a pendulum is given by

where
L is the length of the pendulum
g is the acceleration due to gravity
From the formula, we see that:
1) the period of the pendulum depends only on its length, L, and it is proportional to the square root of the length
2) the period does not depend neither on the mass of the pendulum, nor on its amplitude of oscillation
So, the only correct statements are
f.The period is independent of the suspended mass.
Note: statement "e.The period is proportional to the length of the wire" is also wrong, because the period is NOT proportional to the length of the wire, but it is proportional to the square root of it.
Temperature is how hot or cold something is,barometric is air pressure(when the pressures high,the weather is dry)humidity is how moist the air is and how many water particles are there, wind speed and direction is how the hot and cold air is moving and how fast,which makes wind (high to low, hot to cold) precipitation is how much it is raining at that point.
They were formed in the nuclear<span> fusion reaction inside older </span><span>stars.
As a star burns, fusion reactions inside its core create heavier elements. Those materials are released when the star dies of old age in an explosion.</span>
Answer:
The minimum value of width for first minima is λ
The minimum value of width for 50 minima is 50λ
The minimum value of width for 1000 minima is 1000λ
Explanation:
Given that,
Wavelength = λ
For D to be small,
We need to calculate the minimum width
Using formula of minimum width


Where, D = width of slit
= wavelength
Put the value into the formula

Here,
should be maximum.
So. maximum value of
is 1
Put the value into the formula


(b). If the minimum number is 50
Then, the width is


(c). If the minimum number is 1000
Then, the width is


Hence, The minimum value of width for first minima is λ
The minimum value of width for 50 minima is 50λ
The minimum value of width for 1000 minima is 1000λ