On planet X, an object weighs 7.04 N. Onplanet B where the magnitude of the free-fallacceleration is 1.91g(whereg= 9.8 m/s2isthe
1 answer:
You might be interested in
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
Answer:
The final acceleration becomes (1/3) of the initial acceleration.
Explanation:
The second law of motion gives the relationship between the net force, mass and the acceleration of an object. It is given by :
m = mass
a = acceleration
According to given condition, if the mass of a sliding block is tripled while a constant net force is applied. We need to find how much does the acceleration decrease.
Let a' is the final acceleration,
m' = 3m
So, the final acceleration becomes (1/3) of the initial acceleration. Hence, this is the required solution.