A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of
1 answer:
Answer:
Change in momentum is 1.1275 kg-m/s
Explanation:
It is given that,
Mass of the ball, m = 274 g = 0.274 kg
It hits the floor and rebounds upwards.
The ball hits the floor with a speed of 2.40 m/s i.e. u = -2.40 m/s (-ve because the ball hits the ground)
It rebounds with a speed of 1.7 m/s i.e. v = 1.7 m/s (+ve because the ball rebounds in upward direction)
We have to find the change in the ball's momentum. It is given by :
So, the change in the momentum is 1.1275 kg-m/s
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Answer:
2
Explanation:
We know that in the Fraunhofer single-slit pattern,
maxima is given by
Given values
θ=2.12°
slit width a= 0.110 mm.
wavelength λ= 582 nm
Now plugging values to calculate N we get
Solving the above equation we get
we N= 2.313≅ 2
By holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. Therefore, the direction of the force on the charge you are holding will be to the southwest.
Let I hold the charge , q at the centre of given co-ordinate system and two positive charge of equal magnitude Q are placed 1 m to my North and 1 m to my South .
now, both the charge are same nature e.g., positive . Let my charge is also positive (well, you can assume negative too , I am considering positive because it makes me easy to solve) then, both charge repel to my charge.
charge Q placed on east is repelling my charge q toward west . similarly charge Q placed on North is repelling my charge q toward south.
Now , use vector for solve it.
vector = vector Fe + vector Fn,
⇒ || =
⇒ Fe = Fs = KqQ/(1m)² = KqQ
⇒ = √{Fe² + Fs²} = √{(kqQ)²+(KqQ)²}
⇒ = √2KqQ
Hence, net force act on q {my charge } is √2KqQ and the direction of force is S - W (southwest )direction.
To learn more about positive charges here
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