Answer:
(a) The speed of the target proton after the collision is:
, and (b) the speed of the projectile proton after the collision is: 
.
Explanation:
We need to apply at the system the conservation of the linear momentum on both directions x and y, and we get for the x axle:
, and y axle:
. Now replacing the value given as: 
, 
for the projectile proton and according to the problem 
are perpendicular so 
, and assuming that 
, we get for x axle:
and y axle: 
, then solving for 
, we get:
and replacing at the first equation we get:
, now solving for 
, we can find the speed of the projectile proton after the collision as: and 
, that is the speed of the target proton after the collision.
Answer
B. X
A plate of glass or metal ruled with very close parallel lines, producing a spectrum by diffraction and interference of light.
diffraction grating is an optical component that splits and diffracts light into several beams travelling in different directions. The light emerges with its colours separated.
the water specific heat will remain at 4.184.
The temperature on the Earth would increase.
The temperature of the stratosphere would decrease.
Answer:
weight at height = 100 N .
Explanation:
The problem relates to variation of weight due to change in height .
Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .
At the surface :
Applying Newton's law of gravitation
mg₀ = G Mm / R²
At height h from centre
mg₁ = G Mm /h²
Given mg₀ = 400 N
400 = G Mm / R²
400 = G Mm / (6400 x 10³ )²
G Mm = 400 x (6400 x 10³ )²
At height h from centre
mg₁ = 400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²
= 400 / 4
= 100 N .
weight at height = 100 N
