answer = 33.4 net force.
Ball X has a mass of 8kg and is moving toward ball Y (which is sitting still) at 2m/s. After they collide, ball X is
sitting still. How fast is ball Y moving after the collision if it has a mass of 4kg?
Answer:
v = 4 m/s
Explanation:
Given :
Ball X :
Mass, m1 = 8kg ;
Initial Velocity, u1 = 2 m/s
Final velocity, v1 = 0
Ball Y:
Mass, m2 = 4kg ;
Initial Velocity, u2 = 0 m/s
Final velocity, v2 = v
(m1u1 + m2u2) = (m2v2 + m1v1)
(8*2 + 4*0) = (4*v + 4*0)
16 + 0 = 4v + 0
16 = 4v
v = 16 / 4
v = 4 m/s
Answer:
Explanation:
This is the Law of Momentum Conservation which for us looks like this:
[tex][m_xv_x+m_yv_y]_b=[m_xv_x+m_yv_y]_a[/tex] and that should look familiar to you if this is what you are doing in physics. Filling in our particular info:
[(8.0 × 2.0)+ (4.0 × 0.0)] = [(8.0 × 0.0) + (4.0v)] and
16 + 0 = 0 + 4.0v and
16 = 4.0v so
v = 4.0 in the direction of ball X
with respect to air the refractive index of ice is 1.31 and that of rock salt is 1.54 calculate the refractive index of rock salt with respect to ice
Answer:
1.17
Explanation:
Given that,
The refractive index of ice wrt air = 1.31
The refractive index of rock salt wrt air = 1.54
We need to find the refractive index of rock salt with respect to ice.
We know that,
refractive index = (speed of light in air or vaccum)/( speed of light in that medium)
So,
The speed of light in ice = c/(1.31)
The speed of light in rock salt = c/(1.54)
So, the refractive index of rock salt with respect to ice is :
[tex]\mu=\dfrac{1.54}{1.31}\\\\=1.17[/tex]
So, the required refractive index of rock salt wrt ice is 1.17.