Answer:
v' = 2.4 m/s
Explanation:
Given that,
Mass of one skater, m = 60 kg
Mass of the other's skater, m' = 60 kg
The two skaters push off each other. After the push, the smaller skater has a velocity of 3.0 m/s.
When there is no external force acting on a system, the momentum remains conserved. It means initial momentum is equal to the final momentum. Let v' is the velocity of the larger skater.
mv = m'v'

So, the velocity of the larger skater is 2.4 m/s.
Complete Question
In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?
Answer:
The speed of the helicopter is 
Explanation:
From the question we are told that
The height at which he let go of the brief case is h = 130 m
The time taken before the the brief case hits the water is t = 6 s
Generally the initial speed of the briefcase (Which also the speed of the helicopter )before the man let go of it is mathematically evaluated using kinematic equation as
Here s is the distance covered by the bag at sea level which is zero
=>
=> 
=> 
it’s B (because 20 Is smaller than 50)
Answer:
Explanation:
magnetic field B = (3 i + 8 x 2 j )x 10⁻³ T
= (3 i + 16 j )x 10⁻³ T
L = - i ( unit length of conductor )
Force F = I ( L x B ) , I is current
= 5 [ - i x ( 3i + 16 j ) 10⁻³]
= 5 ( - 16 k ) x 10⁻³
F = - 80 x 10⁻³ k