Shortening the conductor (B) <span> would decrease the resistance
to the flow of an electric current through a body.</span>
Using the conservation of momentum,
ma*va1 + mb*vb1 = ma*va2 + mb*vb2
Let:
ma = mass of the ball
va = velocity of the ball
mb = mass of the man
vb = velocity of the man
The subscript 1 is known as initials while 2 is for finals.
Before the man throws the ball, he starts at rest, meaning the initial velocity of the ball and the initial velocity of the man are zero. So
0 = ma*va2 + mb*vb2
Given ma = 10 kg; va = 20 m/s; mb = 90 kg; vb is unknown, therefore
-(mb*vb2) = ma*va2
vb2 = -(ma*va2)/mb2 = -(10*20)/90 = -2.22 m/s
Notice that his velocity is negative because when he finally throws the ball (say to the right), he moves at the opposite direction (that is to the left) on which he stands on the frictionless surface.
Answer:
v₂ = 63.62 m / s
Explanation:
For this exercise in fluid mechanics we will use Bernoulli's equation
P₁ + ρ g v₁² + ρ g y₁ = P₂ + ρ g v₂² + ρ g y₂
where the subscript 1 refers to the inside of the wing and the subscript 2 to the top of the wing.
We will assume that the distance between the two parts is small, so y₁ = y₂
P₁-P₂ = ρ g (v₂² - v₁²)
pressure is defined by
P = F / A
we substitute
ΔF / A = ρ g (v₂² - v₁²)
v₂² = 
suppose that the area of the wing is A = 1 m²
we substitute
v₂² =
v₂² = 79.10 + 3969
v₂ = √4048.1
v₂ = 63.62 m / s