The final velocity of the ball that is dropped from a height of 200m is v = 44.73 m/s .
<h3>What is velocity with example?</h3>
The rate at which an object is travelling in one direction is referred to as its velocity. an automobile traveling north on a highway, or a rocket taking off. Its velocity vector's absolute value always is equal to the motion's speed because it is a scalar.
<h3>Briefing:</h3>
Given the initial velocity of the ball (u) = 0
Distance travelled by the ball (s) = 200m
Acceleration (a) = 10 m/s²
As we know:
v² = u² + 2as
Putting values:
v² = 0+2 × (10 m/s²) × (200 m)
v = 44.73 m/s.
To know more about Velocity visit:
brainly.com/question/18084516
#SPJ9
Answer:
When balloon moves in the downward direction two forces acts on it.
i) Force exerted by air in the upward direction
ii) Weight
According to newton’s second law of motion:
Sum of forces = Ma
W – F = Ma
Mg – F = Ma …….. (i)
when some of the mass m is dropped and balloon is moving in upward direction with acceleration a/2 then,
F – W = (M-m)a/2
F – (M-m)g = (M-m)a/2
F – Mg + mg = Ma/2 – ma/2 ….. (ii)
Adding equation (i) and (ii)
mg = M(3a/2) – ma/2
m(g + a/2) = M(3a/2)
m = M(3a/2)/(g + a/2)
The height reached by the two carts after collision is determined as 5.34 m.
<h3>
Initial velocity of Cart A</h3>
Apply the principle of conservation of mechanical energy.
K.E = P.E
v = √2gh
v = √(2 x 9.8 x 12)
v = 15.34 m/s
<h3>Final velocity of the two carts after the collision</h3>
Apply the principle of conservation of linear momentum for inelastic collision.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
8(15.34) + 4(0) = v(8 + 4)
122.72 = 12v
v = 10.23 m/s
<h3>Height reached by both carts</h3>
Apply the principle of conservation of mechanical energy.
P.E = K.E
mgh = ¹/₂mv²
h = v²/(2g)
h = (10.23²) / (2 x 9.8)
h = 5.34 m
Learn more about linear momentum here: brainly.com/question/7538238
#SPJ1
Drier air Overland then water BR &/BR are the answers so the second one the second I’m in the first option
An object's gravitational potential energy is
(mass) x (gravity) x (height above ground) .
I don't see the object's speed anywhere in that formula, do you ?
An object's speed has no effect whatsoever on its potential energy ... only if it changes the object's height above ground.