Answer:
W / A = 39200 kg / m²
Explanation:
For this problem let's use the equilibrium equation of / newton
F = W
Where F is the force of the door and W the weight of water
W = mg
We use the concept of density
ρ = m / V
m = ρ V
The volume of the water column is
V = A h
We replace
W = ρ A h g
On the other side the cylinder cover has a pressure
P = F / A
F = P A
We match the two equations
P A = ρ A h g
P = ρ g h
P = 39200 Pa
The weight of the water column is
W = 1000 9.8 4 A
W / A = 39200 kg / m²
It doesn't because when u threw it the first time, u notice that the ball eventually came to a stop because of the force that was acting upon it. Although when u throw it harder it will start out faster than the first time u threw it because u put more kinetic energy onto the ball. But the same thing happens with this ball that happened to the second ball, they both have a type of force acting upon them.
Answer:
424.26 m/s
Explanation:
Given that Two air craft P and Q are flying at the same speed 300m/s. The direction along which P is flying is at right angles to the direction along which Q is flying. Find the magnitude of velocity of the air craft P relative to air craft Q
The relative speed will be calculated by using pythagorean theorem
Relative speed = sqrt(300^2 + 300^2)
Relative speed = sqrt( 180000 )
Relative speed = 424.26 m/s
Therefore, the magnitude of velocity of the air craft P relative to air craft Q is 424.26 m/s
mass of the ball m = 0.63 kg
initial height h = 1.8 m
final height h ' = 3.03 m
initial speed v = 7.09 m / s
final speed v ' = 4.21 m / s
Let the work done on the ball by air resistance W = ?
we know from law of conservation of energy ,
total energy at height h + work done by air = total energy at height h '
mgh + ( 1/ 2) mv^ 2 + W = mgh ' + ( 1/ 2) mv'^ 2
0.630*9.8*1.8 + 0.63*7.09^2 + W = mgh ' + ( 1/ 2) mv'^ 2
From there you can find W
if there is negative sign indicates it work opposite direction to motion
