when the ball hits the floor and bounces back the momentum of the ball changes.
the rate of change of momentum is the force exerted by the floor on it.
the equation for the force exerted is
f = rate of change of momentum

v is the final velocity which is - 3.85 m/s
u is initial velocity - 4.23 m/s
m = 0.622 kg
time is the impact time of the ball in contact with the floor - 0.0266 s
substituting the values

since the ball is going down, we take that as negative and ball going upwards as positive.
f = 189 N
the force exerted from the floor is 189 N
Answer:
<em>20 m/s in the same direction of the bus.</em>
Explanation:
<u>Relative Motion
</u>
Objects movement is always related to some reference. If you are moving at a constant speed, all the objects moving with you seem to be at rest from your reference, but they are moving at the same speed as you by an external observer.
If we are riding on a bus at 10 m/s and throw a ball which we see moving at 10 m/s in our same direction, then an external observer (called Ophelia) will see the ball moving at our speed plus the relative speed with respect to us, that is, at 20 m/s in the same direction of the bus.
Exothermich cools and endo heats so it heats
Answer:
0.56 atm
Explanation:
First of all, we need to find the number of moles of the gas.
We know that
m = 1.00 g is the mass of the gas
is the molar mass of the carbon dioxide
So, the number of moles of the gas is

Now we can find the pressure of the gas by using the ideal gas equation:

where
p is the pressure
is the volume
n = 0.023 mol is the number of moles
is the gas constant
is the temperature of the gas
Solving the equation for p, we find

And since we have

the pressure in atmospheres is

(a) 5.66 m/s
The flow rate of the water in the pipe is given by

where
Q is the flow rate
A is the cross-sectional area of the pipe
v is the speed of the water
Here we have

the radius of the pipe is
r = 0.260 m
So the cross-sectional area is

So we can re-arrange the equation to find the speed of the water:

(b) 0.326 m
The flow rate along the pipe is conserved, so we can write:

where we have

and where
is the cross-sectional area of the pipe at the second point.
Solving for A2,

And finally we can find the radius of the pipe at that point:
