Answer:
hope this helps!
Explanation:
Volume of the air bubble, V1=1.0cm3=1.0×10−6m3
Bubble rises to height, d=40m
Temperature at a depth of 40 m, T1=12oC=285K
Temperature at the surface of the lake, T2=35oC=308K
The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa
The pressure at the depth of 40 m: P1=1atm+dρg
Where,
ρ is the density of water =103kg/m3
g is the acceleration due to gravity =9.8m/s2
∴P1=1.103×105+40×103×9.8=493300Pa
We have T1P1V1=T2P2V2
Where, V2 is the volume of the air bubble when it reaches the surface.
V2=
They travel the way the wind is blowing and also towards the shoreline
Answer:
17.6 N
Explanation:
The force exerted by the punter on the football is equal to the rate of change of momentum of the football:
![F=\frac{\Delta p}{\Delta t}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7B%5CDelta%20p%7D%7B%5CDelta%20t%7D)
where
is the change in momentum of the football
is the time elapsed
The change in momentum can be written as
![\Delta p = m(v-u)](https://tex.z-dn.net/?f=%5CDelta%20p%20%3D%20m%28v-u%29)
where
m = 0.55 kg is the mass of the football
u = 0 is the initial velocity (the ball starts from rest)
v = 8.0 m/s is the final velocity
Combining the two equations and substituting the values, we find the force exerted on the ball:
![F=\frac{m(v-u)}{\Delta t}=\frac{(0.55)(8.0-0)}{0.25}=17.6 N](https://tex.z-dn.net/?f=F%3D%5Cfrac%7Bm%28v-u%29%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B%280.55%29%288.0-0%29%7D%7B0.25%7D%3D17.6%20N)
The momentum of ball is given by:
Since both have the same momentum, we have:
Number 3If you notice any mistake in my english, please let me know, because i am not native.