Answer:
new atmospheric pressure is 0.9838 × 
Pa
Explanation:
given data
height = 21.6 mm = 0.0216 m
Normal atmospheric pressure = 1.013 ✕ 10^5 Pa
density of mercury = 13.6 g/cm³
to find out
atmospheric pressure
solution
we find first height of mercury when normal pressure that is
pressure p = ρ×g×h
put here value
1.013 × = 13.6 × 10³ × 9.81 × h
h = 0.759 m
so change in height Δh = 0.759 - 0.0216
new height H = 0.7374 m
so new pressure = ρ×g×H
put here value
new pressure = 13.6 × 10³ × 9.81 × 0.7374
atmospheric pressure = 98380.9584
so new atmospheric pressure is 0.9838 × Pa
I am sitting in my seat.
I am listening to my mp3 and reading my book.
My eyes are getting heavy. They start to close.
I try to stay awake, but it's no use.
I am so warm and comfortable and sleepy,
and I have just finished my dinner.
Finally I can't help it. Resistance is futile.
I give up, and fall deep asleep.
My head rests back against my soft, comfy seat.
My seat is in row 26 on the airplane I'm flying in
to visit my grandmother on the coast.
We are cruising at 560 miles an hour, bearing 280°,
at flight level 320 .
The temperature outside my window is -60°F .
A) Agreed.
<span>b) Value agreed but units should be W (watts). </span>
<span>c) Here's one method... </span>
<span>15 miles = 24140 m </span>
<span>1 gallon of gasoline contains 1.4×10⁸ J. </span>
<span>So moving a distance of 24140m requires gasoline containing 1.4×10⁸ J </span>
<span>Therefore moving a distance of 1m requires gasoline containing 1.4×10⁸/24140 = 5800 J </span>
<span>Overcoming rolling resitance for 1m requires (useful) work = force x distance = 1000x1 = 1000J </span>
<span>So 5800J (in the gasoline) provides 1000J (overcoming rolling resistance) of useful work for each metre moved. </span>
<span>Efficiency = useful work/total energy supplied </span>
<span>= 1000/5800 </span>
<span>= 0.17 (=17%) </span>
solution:
radius of steel ball(r)=5cm=0.05m
density of ball =8000kgm
terminal velocity(v)=25m/s^2
density of air( d) =1.29 kgm
now
volume of ball(V)=4/3pir^3=1.33×3.14×0.05^3=0.00052 m^3
density of ball= mass of ball/Volume of ball
or, 8000=m/0.00052
or, m=4.16 kg
weight of the ball (W)= mg=4.16×10=41.6 N
viscous force(F)=6 × pi × eta × r × v
=6×3.14×eta×0.05×25
=23.55×eta
To attain the terminal velocity,
Fiscous force=Weight
or, 23.55× eta = 41.6
or, eta = 1.76
whete eta is the coefficient of viscosity.
