Answer:
By decreasing the surface area exposed to the air. Vertical body position instead of horizontal.
Explanation:
So the terminal velocity formula is as follows.
v = sqrt[(2*m*g)/(p*A*c)]
m = mass
g = 9.81 m/s^2
p = density of air
A = surface area of object
c = Drag coefficient
So the only thing you can change in mid-air is surface area.
Work = (weight) x (distance)
Work = (50 lb) x (1 kg / 2.20462 lb) x (9.81 newton/kg)
x (4 feet) x (1 meter / 3.28084 feet)
= (50 x 9.81 x 4) / (2.20462 x 3.28084) newton-meter
= 271.3 joules .
We don't need to know how long the lift took, unless we
want to know how much power he was able to deliver.
Power = (work) / (time)
= (271.3 joule) / (5 sec) = 54.3 watts .
________________________________________
The easy way:
Work = (weight) x (distance)
= (50 pounds) x (4 feet) = 200 foot-pounds
Look up (online) how many joules there are in 1 foot-pound.
There are 1.356 joules in 1 foot-pound.
So 200 foot-pounds = (200 x 1.356) = 271.2 joules.
That's the easy way.
Explanation:
For the reaction.......
H
2
O
(
l
)
+
Δ
→
H
2
O
(
g
)
WHERE BOTH PRODUCT AND REACTANT ARE AT
100
∘
C
,
..........
Δ
H
∘
vaporization
=
40.66
⋅
k
J
⋅
m
o
l
−
1
And thus we need to assess the molar quantity of the water vaporized......
Δ
H
rxn
=
5.00
⋅
g
18.01
⋅
g
⋅
m
o
l
−
1
×
40.66
⋅
k
J
⋅
m
o
l
−
1
=11.3 kJ
Answer:
I don't know this answer questions
