While hiking in the mountains, tanya observed a narrow and steep valley with a stream at the bottom of it. what most likely caus
2 answers:
You might be interested in
Answer:
a) E = 1.58 10²¹ J , b) Oil = 4,236 107 liter , e) T = 54.3 C
Explanation:
a) To calculate the energy that reaches Earth, let us combine that the power emitted by the Sun is distributed uniformly on a spherical surface
I = P / A
A = 4π r²
in this case the radius of the sphere is the distance from the Sun to Earth r = 1.5 10¹¹ m
I = P / A
I = P / 4π r²
let's calculate
I = 3,828 10²⁵/4 pi (1.5 10¹¹)²
I = 1.3539 10²W / m² = 135.4 W / m2
the energy that reaches the disk of the Earth is
E = I A
the area of a disc
A = π r²
E = I π r²
where r is the radius of the Earth 6.37 10⁶ m
E = 135.4 π(6.37 10⁶)
E = 1,726 10¹⁶ W
This is the energy per unit of time that reaches Earth
t = 1 dai (24h / 1day) (3600s / 1h) = 86400 s
E = 1,826 10¹⁶ 86400
E = 1.58 10²¹ J
b) for this part we can use a direct proportions rule
Oil = 1.58 10²¹ (1 / 37.3 10⁶)
Oil = 4,236 10⁷ liter
c) to silence the surface temperature of the Earth we use the Stefan-Bolztman Law
P = σ A e T⁴
T =
nos indicate the refect, therefore the amount of absorbencies
P_absorbed = 0.7 P
let's calculate
T = REA (0.7 1.58 1021 / [pi (6.37 106) 2 1)
T = RER (8,676 106)
T = 54.3 C
b) Among the other factors that must be taken into account is the greenhouse effect, due to the absorption of gases from the atmosphere
The mass of water that must be raised is
Explanation:
Since the process is 70% efficiency, the power in output to the turbine can be written as
where is the power in input.
The power in input can be written as
where
W is the work done in lifting the water
t = 3 h = 10,800 s is the time elapsed
The work done in lifting the water is given by
where
m is the mass of water
is the acceleration of gravity
h = 45 m is the height at which the water is lifted
Combining the three equations together, we get:
Where
And solving for m, we find:
Learn more about power:
#LearnwithBrainly
Answer:
a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.
b) Average power
P(w)= 1062.07 [w]
P(hp)=1.42 [hp]
c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.
Explanation:
First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)
Work:
Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )
Converting from Joules to Kcals:
Now lets take into account the efficiency of the human body (20%)
2.537 ---> 20%
x ---> 100%
So the student is consuming 12.685 KCals each time he runs up the stairs.
Now,
1 g --> 9 Kcals
1000 g --> 9000KCals
Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:
He must run up the stairs 709.5 times, to burn 1 Kg of fat.
********************
For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)
*****