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Nutka1998 [239]
2 years ago
8

How many days of a diet of 1910 large calories are equivalent to the gravitational energy change from sea level to the top of Mo

unt Everest, 8848 m above sea level? Assume your mass is 53 kg. (The body is not anywhere near 100% efficient in converting chemical energy into change in altitude. Also note that this is in addition to your basal metabolism.)
Physics
1 answer:
andreyandreev [35.5K]2 years ago
3 0

Answer:

0.58 days is needed

Explanation:

Gravitational energy change from a height change equal to 8848 m for a mass of 53 kg is:

m*g*h = 53*9.81*8848 = 4600340.64 J = 4600.34 kJ

4.184 kJ are equivalent to 1 kcal, then:

4.184 kJ / 4600.34 kJ = 1 kcal / x kcal

x = 4600.34/4.184

x = 1099.5 kcal or large calories

The diet is 1910 kcal/day, then you need 1099.5/1910 = 0.58 days

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What has been happening to the cosmic microwave background radiation since the Big Bang?​
kondor19780726 [428]

Answer:

Explanation:

Cosmologists refer to a "surface of last scattering" when the CMB photons last hit matter; after that, the universe was too big. So when we map the CMB, we are looking back in time to 380,000 years after the Big Bang, just after the universe was opaque to radiation. But the CMB was first found by accident.

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3 years ago
1) Should I take vitamin D supplements in the winter?
iren2701 [21]
Yes you need the light or just go outside to get it from the sun
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How does upwelling affect the weather of a coastal region?
SIZIF [17.4K]

c.The warm surface water results in moist air and more rainfall.

Explanation:

  • During upwelling, cold water in the ocean is stirred up and brought to the surface.
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3 years ago
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How to solve 30(a)<br><br> Give solution asap.
expeople1 [14]

Answer:

They will not meet

h-hX=1.2*g*t²

hX=v0*t-(1/2*g*t²)

Explanation:

fall h=1/2*g*t²

elevation time if v0=20 m/s  te=v0/g=20 m/s /9.81 m/s²=2.0387s

hmax=v0²/(2*g)=(400 m²/s²)/19.62 m/s²2=20.387 m

free fall

t=2.0387s yields hX=1/2*g*t²=20.387 m

h-hX=200m - 20.387 m=179,613 m.

so, the second body has not enough initianoal speed to reach a meeting point

5 0
3 years ago
A piano string having a mass per unit length equal to 4.70 10-3 kg/m is under a tension of 1 400 N. Find the speed with which a
sweet [91]

Answer:

The speed of the sound wave on the string is 545.78 m/s.

Explanation:

Given;

mass per unit length of the string, μ = 4.7 x 10⁻³ kg/m

tension of the string, T = 1400 N

The speed of the sound wave on the string is given by;

v = \sqrt{\frac{T}{\mu} }

where;

v is the speed of the sound wave on the string

Substitute the given values and solve for speed,v,

v = \sqrt{\frac{T}{\mu} }\\\\v = \sqrt{\frac{1400}{4.7*10^{-3}} }\\\\v = \sqrt{297872.34}\\\\v = 545.78 \ m/s

Therefore, the speed of the sound wave on the string is 545.78 m/s.

3 0
2 years ago
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