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Nutka1998 [239]
3 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]3 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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Answer:

F₂= 210 pounds

Explanation:

Conceptual analysis

Hooke's law

Hooke's law establishes that the elongation (x) of a spring is directly proportional to the magnitude of force (F) applied to it, provided that said spring is not permanently deformed:

F= K*x   Formula (1)

Where;

F  is the magnitude of the force applied to the spring in Newtons (Pounds)

K is the elastic spring constant, which relates force and elongation. The higher its value, the more work it will cost to stretch the spring. (Pounds/inch)

x the elongation of the spring (inch)

Data

The data given is incorrect because if we apply them the answer would be illogical.

The correct data are as follows:

F₁ =80 pounds

x₁= 8 inches

x₂= 21  inches

Problem development

We replace data in formula 1 to calculate  K :

F₁= K*x₁

K=( F₁) / (x₁)

K=( 80) / (8) = 10 pounds/ inche

We apply The formula 1 to calculate  F₂

F₂= K*x₂

F₂= (10)*(21)

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8 0
3 years ago
A 3.50 cm tall object is held 24.8 cm from a lens of focal length 16.0 cm. What is the image height?
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Answer:

<h2>6.36 cm</h2>

Explanation:

Using the formula to first get the image distance

1/f = 1/u+1/v

f = focal length of the lens

u = object distance

v = image distance

Given f = 16.0 cm, u = 24.8 cm

1/v = 1/16 - 1/24.8

1/v = 0.0625-0.04032

1/v = 0.02218

v = 1/0.02218

v = 45.09 cm

To get the image height, we will us the magnification formula.

Mag = v/u = Hi/H

Hi = image height = ?

H = object height = 3.50 cm

45.09/24.8 = Hi/3.50

Hi = (45.09*3.50)/24.8

Hi = 6.36 cm

The image height is 6.36 cm

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