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Sliva [168]
3 years ago
8

A typical human consumes 2500 Kcal of energy during a day. This is the equivalent to 10,450,000 J! Say you decided to run stairs

all day. Given that there are 86,400 seconds / day, how much energy, in Joules, would you burn in climbing stairs all day?
Physics
1 answer:
defon3 years ago
8 0

Answer:

2940.1 joules  would you burn in climbing stairs all day.

Explanation:

Work = W = F\times d

going up stairs  would be against force of gravity

W = mgh

where h is the  height

the question is not complete because we need speed or distance

h =  v \times t

so assuming 1 step per second

h = 86,400 steps  \times 7inchs/step \times 0.0254 m/inch

h = 15362 m

so from this    

W = 800 N \times 15362

   = 12289600 J  

that means YOU  need 12289600 J to walk 1 step per second all day

divide that by 4180 J /Kcal  

Kcal =  \frac{W}{(J/Kcal)}

       = \frac{12289600}{4180}

       = 2940.1 Kcal

if you ran faster you would use more energy  2 steps per second would mean 5880 Kcal.

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Harlamova29_29 [7]

Answer:

This could represent something like sliding a small rock across an icy lake.

Explanation:

A 20N force of gravity (weight), and 20N normal force exerted back onto the object imply it is on the ground and has no vertical motion. There is a net force of 0N

An 80N force to the left and a 5N force to the right create a net force of 75N to the left. This means that there is a force acting on the object that makes it accelerate to the left. 80N represents a push or pull force and 5N represents a relatively small frictional force due to the object being slid on a surface such as steel or in this case ice.

8 0
3 years ago
Suppose you have three springs with force constants of k1 = k2 = k3 = 3.70 x 10^3 N/m. What is their effective force constant if
disa [49]

Answer:

The effective force constant is 1233.33 N/m.

Explanation:

It is given that,

Force constant 1, k_1=3.7\times 10^3\ N/m

Force constant 2, k_2=3.7\times 10^3\ N/m

Force constant 3, k_3=3.7\times 10^3\ N/m

The effective force constant if one is hung from the other in series is given by :

\dfrac{1}{K_{eff}}=\dfrac{1}{k_1}+\dfrac{1}{k_2}+\dfrac{1}{k_3}

\dfrac{1}{K_{eff}}=\dfrac{1}{3.7\times 10^3}+\dfrac{1}{3.7\times 10^3}+\dfrac{1}{3.7\times 10^3}

K_{eff}=1233.33\ N/m

So, the effective force constant is 1233.33 N/m. Hence, this is the required solution.

6 0
3 years ago
You drop your cell phone. Prior to hitting the ground, the phone's kinetic energy will ________ and its potential energy will __
SVETLANKA909090 [29]

Answer:

The kinetic energy of the phone would increase. The gravitational potential energy of the phone would decrease.

Explanation:

The kinetic energy {\rm KE} of an object is proportional to the square of the speed of that object. If air resistance is negligible, the phone would accelerate under gravitational pull and speed up. Hence, the kinetic energy of the phone would increase.

The gravitational field near the surface of the earth is approximately constant. Hence, the gravitational potential energy {\rm GPE} of the phone would be proportional to its height. As the phone approaches the ground, the height of the phone becomes lower and the gravitational potential energy of the phone would decrease.

5 0
1 year ago
A water molecule consists of an oxygen atom with two hydrogen atoms bound to it. The angle between the two bonds is 106◦ . If ea
balu736 [363]

Answer:

easy 16

Explanation:

because i love jesus

3 0
3 years ago
Biologists have studied the running ability of the northern quoll, a marsupial indigenous to Australia. In one set of experiment
Drupady [299]

Answer:

u_K=0.862

Explanation:

The force of friction between the quails feet and the ground is:

F=m*a

F_K=m*a

F_K=u_k*m*g

u_K*m*g=m*a_c

u_K*g=a

u_K=\frac{a_c}{g}

a_c=\frac{v^2}{r}

So the coefficient of static is solve

u_K=\frac{\frac{v^2}{r}}{g}

u_K=\frac{v^2}{r*g}=\frac{(2.6m/s)^2}{0.80m*9.8m/s^2}

u_K=0.862

4 0
3 years ago
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