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laila [671]
3 years ago
7

The USDA recommends that women consume about 2000 Calories per day and men consume about 2500 Calories per day. How much average

power use (in Watts) does this imply for a human body
Physics
1 answer:
professor190 [17]3 years ago
4 0

Answer:

   1500 W to 2200 W

Explanation:

Every person does work in his daily day to day life. A person needs energy in order to perform work. And the energy consumed by an individual while performing a daily work is directly responsible to the mount of oxygen consumed by the person.

The USDA is the federal agency which looks after the food and agriculture matters of the US government. It deals with and formulates different policies and laws for the country and it s people. It recommends about 2000 calories per day for women and for men it recommends about 2500 calorie per days of food intake.

Accordingly, the average power required by a human body for doing regular work is in the range of 1500 W to 2200 W.

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Which image shows an example of the strong nuclear force in action?
dusya [7]

i think this is the incomplete page that you are showing but the answer is

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<h2><u>B</u></h2>
6 0
3 years ago
A projectile is fired vertically from Earth's surface with
scoray [572]

Answer:

h=25.52\times10^6 m

Explanation:

Initial speed, v = 10 x 10^3 m/s

Mass of the earth, M = 6 x 10^24 kg

Radius of the earth, R = 6.4 x 10^6 m

Maximum from the surface of earth, h = ?

Let  m = Mass of the projectile

Solution:

Potential energy at maximum height =  ( Potential + Kinetic energy ) at the surface

-G M m / ( R + h )=- G M m / R + (1/2) m v^2

- G M / ( R + h ) = - G M / R + (1/2) v^2

-2\times G M / ( R + h ) = ( - 2 G M / R ) + v^2

-2\times6.67\times10^{-11}\times6\times10^{24}/ ( R + h )

=( (- 2\times 6.67\times10^{-11}\times6\times10^{24}) /(6.4\times10^6)} +10000^2

=-2.50625\times10^7 J

=- 8\times10^{14} / ( R + h )=-2.50625\times 10^7

R+h=31.92\times10^{6}

h=31.92\times10^{6}-6.4\times10^6

h=25.52\times10^6 m

5 0
3 years ago
An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef
navik [9.2K]

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.  

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

\sum F_x = \sum F_y = 0

\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }

  • So; On the L.H.S

\mathbf{mg sin \theta =f_s}

\mathbf{mg sin \theta =\mu_s N}

  • On the R.H.S

N = mg cos θ

Equating both force component together, we have:

\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}

\mathbf{sin \theta =\mu_s \ \ cos \theta}

\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}

From trigonometry rule:

\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}

∴

\mathbf{\mu_s =\tan \theta}}

\mathbf{\mu_s =\tan 35^0}}

\mathbf{\mu_s = 0.700}}

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

8 0
3 years ago
On the Moon, the acceleration due to the effect of gravity is only about 1/6 of that on Earth. An astronaut whose weight on Eart
Ulleksa [173]
A = 94.22 Newtons

b = 58.16 kg

Gravity on the moon is 1.62 m/s^2
8 0
4 years ago
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Andrew [12]

Answer:

AAAAAAAAHHHDHSFJSBDKFANEDASNXDEHAHDBWEEEEHJDBHBAH

AAHHH YES THEY DOO AAAHHH EEHSFJWEHASAAAAHJEHER

7 0
3 years ago
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