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1 year ago

7

Jenny puts a book on her desk. Jenny’s book has an area of 200 cm2.It exerts a pressure of 0.05 N/cm2 on the desk. What is the w

eight of the book? *

1 answer:

Luden [163]1 year ago

8 0

The weight is 10N :))

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What is the magnitude of the gravitational force acting on the sun due to the earth?The earth does not exert any gravitational f

Sholpan [36]

I haven't worked on Part-A, and I don't happen to know the magnitude of the gravitational force that the Sun exerts on the Earth.

But whatever it is, it's exactly, precisely, identical, the same, and equal to the magnitude of the gravitational force that the Earth exerts on the Sun.

I think that's the THIRD choice here, but I'm not sure of that either.

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2 years ago

A ray of light has a wavelength of

MArishka [77]

Answer:

The wavelength in vacuum is equal to 428.8 nm.

Explanation:

Given that,

The wavelength of light, $\lambda=284\ nm$

The refractive index of glass, n = 1.51

We need to find the wavelength in vacuum. The relation between wavelength and refractive index is given by :

$n=\dfrac{\lambda_v}{\lambda}\\\\\lambda_v=n\times \lambda\\\\\lambda_v=1.51\times 284\\\\\lambda_v=428.8\ nm$

So, the wavelength in vacuum is equal to 428.8 nm.

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2 years ago

Individuals with favorable traits are ________ in an environment

77julia77 [94]

Succesful is correct.

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2 years ago

Larger animals use proportionately less energy than smaller animals; that is, it takes less energy per kg to power an elephant t

marin [14]

Answer:

Part a)

$Q = 952 cal/day$

Part b)

$P = 46 Watt$

Part c)

$\Delta P = 54 W$

Explanation:

As we know that 5000 kg African elephant requires 70,000 Cal for basic needs per day

so we will have

$m = 5000 kg$

$Q = 70,000 Cal$

so we have energy required per kg

$E = \frac{70,000}{5000}$

$E = 14 cal/kg$

Part a)

now we know that per kg the energy required will be same

so we have mass of the human is 68 kg

so energy required per day is given as

$Q = 68 \times 14$

$Q = 952 cal/day$

Part b)

Resting power is the rate of energy in Joule required per sec

so it is given as

$P = \frac{952 \times 4186}{24\times 3600}$

$P = 46 Watt$

Part c)

resting power given in the book is

$P' = 100 W$

so this is less than the power given

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$\Delta P = 54 W$

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2 years ago

Use the ratio version of Kepler’s third law and the orbital information of Mars to determine Earth’s distance from the Sun. Mars

zhuklara [117]

Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.

The distance of the earth from the sun is $1.50 \times 10^{11}\;\rm m$ .

<h3>What is Kepler's third law?</h3>

Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

$T^2 \propto R^3$

Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.

By using Kepler's third law, this can be written as,

$T^2 \propto R^3$

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Substituting the values, we get the value of constant k for mars.

$687^2 = k\times (2.279 \times 10^{11})^3$

$k = 3.92 \times 10^{-29}$

The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.

$365^3 = 3.92\times 10^{-29} R^3$

$R^3 = 3.39 \times 10^{33}$

$R= 1.50 \times 10^{11}\;\rm m$

Hence we can conclude that the distance of the earth from the sun is $1.50 \times 10^{11}\;\rm m$ .

To know more about Kepler's third law, follow the link given below.

brainly.com/question/7783290.

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1 year ago