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2 years ago

Can someone please help

Physics

1 answer:

Anna11 [10]2 years ago

3 0

The answer is B. Mono and polyunsaturated fats because the other two are really harmful and they are not healthy whatsoever.

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5) What is the net force caused by the moon acting on earth when the moon is 3.86x10^8 m away? The moon has a mass of 7.46x10^23

Dimas [21]

Answers:

5) $1.99(10)^{21} N$

6) $1.37(10)^{18} N$

7) $1.64(10)^{21} N$

8) $4.29(10)^{10} N$ more than Venus force of gravity on Pluto

Explanation:

According to Newton's law of Universal Gravitation, the force $F$ exerted between two bodies of masses $M$ and $m$ and separated by a distance $R$ is equal to the product of their masses and inversely proportional to the square of the distance:

$F=G\frac{Mm}{R^{2}}$ (1)

Where $G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

This is the equation we will use to solve each question in this problem.

<h3>5) Gravitational force between Earth and Moon</h3>

In this case we have:

$F_{earth-moon}$ is the gravitational force between Earth and Moon

$M=5.97(10)^{24} kg$ is the mass of the Earth

$m=7.46(10)^{23} kg$ is the mass of the Moon

$R=3.86(10)^{8} m$ is the distance between Earth and Moon

Solving:

$F_{earth-moon}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(5.97(10)^{24} kg)(7.46(10)^{23} kg)}{(3.86(10)^{8} m)^{2}}$ (2)

$F_{earth-moon}=1.99(10)^{21} N$ (3)

<h3>6) Gravitational force between Jupiter and Venus</h3>

Assuming for a moment that the planets are perfectly aligned and all are in the same orbital period, we can make a rough estimation of the distance between Jupiter and Venus, knowing the distance of each to the Sun:

distance between Sun and Jupiter - distance between Sun and Venus=distance between Jupiter and Venus= $R_{jupiter-venus}$ (4)

$R_{jupiter-venus}=778.3(10)^{9} m - 108(10)^{9} m=6.703(10)^{11} m$ (5)

Using this value in the Law of Universal Gravitation equation:

$F_{jupiter-venus}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.90(10)^{27} kg)(4.87(10)^{24} kg)}{(6.703(10)^{11} m)^{2}}$ (6)

$F_{jupiter-venus}=1.37(10)^{18} N$ (7)

<h3>7) Gravitational force between Saturn and Mars</h3>

Using the same assumption we made in the prior question:

distance between Sun and Saturn - distance between Sun and Mars=distance between Saturn and Mars= $R_{saturn-mars}$ (8)

$R_{saturn-mars}=1427(10)^{9} m - 227.9(10)^{9} m=227.9(10)^{9} m$ (9)

Using this value in the Law of Universal Gravitation equation:

$F_{saturn-mars}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(1.989(10)^{30} kg)(6.42(10)^{23} kg)}{(227.9(10)^{9} m)^{2}}$ (10)

$F_{saturn-mars}=1.64(10)^{21} N$ (11)

<h3>8) How much more is earths force of gravity on Pluto than Venus force of gravity on Pluto?</h3>

Firstly, we need to find $F_{earth-pluto}$ and then find $F_{venus-pluto}$ in order to find the difference.

<u>For $F_{earth-pluto}$ :</u>

$M=5.97(10)^{24} kg$ is the mass of the Earth

$m=1.46(10)^{22} kg$ is the mass of Pluto

$R_{earth-pluto}=5.7504(10)^{12} m$ is the distance between Earth and Pluto

$F_{earth-pluto}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(5.97(10)^{24} kg)(1.46(10)^{22} kg)}{(5.7504(10)^{12} m)^{2}}$ (12)

$F_{earth-pluto}=1.759(10)^{11} N$ (13) Force between Earth and Pluto

<u></u>

<u>For $F_{venus-pluto}$ :</u>

$M=4.87(10)^{24} kg$ is the mass of Venus

$m=1.46(10)^{22} kg$ is the mass of Pluto

$R_{venus-pluto}=5.792(10)^{12} m$ is the distance between Venus and Pluto

$F_{venus-pluto}=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}\frac{(4.87(10)^{24} kg)1.46(10)^{22} kg)}{(5.792(10)^{12} m)^{2}}$ (14)

$F_{venus-pluto}=1.33(10)^{11} N$ (15) Force between Venus and Pluto

Calculating the difference:

$F_{earth-pluto}-F_{venus-pluto}=1.759(10)^{11} N-1.33(10)^{11} N$

Finally:

$F_{earth-pluto}-F_{venus-pluto}=4.29(10)^{10} N$ (16)

Hence:

Earths force of gravity on Pluto is $4.29(10)^{10} N$ than Venus force of gravity on Pluto.

3 0

2 years ago

A person standing for a long time gets tired when he does not appear to do any work .why?

Keith_Richards [23]

Explanation:

A person standing still for a long time feels tired because the force of gravity acts on our body and puts stress on our muscles. so our muscles need energy to do work and keep body balanced and help to stand upright.

6 0

2 years ago

suppose the circumference of a bicycle wheel is 2 meters. if it rotates at 1 revolution per second when you are riding the bicyc

kirill115 [55]

Then the speed of the rider is, v = 2m/s

The linear velocity of an object in a straight line is its speed.

The following is the relationship between linear and angular velocity:

v = rω

The circumference of the bicycle wheel is,

2 $\pi$ r = 2

r = 0.318m

The angular velocity of the wheel is,

ω = 1rev/sec = 2 $\pi$ rad/s

Then the speed of the rider is,

v = rω

v = 0.318×2π

v = 2m/s

Learn more about angular velocity here:

brainly.com/question/12446100

#SPJ4

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1 year ago

Which test was conducted on the skin found under the victim’s nails

DENIUS [597]

. Underclothes; Bedding; Dirty laundry; Fingernail scrapings; Cups/bottles ... A cigarette butt found at a crime scene may contain valuable DNA material in the dried ... DNA evidence from both the victim's blood and the perpetrator's skin cells may ... DNA testing must be conducted in a laboratory with dedicated facilities and ...

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3 years ago

Read 2 more answers

Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1.38 m and finds that it makes 441

Tasya [4]

The period of a simple pendulum is given by:
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
$g= \frac{4 \pi^2}{T^2}L$ (1)

We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.

We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
$f= \frac{N}{t}= \frac{441}{1090 s}=0.40 Hz$
And its period is the reciprocal of its frequency:
$T= \frac{1}{f}= \frac{1}{0.40 Hz}=2.47 s$

So now we can use eq.(1) to find the gravitational acceleration of the planet:
$g= \frac{4 \pi^2}{T^2}L = \frac{4 \pi^2}{(2.47 s)^2} (1.38 m) =8.92 m/s^2$

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3 years ago