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lara31 [8.8K]
3 years ago
5

Suppose that the sun shrank in size but that its mass remained the same. What would happen to the orbit of the earth?

Physics
1 answer:
Minchanka [31]3 years ago
6 0

Answer:

Nothing

Explanation:

The radius of the orbit of the Earth does not depend on the radius of the sun.

In fact, the gravitational attraction between the Earth and the Sun provides the centripetal force that keeps the Earth in orbit:

G\frac{Mm}{r^2} = m\frac{v^2}{r}

where

G is the gravitational constant

M is the mass of the sun

m is the mass of the Earth

r is the radius of the orbit of the Earth

v is the orbital speed of the earth

Re-arranging the equation for r:

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

Also,

v=\omega r

where \omega is the angular velocity of the Earth's orbit. So we can rewrite the equation as

r=\frac{GM}{\omega^2 r^2}\\r^3 = \frac{GM}{\omega^2}

As we see, the radius of the orbit of the Earth, r, does not depend on the mass of the Sun, so if the sun shrank in size, the orbit remains the same.

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The recommended adult dose of elixophyllin, a drug used to treat asthma, is 6.00 mg/kg of body mass. calculate the dose in milli
andreev551 [17]

For a 166-lb person, a dose of 452 milligrams is prescribed to treat asthma.

<h3>How would you define asthma?</h3>

Lung damage is a side effect of asthma. Repeated episodes of coughing, dyspnea, chest tightness, and wheezing are also brought on by it. By taking medication and avoiding the triggers that can set off an attack, asthma can be managed. Outdoor allergens, such as pollen from grass, trees, and weeds, are common asthma triggers. Different allergens and irritants might function as triggers for different people. Dust mites, cockroaches, mold, and pet dander are examples of indoor allergies. Air irritants like smoke, chemical fumes, and powerful scents.

Mass = 166 lb×(453.59 g/1 lb)×(1kg/1000g)=75.30 kg

Dose = 75.30 kg ×(6.00 mg/1 kg)=452 mg

Elixophyllin has a 452 mg dose for a 166 lb person with asthma.

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8 0
2 years ago
A robot is exploring​ charon, the dwarf planet​ pluto's largest moon. Gravity on charon is 0.278 meters per second ​[m divided b
DochEvi [55]

In order to find the efficiency first we will find the Change in Potential energy of the small stone that robot picked up

First we will find the mass of the stone

As it is given that stone is spherical in shape so first we will find its volume

V = \frac{4}{3}\pi r^3

V = \frac{4}{3}\pi *(\frac{0.06}{2})^3

V = 1.13 * 10^{-4} m^3

Now it is given that it's specific gravity is 10.8

So density of rock is

\rho = 10.8 * 10^3 kg/m^3

mass of the stone will be

m = \rho V

m = 10.8* 10^3 * 1.13 * 10^{-4}

m = 1.22 kg

now change in potential energy is given as

\Delta U = mgH

here

g = gravity on planet = 0.278 m/s^2

H = height lifted upwards = 15 cm

\Delta U = 1.22* 0.278 * 0.15

\Delta U = 0.051 J

Now energy supplied by internal circuit of robot is given by

E = Vit

V = voltage supplied = 10 V

i = current = 1.83 mA

t = time = 12 s

E = 10* 1.83 * 10^{-3} * 12

E = 0.22 J

Now efficiency is defined as the ratio of output work with given amount of energy used

\eta = \frac{\Delta U}{E}*100

\eta = \frac{0.051}{0.22} = 0.23

so efficiency will be 23 %

5 0
3 years ago
If earth increase the distance from the sun, what will happen to the period of orbi t(the time it takes to complete one revoluti
Mandarinka [93]

The period of the orbit would increase as well

Explanation:

We can answer this question by applying Kepler's third law, which states that:

"The square of the orbital period of a planet around the Sun is proportional to the cube of the semi-major axis of its orbit"

Mathematically,

\frac{T^2}{a^3}=const.

Where

T is the orbital period

a is the semi-major axis of the orbit

In this problem, the question asks what happens if the distance of the Earth from the Sun increases. Increasing this distance means increasing the semi-major axis of the orbit, a: but as we saw from the previous equation, the orbital period of the Earth is proportional to a, therefore as a increases, T increases as well.

Therefore, the period of the orbit would increase.

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HElP I NEED TO turn this in
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