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scZoUnD [109]
3 years ago
15

While planets are smaller than stars, planets are generally larger than which of the following?

Physics
2 answers:
wel3 years ago
6 0

Answer:

The other bodies in a solar system

Explanation:

Stars are generally bigger in size than the objects orbiting around them. Star is the object that holds all the object of a star system in a definite shape and makes all the objects orbit it due to its gravity. Of all these objects few are designated as planets as per the criteria laid down by the International Astronomical Union (IAU). One of the criteria is: "The planets orbit around the star and clear their orbit i.e. they are the biggest object in that region".

By this logic it is pretty evident that they are bigger than other objects in the star system (or in our case the solar system).

Sauron [17]3 years ago
3 0

the answer was B

hope that helps you

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List out the fundamental and derived units​
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3 years ago
You drop an ice cube into an insulated flask full of water and wait for the ice cube to completely melt. The ice cube initially
3241004551 [841]

Answer:T=12.69^{\circ}C

Explanation:

Given

Mass of ice m=60 gm

mass of water M=760 gm

Initial Temperature of water T_i=20^{\circ}C

Let T be the Final Temperature of mixture

Latent heat of Fusion L=334 J/gm

heat required to melt ice completely is

Q_1=60\times 334=20.04 kJ

Heat released by water is taken by ice thus

Mc_{water}(20-T)=mL+mc_{water}(T-0)

0.76\times 4.184\cdot (20-T)=20.04+0.06\times 4.184\cdot (T)

T=12.69^{\circ}C

                       

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3 years ago
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7 0
3 years ago
Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.
Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge q_{1}= 3.8\times10^{-6}\ C

Second charge q_{2}=3.2\times10^{-6}\ C

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

E_{1}=E_{2}

\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}

\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}

\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}

x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}

Put the value into the formula

x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}

x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}

x=1.69\ m

Hence, The distance is 1.69 m.

5 0
3 years ago
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