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nordsb [41]
3 years ago
7

Alkaline earth metals have a low density true false

Physics
1 answer:
Marta_Voda [28]3 years ago
6 0

true

Explanation:

this is because melting point and boiling point decreases down the group because they are held together by attractions between positive nuclei and delocalised electrons

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minimum speed v=\sqrt({Fr}/m)}

Explanation:

Recall the formula for centripetal force;

Centripetal force is the force that is required to keep an object moving in circular part

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where;

F=centripetal force

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What is modulus of elasticity for shearing stress and strain called?
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The center of a moon of mass m is a distance D from the center of a planet of mass M. At some distance x from the center of the
nataly862011 [7]

Answer with Explanation:

Let  rest mass m_0 at point P  at  distance x from center of the planet, along a line connecting the centers of planet and the moon.

Mass of moon=m

Distance between the center of moon and center of planet=D

Mass of planet=M

We are given that net force on an object will be zero

a.We have to derive an expression for x in terms of m, M and D.

We know that gravitational force=\frac{GmM}{r^2}

Distance of P from moon=D-x

F_m=Force applied on rest mass due to m

F_m=Force on rest mass due to mas M

F_M=F_m because net force is equal to 0.

F_m=F_M

\frac{Gm_0m}{(D-x)^2}=\frac{Gm_0M}{x^2}

\frac{m}{(D-x)^2}=\frac{M}{x^2}

\frac{x^2}{(D-x)^2}=\frac{M}{m}

\frac{x}{D-x}=\sqrt{\frac{M}{m}}

Let R=\sqrt{\frac{M}{m}}

Then, \frac{x}{D-x}=R

x=DR-xR

x+xR=DR

x(1+R)=DR

x=\frac{DR}{1+R}

b.We have to find the ratio R of the mass of the mass of the planet to the mass of the moon when x=\frac{2}{3}D

Net force is zero

F_m=F_M

\frac{Gm_0m}{(D-\frac{2}{3}D)^2}=\frac{Gm_0M}{\frac{4}{9}D^2}

\frac{m}{\frac{D^2}{9}}=\frac{9M}{4D^2}

\frac{M}{m}=4

Hence, the ratio R of the mass of the planet to the mass of the moon=4:1

8 0
4 years ago
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