Question

State any two limitations of a computer system.​

Answer 1

Answer:

A planet 'x' has mass 2 times and radius 3 times than that of Earth. What is the acceleration due to gravity on the planet,if acceleration due to gravity on Earth is 10m/s²?

EXPLANATION-:

We know that-:

\leadsto \boxed{\underline{\tt\dag\;\; g_{(earth)}=\frac{G\;\times m_{(earth)}}{R_{(earth)}^2\;\;}\dag}}⇝

†g

(earth)

=

R

(earth)

2

G×m

(earth)

†

Now in the planet "x" -:

\begin{gathered}\bf\; g=g'\\\\mass=2m_{(earth)}\\\\\bf\; Radius=3R_{(earth)}\end{gathered}

g=g

′

mass=2m

(earth)

Radius=3R

(earth)

Now putting values in the formula we have-:

\begin{gathered}\rightarrow \bf\; g'=\frac{G2m_{(earth)}}{(3R_{(earth)})^2}\\\\\rightarrow \bf\; 'g'=\frac{2}{9}\frac{Gm_{(earth)}}{R_{(earth)}^2}\\\\\bf\; We\;know\;that\;\frac{Gm_{(earth)}}{R_{(earth)}^2}\;is;g_{(earth)}\\\\\rightarrow \bf\; g'=\frac{2g}{9}\\\\\rightarrow \bf\; \boxed{\tt g'=2.22\;m/s}\end{gathered}

→g

′

=

(3R

(earth)

)

2

G2m

(earth)

→

′

g

′

=

9

2

R

(earth)

2

Gm

(earth)

Weknowthat

R

(earth)

2

Gm

(earth)

is;g

(earth)

→g

′

=

9

2g

→

g

′

=2.22m/s

So acc. due to gravity is 2.22 m/s²

Answer 2

Hqi how are you have a nice day