Question

two object of masses 40kg are placed at a distance of 4m calculated the gravitation form of attraction​

Answer 1

Answer:

Given : m

1

=80 kg m

2

=1200 kg d=10 m

Force of gravitation F=

d

2

Gm

1

m

2

Or F=

10

2

6.67×10

−11

×80×1200

=6.4×10

−8

N

Answer 2

Required Solution:

Given:

• Mass of the 1st object $\sf {(m_1)}$ = 40kg
• Mass of the 2nd object $\sf {(m_2)}$ = 40kg
• Distance (r) = 4m

To calculate:

• Force between them (F)

Calculation:

We know that,

• ${\underline {\boxed {\Large {\sf { F = \dfrac{Gm_1m_2}{ {r}^{2} } } }}}}$

Where,

→ F = Force of attraction

→ G = Universal gravitation constant

(Value of G is $\sf { 6.67 \times {10}^{-11} }$)

→$\sf {(m_1)}$ = Mass of the 1st object

→$\sf {(m_2)}$ = Mass of the 2nd object

Now, substitute the values:

$\sf {⇢ \dfrac{6.67 \times {10}^{-11} \times 40 \times 40}{ {(4)}^{2} } }$

$\qquad$

$\sf {⇢ \dfrac{6.67 \times {10}^{-11} \times \cancel{40} \times \cancel{40}}{ \cancel{4} \times \cancel{4} } }$

$\qquad$

$\sf {⇢ 6.67 \times {10}^{-11} \times 10 \times 10 }$

$\qquad$

$\sf {⇢ 6.67 \times {10}^{-11} \times {10}^{2} }$

$\qquad$

$\sf {⇢ 6.67 \times {10}^{(-11+2)} }$

$\qquad$

$\sf {⇢ 6.67 \times {10}^{-9} }$ $\purple { \bigstar}$

$\qquad$

Now, we can also write it in usual form:

$\sf {⇢ 6.67 \times {10}^{-9}N }$

$\qquad$

$\sf {⇢ \dfrac{667}{100} \times \dfrac{1}{ {10}^{-9}} }$

$\qquad$

$\sf {⇢ \dfrac{667}{ {10}^{2} \times {10}^{9} } }$

$\qquad$

$\sf {⇢ \dfrac{667}{ {10}^{(2+9)} } }$

$\qquad$

$\sf {⇢ \dfrac{667}{ {10}^{11} } }$

$\qquad$

$\sf {⇢ \dfrac{667}{ 100000000000 } }$

$\qquad$

$\boxed {\mathfrak \pink {⇢ 0.00000000667 N }}$

$\qquad$

Therefore, $\sf { 0.00000000667 N }$ or $\sf { 6.67 \times {10}^{-9}N }$ is the gravitation form of attraction between them.

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