6-7 - 10 - 11 6 - 1024 7-4 -1022- 3-84 - 1082

( 6.7 ×10^ 11)( 6 ×10^24)( 7.4 ×10^22)( 3.84 ×10^8)^2 = ( 6.7 × 6 × 7.4)( 10^ 11×10^24×10^22)3.8 × 3.8 ×10^8×10^8 = 297.48 ×10^3514.44 ×10^16 ...

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Byju's Answer

Standard VI

Mathematics

Addition of Fractions

6.7 × 10 - 11...

Question

$\frac{(6.7 \times 10^{- 11}) (6 \times 10^{24}) (7.4 \times 10^{22})}{{(3.84 \times 10^{8})}^{2}}$

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Solution

$\frac{(6.7 \times 10^{- 11}) (6 \times 10^{24}) (7.4 \times 10^{22})}{{(3.84 \times 10^{8})}^{2}}$

$= \frac{(6.7 \times 6 \times 7.4) (10^{- 11} \times 10^{24} \times 10^{22})}{3.8 \times 3.8 \times 10^{8} \times 10^{8}}$

$= \frac{297.48 \times 10^{35}}{14.44 \times 10^{16}}$

$= 20.6011 \times 10^{35 - 16}$

$= 20.6011 \times 10^{19}$

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Similar questions

Solve

$\frac{(6.7 \times 10^{- 11}) (6 \times 10^{24}) (7.4 \times 10^{22})}{{(3.84 \times 10^{8})}^{2}}$

Q. Simplify

\frac{(6.7 \times 10^{- 11}) (6 \times 10^{24}) (7.4 \times 10^{22})}{(3.84 \times 10^{8})^{2}}

Q. Solve:

\frac{6.7 \times 10^{- 11} \times 7.4 \times 10^{22} \times 1}{(1.74 \times 10^{6})^{2}}

The mass of the earth is $6 \times 10^{24}$ kg and that of the moon is $7.4 \times 10^{22}$ kg. If distance between the earth and the moon be $3.84 \times 10^{5}$ km, calculate the force exerted by the earth on the moon. $(G = 6.7 \times 10^{- 11} N m^{2} k g^{- 2})$ .

6.7 into 10 power Minus 11into six into 10 power 24 KG into 7.4 into 10 power 22 KG by 3.84 into 10 power 8 m whole square

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(6.7×10−11)(6×1024)(7.4×1022)(3.84×108)2

(6.7×10−11)(6×1024)(7.4×1022)(3.84×108)2=(6.7×6×7.4)(10−11×1024×1022)3.8×3.8×108×108=297.48×103514.44×1016=20.6011×1035−16=20.6011×1019

$\frac{(6.7 \times 10^{- 11}) (6 \times 10^{24}) (7.4 \times 10^{22})}{{(3.84 \times 10^{8})}^{2}}$