Physics, asked by serinawangjam1, 4 months ago

Find the weight of a body on the earth's surface whose mass is 3 kg. [hint:W=mg]​

Answers

Answered by MystícPhoeníx
18

Answer:

  • 29.4 Newton is the required Weight

Explanation:

Given:-

  • Mass ,m = 3 kg

To Find:-

  • Weight of body on the earth's surface ,W

Solution:-

As we have to calculate the weight of the body on the earth's surface.

Weight of a body on any planet can be calculated by the product of mass and acceleration due to gravity at the surface of that planet.

Here, the acceleration due to gravity on earth's surface is 9.8 m/s².

W = mg

where,

  • W denote Weight
  • m denote mass
  • g denote acceleration due to gravity

Substitute the value we get

→ W = 3 × 9.8

→ W = 29.4 N

  • Hence, the weight of body on the earth's surface is 29.4 Newton.


Anonymous: Wonderful :D
Answered by Anonymous
5

\; \; \; \; \; \; \;{\Large{\bf{\underbrace{Required \; answer}}}}

{\large{\bold{\rm{\underline{Let's \; understand \; the \; question \; 1^{st}}}}}}

✴ This question says that we have to find the weight of the body on the surface of the earth whose mass is 3 kg. A hint is given as W = mg. In short we have to use this formula to find the weight.

{\large{\bold{\rm{\underline{Given \; that}}}}}

⋆ Use = W = mg

⋆ Mass of the body on earth's surface = 3 kg

⋆ Acceleration due to gravity = 9.8 m/s²

Note : We take acceleration due to gravity here because the question says that the body is on the earth's surface so we already know that when this is given then we have to use acceleration due to gravity as 9.8 m/s²

{\large{\bold{\rm{\underline{To \; find}}}}}

⋆ Weight of body (earth surface)

{\large{\bold{\rm{\underline{Solution}}}}}

⋆ Weight of body (earth surface) = 29.4 Newton

{\large{\bold{\rm{\underline{Full \; Solution}}}}}

↝ W = mg

Where,

➙ W denotes weight

➙ m denotes mass

➙ g denotes acceleration due to gravity

↝ W = (3)(9.8)

↝ W = 3 × 9.8

↝ W = 29.4 Newton

{\frak{Henceforth, \: 29.4 \: N \: is \: weight}}

{\large{\bold{\rm{\underline{Additional \; knowledge}}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Number \: of \: SI \: units \: are \: 7}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Ampere \: is \: the \: unit \: of \: current \: electricity}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: Young's \: modulus \: of \: elasticity \: is \: Newton/m^{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: pressure \: is \: Pascal}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Curie \: is \: the \: unit \: of \: radio \: activity}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Decibel \: is \: the \: unit \: of \: intensity \: of \: sound}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: electric \: charge \: is \: coulomb}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: resistance \: is \: ohm}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: acceleration \: is \: ms^{-2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Maxwell \: is \: unit \: of \: magnetic \: flux}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: magnetic \: flux \: is \: Weber}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: surface \: tension \: is \: \dfrac{N}{m}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto SI \: unit \: of \: mechanical \: power \: is \: Watt}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto 1 \: horsepower \: = \: approx \: 746 \: watts}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Momentum \: is \: measured \: as \: the \: product \: of \: Mass \: and \: velocity}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto \pi 'pi' \: is \: calculated \: by \: Aryabhatta}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto One \: J \: = \: 0.24 \: cal}}}


Anonymous: Good work, as always.
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