Physics, asked by vaibhavraghav76, 4 months ago

Please tell me the answer

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Answered by ItzArchimedes
5

Kinetic energy = 0.16 J

As we know that ,

  • Momentum = Mass × velocity

Now ,

✧ p = 0.5 × v

✧ 0.4/0.5 = v

✧ (4/10)/(5/10) = v

✧ 4/5 = v

Velocity = 0.8 m/s

As we know that ,

  • Kinetic energy = 1/2 mv²

☆ K.E = 1/2 × 0.5 × 0.8²

☆ K.E = 1/4 × 4/5 × 4/5

☆ K.E = 1/5 × 4/5

☆ K.E = 4/25

K.E = 0.16 J


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Answered by Anonymous
6

Question -

✠ A body of mass 0.5 kg has a momentum of 0.4 kgm/s. Find it's kinetic energy.

Given that -

✠ Mass of the body = 0.5 kg

✠ Momentum = 0.4 kgm/s

To find -

✠ Kinetic energy.

Solution -

✠ Kinetic energy = 0.16 Joules

Using concept -

✠ Formula to find kinetic energy

✠ Formula to find momentum (to find velocity) because in kinetic energy formula we need to use velocity.

Using formula -

✠ Kinetic energy = ½ mv²

✠ Momentum = m × v

Where,

★ m denotes mass

★ v denotes velocity

Full solution -

~ Let's find the velocity by using formula to find momentum.

⇢ Momentum = m × v

⇢ 0.4 = 0.5 × v

⇢ 0.4/0.5 = v

⇢ (4/10)/(5/10) = v

⇢ 4/5 = v

⇢ 0.8 = v

⇢ v = 0.8 m/s

  • Henceforth, 0.8 m/s is the velocity.

~ Let us find kinetic energy now,

⇢ Kinetic energy = ½ mv²

⇢ Kinetic energy = ½ × 0.5 × (0.8)²

⇢ Kinetic energy = ½ × 0.5 × 0.8 × 0.8

⇢ Kinetic energy = ½ × 0.5 × 0.64

⇢ Kinetic energy = ½ × 0.320

⇢ Kinetic energy = ½ × 320/1000

⇢ Kinetic energy = ½ × 32/100

⇢ Kinetic energy = 16 / 100

⇢ Kinetic energy = 8 / 50

⇢ Kinetic energy = 4/25

⇢ Kinetic energy = 0.16 J

More knowledge -

\; \; \; \; \; \; \;{\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}}}

\; \; \; \; \; \; \;{\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 Kinetic \: energy \: is \: given \: by \: \dfrac{1}{2}mv^{2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Value \: of \: G \: is \: 6.673 \times 10^{-11}Nm^{2}kg{-3}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Dimensional \: formula \: for \: universal \: gravitational \: constant \: is \: M^{-1} L^{3} T^{-2}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto The \: unit \: of \: force \: constant \: k \: of \: a \: spring \: is \: \dfrac{N}{m}}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Sir \: Cavendish \: was \: the \: first \: to \: gave \: value \: of \: G \: experimentally}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto The \: Young's \: modulus \: for \: perfect \: rigid \: body \: is \: infinite}}}

\; \; \; \; \; \; \;{\sf{\bold{\leadsto Density \: is \: the \: ratio \: of \: \dfrac{Volume}{Mass}}}}


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