Question

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Answer 1

Answer:

$\large\bold\red{\triangle{K.E} = 40.4 \: Kg \: {m}^{2} {s}^{ - 2} }$

Explanation:

Given,

Mass of the body, m = 0.8 kg

Initial velocity, u = ( 4i + 3j ) m/s

Final velocity, v = ( -6j + 2k ) m/s

Therefore,

Change in velocity, ∆v = v - u

=> ∆v =[ (-6j + 2k ) - ( 4i + 3j ) ] m/s

=> ∆v = (-6j + 2k - 4i - 3j ) m/s

=> ∆v = ( -4i - 9j + 2k ) m/s

Since,

the velocity is given in vector form,

we have to find it's magnitude,

Therefore,

$= > | \triangle{v}| = \sqrt{ {4}^{2} + {9}^{2} + {2}^{2} } \\ \\ = > | \triangle{v}| = \sqrt{16 + 81 + 4} \\ \\ = > | \triangle{v}| = \sqrt{101} \: m {s}^{ - 1}$

Now,

we know that,

Change in Kinetic Energy ,

$\triangle{K.E} \: = \frac{1}{2} m {| \triangle{v}|}^{2}$

pUtting the respective values,

we get,

$\triangle{K.E} = \frac{1}{2} \times 0.8 \times { (\sqrt{101} )}^{2} \\ \\ = > \triangle{K.E} = 0.4 \times 101 \\ \\ = > \bold{\triangle{K.E} = 40.4 \: Kg \:{ m}^{2} {s}^{ - 2} } \:$

Answer 2

Answer:

here is ur ans..

m = 0.8 kg

Initial velocity, u = ( 4i + 3j ) m/s

Final velocity, v = ( -6j + 2k ) m/s

Therefore,

Change in velocity, ∆v = v - u

=> ∆v =[ (-6j + 2k ) - ( 4i + 3j ) ] m/s

=> ∆v = (-6j + 2k - 4i - 3j ) m/s

=> ∆v = ( -4i - 9j + 2k ) m/s

Since,

the velocity is given in vector form,

we have to find it's magnitude,

Therefore,

=>∣△v∣=42+92+22=>∣△v∣=16+81+4=>∣△v∣=101ms−1\begin{lgathered}= > | \triangle{v}| = \sqrt{ {4}^{2} + {9}^{2} + {2}^{2} } \\ \\ = > | \triangle{v}| = \sqrt{16 + 81 + 4} \\ \\ = > | \triangle{v}| = \sqrt{101} \: m {s}^{ - 1}\end{lgathered}

=>∣△v∣=

4

2

+9

2

+2

2

=>∣△v∣=

16+81+4

=>∣△v∣=

101

ms

−1

we know

Change in Kinetic Energy ,

△K.E=12m∣△v∣2\triangle{K.E} \: = \frac{1}{2} m {| \triangle{v}|}^{2}△K.E=

2

1

m∣△v∣

2

pUtting the respective values,

we get,

△K.E=12×0.8×(101)2=>△K.E=0.4×101=>△K.E=40.4Kgm2s−2\begin{lgathered}\triangle{K.E} = \frac{1}{2} \times 0.8 \times { (\sqrt{101} )}^{2} \\ \\ = > \triangle{K.E} = 0.4 \times 101 \\ \\ = > \bold{\triangle{K.E} = 40.4 \: Kg \:{ m}^{2} {s}^{ - 2} } \:\end{lgathered}

△K.E=

2

1

×0.8×(

101

)

2

=>△K.E=0.4×101

=>△K.E=40.4Kgm

2

s

−2