Math, asked by Anonymous, 16 days ago

A stone of mass 2 kg moving with a velocity of 4 m/s collides to another stone of mass 4 kg which is at rest. If after the collision they move with the same velocity, then what would be the combined kinetic energy of two stones after the collision?
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Answers

Answered by darshusuvarna
1

Answer:

i hope it will help you ❣️ plz make me brainlist ❣️

Step-by-step explanation:

a) By momentum conservation,

2(4)−4(2)=2(−2)+4(ν

2

)

⟹ν

2

=1m/s.

b) e=

Velocityofapproach

Velocityofseparation

=

4−(−2)

1−9−2)

=

6

3

=0.5

c) At the maximum deformed state, by conservation of momentum, common velocity is ν=0

J

D

=m

2

(ν−u

2

)=4[(0−(−2)]=8Ns

d) Potential energy at the maximum deformed state,

U= loss in kinetic energy during defirmation

or U=(

2

1

m

1

u

1

2

+

2

1

m

2

u

2

2

)−

2

1

(m

1

+m

2

2

=(

2

1

2(4)

2

+

2

1

4(2)

2

)−

2

1

(2+4)(0)

2

=24J

e) J

R

=m

2

2

−ν)=4(1−0)=4Ns

also J

R

=eJ

D

v

=0.5×8=4Ns thank you bro

Answered by BangtanXArmy0t7
9

Answer:

Mawsynram village of Meghalaya receives rainfall over 400cm.

Northeastern regions and the windward side of the Western ghats experience an average of 400cm of annual rainfall. Areas like Assam, Meghalaya, Arunachal Pradesh and hilly tracts of the Western Ghats are host to tropical rainforests.

Jaisalmer is the place in India which receives the lowest rainfall.

Some parts of Jammu & Kashmir such as the Ladakh plateau are also included in this zone as cold deserts.

Same problem Haha

I am fine. thanks

I will go hospital tomorrow for vaccination.

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