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3 years ago

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An inelastic collision occurs in one dimension, in which a 10 kg block traveling at 5 m/s collides with a 5 kg block traveling a

t 3 m/s in the same direction, and they stick together. What are the velocities of the blocks immediately after the collision?

1 answer:

gulaghasi [49]3 years ago

7 0

Why’d Kahn’s his Judd hue
us jibe Japheth JSU’s

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What is the distance of a car travels is its velocity is 30 mph and its time is 2 hours ?

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The rod of the fixed hydraulic cylinder is moving to the left with a constant speed vA = 25 mm/s. Determine the corresponding ve

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Explanation:

let vertical distance from A to C be $h=250mm$ the constraint equation is:

$l_{ac} +l_{ab} =L$

we want to find $l_{bc}$ distance can be written as $l_{bc} =h_{1}+h_{2}$ which we will find by using Pythagorean theorem h 1 is a constant and can be written as h:

$h_{2} =\sqrt{l_{ab}^2- s_{A} ^2}$

$l_{ac} =\sqrt{s_{A}^2+ h^2 }$

$l_{ab} =L-\sqrt{s_{A} ^2+h^2 }$

taking derivative w.r.t time we get

$h^._{2} =-v_{B =(l_{ab} l_{ab} ^.-s_{A} v_{A} )/h_{2}$

$l_{ab} ^.=(s_{A} v_{A} )/l_{ab} -L$

given $s_{A} =425mm$ which gives us

$l_{ab} =1050-\sqrt{450^2+250^2} =556.923mm\\\\h_{2} =\sqrt{556.923^2-425^2}=359.914mm$

$l_{ab} ^.=(425.25)/(556.923-1050)=-21.548mm\\\\-v_{B} =(-556.923*21.548-425.25)/(359.914)\\\\v_{B} =62.864mm/s$

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4 years ago

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3 years ago

A rocket initially at rest accelerates at a rate of 99.0 meters/second2. Calculate the distance covered by the rocket if it atta

Jobisdone [24]

Answer: The correct answer is option b.

Explanation: We are given that the rocket is at rest initially final velocity is 445m/s.

The acceleration of the rocket is $99.0m/s^2$

To calculate the distance of rocket, we use third equation of motion, which is:

$v^2-u^2=2as$

where, v = final velocity = 445m/s

u = initial velocity = 0m/s

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s = distance = ? m

Putting values in above equation, we get:

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