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Papessa [141]
2 years ago
9

In a closed system, the loss of momentum of one object_____ the gain in momentum of another object.

Physics
1 answer:
densk [106]2 years ago
6 0

In a closed system, the loss of momentum of one object  is same as________ the gain in momentum of another object

according to law of conservation of momentum, total momentum before and after collision in a closed system in absence of any net external force, remains conserved . that is

total momentum before collision = total momentum after collision

P₁ + P₂ = P'₁ + P'₂

where P₁ and P₂ are momentum before collision for object 1 and object 2 respectively.

P'₁ - P₁  = - (P'₂ -  P₂)

so clearly gain in momentum of one object is same as the loss of momentum of other object

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"The two equations below express conservation of energy and conservation of mass for water flowing from a circular hole of radiu
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The two equations below express conservation of energy and conservation of mass for water flowing from a circular hole of radius 3 centimeters at the bottom of a cylindrical tank of radius 10 centimeters. In these equations, delta m is the mass that leaves the tank in time delta t, v is the velocity of the water flowing through the hole, and h is the height of the water in the tank at time t. g is the acceleration of gravity, which you should approximate as 1000 cm/s2.

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The 500 pages of a book have a mass of 2.50 kg. What is the mass of each page A in kg B in mg?
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Answer:

A.

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B.

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Newton’s law of cooling states that dx dt = −k(x−A) where x is the temperature,t is time, A is the ambient temperature, and k &g
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Answer:

a)X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}

b)Does not affect the long term.

Explanation:

Given that

\dfrac{dx}{dt}=-k(x-A)

A = A0 cos(ωt)

\dfrac{dx}{dt}=-k(x-A_o cos(\omega t))

\dfrac{dx}{dt}+kx=kA_o cos(\omega t)

This is linear equation so integration factor ,I

I=e^{\int kdt}

I=e^{kt}

Now by using linear equation property

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e^{kt} X= kA_o \dfrac{e^{kt}}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+C

X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+Ce^{-kt}

b)

at t= 0

X(0)=\dfrac{k^2A_o}{\omega^2+k^2}+C

X= kA_o\dfrac{1}{k^2+\omega^2}\left ( kcos\omega t+\omega sin\omega t \right )+e^{-kt}\times \left ( X(0)-\dfrac{k^2A_o}{\omega^2+k^2} \right )

So the initial condition does not affect the long term.

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2 years ago
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