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3 years ago

7

Under what conditions is the conservation of momentum applicable

1 answer:

KiRa [710]3 years ago

6 0

<span>The momentum of an isolated system is a constant. The vector sum of the momenta mv of all the objects of a system cannot be changed by interactions within the system. This puts a strong constraint on the types of motions which can occur in an isolated system. If one part of the system is given a momentum in a given direction, then some other part or parts of the system must simultaneously be given exactly the same momentum in the opposite direction. As far as we can tell, conservation of momentum is an absolute symmetry of nature. That is, we do not know of anything in nature that violates it. </span>

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Read 2 more answers

A system gains 757 kJ757 kJ of heat, resulting in a change in internal energy of the system equal to +176 kJ.+176 kJ. How much w

Alecsey [184]

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581 kJ, work was done by the system

Explanation:

According to the first law of thermodynamics:

$\Delta U = Q-W$

where

$\Delta U$ is the change in internal energy of the system

Q is the heat absorbed by the system (positive if absorbed, negative if released)

W is the work done by the system (positive if done by the system, negative if done by the surrounding)

In this problem,

$Q=+757 kJ$

$\Delta U = +176 kJ$

Therefore the work done by the system is

$W=Q-\Delta U=757 kJ-176 kJ=+581 kJ$

And the positive sign means the work is done BY the system.

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3 years ago

A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500

jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5 ×10 ⁻³ kg

initial velocity of stone ( $u_{stone}$ ) = 0 m/s

Initial velocity of bullet ( $u_{bullet}$ ) = (500 m/s)i

Speed of the bullet after collision ( $v_{bullet}$ ) = (300 m/s) j

Suppose we represent $(v_{stone})$ to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

$m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}$

$(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}$

$(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j= (0.100 \ kg)v_{stone}$

$v_{stone}= (1.25\ kg.m/s)i-(0.75\ kg m/s)j$

$v_{stone}= (12.5\ m/s)i-(7.5\ m/s)j$

∴

The magnitude now is:

$v_{stone}=\sqrt{ (12.5\ m/s)^2-(7.5\ m/s)^2}$

$\mathbf{v_{stone}= 14.6 \ m/s}$

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

$\theta = tan^{-1} (\dfrac{-7.5}{12.5})$

$\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}$

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3 years ago