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3 years ago

Which statement best describes an isolated system?

Physics

2 answers:

NARA [144]3 years ago

4 0

Answer:

C. Neither energy nor matter are exchanged with the surroundings.

Explanation:

Isolated system is a system that is completely cut off from its surroundings.

In an isolated system, both energy and matter are said to be conserved.

It cannot release energy to its surroundings or it cannot take in energy from the surroundings. Apart from energy, even matter cannot be exchanged .

For example if there is a hot box which is insulated and some contents like hot water or milk are poured into it, the heat energy of the water/milk cannot be lost to the surroundings and also it will not take in any heat energy from the surroundings; and the actual amount of water/milk content doesn't change. Such a system is called an isolated system.

charle [14.2K]3 years ago

3 0

It is c
because Mrs.Brown told us in class that Neither energy nor matter are exchanged with the surroundings.

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The relative hardness of a mineral can best be tested by

stellarik [79]

The relative hardness of a mineral can be tested by the Moh’s scale or the mineral hardness. The Moh’s scale can be characterized by the ability of the mineral to resist scratch resistance by scratching a harder or softer mineral. It has a scale of 1 to 10 where 1 being the softest and 10 being the hardest. The relative hardness of a mineral can best be tested by scratching the mineral across a glass plate. The answer is letter A.

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3 years ago

A rifle with a weight of 25 N fires a 4.5-g bullet with a speed of 240 m/s. (a) Find the recoil speed of the rifle. m/s (b) If a

asambeis [7]

Answer:

The recoil speed of the man and rifle is $v_{man}=0.016 ms^{-1}$ .

Explanation:

The expression for the force in terms of mg is as follows;

F=mg

Here, m is the mass and acceleration due to gravity.

Rearrange the expression for mass.

$m=\frac{F}{g}$

Calculate the combined mass of the man and rifle.

$m_{man,rifle}=\frac{650+25}{g}$

Put $g=9.8 ms^{-2}$ .

$m_{man,rifle}=\frac{650+25}{9.8}$

$m_{man,rifle}=68.88 kg$

The expression for the conservation of momentum is as follows as;

$m_{man}u_{man}+m_{bullet}u_{bullet}=m_{man}v_{man}+m_{rifle}v_{man,rifle}$

Here, $m_{man,rifle}$ is the mass of the man and rifle, $m_{rifle}$ is the mass of the rifle, $u_{man},u{bullet}$ are the initial velocities of the man and bullet and $v_{man},v{man,rifle}$ are the final velocities of the man and rifle and rifle.

It is given in the problem that a rifle with a weight of 25 N fires a 4.5-g bullet with a speed of 240 m/s.

Convert mass of rifle from gram to kilogram.

$m_{bullet}=4.5 g$

$m_{bullet}=.0045 kg$

Put $m_{bullet}=.0045 kg$ , $m_{man,rifle}=68.88 kg$ , $u_{man,rifle}=0$ , $v_{bullet}= 240 ms^{-1}$ and $u_{bullet}=0$ .

$m_{man}(0)+m_{bullet}(0)=(68.88)v_{man,rifle}+(.0045)(240)$

$0=(68.88)v_{man,rifle}+(.0045)(240)$

$0=(68.88)v_{man,rifle}+1.08$

$(68.88)v_{man,rifle}=\frac{-1.08}{68.88}$

$v_{man,rifle}=-0.016 ms^{-1}$

Therefore, the recoil speed of the man and rifle is $v_{man}=0.016 ms^{-1}$ .

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3 years ago

How would the absence of gravity affect the formation of planets?

Mumz [18]

In the absence of gravity, the rocks and debris
would never accrete into a planet. (B)

Also by the way, it wouldn't matter much, because
there wouldn't be a star to orbit around, AND orbits
wouldn't exist either.

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3 years ago

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The rate at which light energy is radiated from a source is measured in which of the following units?

Over [174]

The correct option is D.
Lumen is used to quantify the amount of total light energy that a source is putting out in all direction, thus, it refers to luminous output of a light source. Initial lumen refers to the luminosity of a light when it was first turned on; the luminosity is highest at this point.

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3 years ago

Read 2 more answers

On the way to the moon, the Apollo astronauts reach a point where the Moon’s gravitational pull is stronger than that of Earth’s

Drupady [299]

Answer:

rm = 38280860.6[m]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

6 0

3 years ago