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3 years ago

What could you do to change the volume of a gas?

1 answer:

8 0

The ONLY way to change the volume of a sample of gas is to transfer it to a container with different volume.
Simply changing its temperature or pressure in the same jar won't do it. Any amount of gas always fills whatever container you keep it in.

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How far away is mars from the sun. why is is red?

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Mars is 139.58 million miles away from the sun. mars is red because its soil has iron oxide (rust particles) in it.

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3 years ago

In a nuclear power plant, _____. energy is released from the nuclei of atoms energy is released from the bonds of molecules ener

oksian1 [2.3K]

In a nuclear power plant, energy is released from the nuclei of atoms. The correct option among all the options given in the question is the first option. Huge amount of thermal energy is released by the breaking of the uranium atoms. This energy is used for turning a turbine that produces electricity. It is a very clean method of producing electricity.

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3 years ago

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A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

Kamila [148]

Answer:

a) 6738.27 J

b) 61.908 J

c) $\frac{4492.18}{v_{car} ^{2} }$

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = 6738.27 J

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

$(I_{1} +I_{2} )w$

where the subscripts 1 and 2 indicates the values first and second flywheels

$(I_{1} +I_{2} )w$ = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = $Iw^{2}$ = 6.655 x $3.05^{2}$ = 61.908 J

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = $\frac{1}{2}mv_{car} ^{2}$

where m is the mass of the car

$v_{car}$ is the velocity of the car

Equating the energy

2246.09 = $\frac{1}{2}mv_{car} ^{2}$

making m the subject of the formula

mass of the car m = $\frac{4492.18}{v_{car} ^{2} }$

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3 years ago

Intro to Springs and Hooke’s law

wel

A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.
Springs come in a huge variety of different forms, but the simple metal coil spring is probably the most familiar. Springs are an essential part of almost all moderately complex mechanical devices; from ball-point pens to racing car engines.
There is nothing particularly magical about the shape of a coil spring that makes it behave like a spring. The 'springiness', or more correctly, the elasticity is a fundamental property of the wire that the spring is made from. A long straight metal wire also has the ability to ‘spring back’ following a stretching or twisting action. Winding the wire into a spring just allows us to exploit the properties of a long piece of wire in a small space. This is much more convenient for building mechanical devices.

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3 years ago

If the plotted points on a speed-time graph do not form a straight line, what do you know about the object's acceleration?

slega [8]

Remember, that while sped is constant, acceleration is not. Acceleration is when velicity changes. So the graph which shows the slop of a velocity vs time describes acceleration.
If we have the straight line on the graph it means that the slope is always the same whereas the non-linear graphs has a variable slope that changes depending on your point in the graph.
To conclude - if your graph is not a straight line it has variable acc at many points.

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3 years ago

Read 2 more answers