As the pressure on a sample of gas increases, the volume of the sample _____. decreases stays the same increases

Answers

Answer 1

Answer:

Answer : As the pressure on a sample of gas increases, the volume of the sample decreases.

Explanation :

According to the Boyle's, law, the pressure of the gas is inversely proportional to the volume of gas at constant temperature and moles of gas.

$P\propto (1)/(V)$ (At constant temperature and moles of gas)

From the we conclude that the higher the pressure of gas, lower will be the volume and lower the pressure of gas, higher will be the volume of gas.

Hence, as the pressure on a sample of gas increases, the volume of the sample decreases.

Answer 2

Answer: By Boyle's law the volume of the sample decreases, provided temperature is constant.

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What is the maximum temperature in a hot whirlpool to be considered safe?

Answers

Answer:

104 degrees

Never let the hot tub's water temperature get above 104 degrees Fahrenheit. For a healthy adult, a temperature of 100 degrees is regarded as safe. Young kids should use extra caution.

When electrons are removed from the outermost shell of a calcium atom, the atom becomes

Answers

A cation has a smaller radius than the atom.

a cation
Cations are positive ions that have lost an electron

A bus slows down uniformly from 21 m/s to a complete stop in 21 seconds. How far did the bus travel before stopping?

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$velocity_(final)-velocity_(initial)=(distance)/(time)\n\n v_(initial)=21(m)/(s)\n v_(final)=0(m)/(s)\n time=21s\n\n \Delta v *time=distance\n\n distance=|(0-21)*21|=441m \n\nBus\ traveled\ 441meters.$

You are on roller blades on top of a small hill. Your potential energy is equal to 1,000.0 joules. The last time you checked your mass was 60.0 kilograms.
a. What is your weight in newtons?
b. What is the height of the hill?
c. If you start skating down this hill, your potential energy will be converted to kinetic energy. At the
bottom of the hill, your kinetic energy will be equal to your potential energy at the top. What will be
your speed at the bottom of the hill?

Answers

a) The weight of a person is given by:
$W=mg$
where m is the mass of the person and g is the gravitational acceleration. In this problem, the mass of the person is m=60.0 kg, while the value of g is $g=9.81 m/s^2$ , therefore the weight of the person is
$W=(60 kg)(9.81 m/s^2)=588.6 N$

b) The gravitational potential energy at the top of the hill is U=1000 J. The potential energy is also given by the product between the weight W and the height of the hill h:
$U=Wh$
If we rearrange the equation, we can calculate the height of the hill, h:
$h= (U)/(W) = (1000 J)/(588.6 N) =1.70 m$

c) At the bottom of the hill, all the potential energy at the top of the hill has converted into kinetic energy:
$U=K$
$U= (1)/(2) mv^2$
where m is the mass and v is the speed. If we rearrange the formula, we can calculate the speed at the bottom of the hill:
$v=\sqrt{(2U)/(m)}=\sqrt{(2\cdot 1000 J)/(60.0 kg)}=5.8 m/s$

I am in summer school for physics and I got this question for home work. A rock climber of mass 55 kg is hanging suspended from a rope tied to another climber of mass 65 kg on a horizontal cliff ledge. If the coefficient of kinetic friction between the climber on the ledge and the ledge 0.45, what is the net acceleration of the climber on the ledge? I have been stuck on this question for 2 hours and I need it for tomorrow can anyone help me. Thanks

Answers

First of all, we have to sort out how this situation is put together. You have 55 kg that's hanging suspended, but it's tied to another 65 kg on a horizontal ledge. Somewhere in there, the rope has to turn a corner. Obviously, it has to go around the edge of the ledge, and we have top assume that there's absolutely no friction against the rope at that point, because we're not told anything different. So we have to treat the edge like a frictionless pulley. And we're also going to ignore the weight of the rope.

OK. The weight of the 55-kg guy who is hanging suspended is (mass) x (gravity) =
(55 x 9.8) = 539 newtons. That force soaks upward through the rope, makes the turn
at the edge, and is exerted horizontally against the 65-kg guy on the ledge.

Hopefully, the suspended guy can hold on for a few minutes longer, while we analyze
the forces around the heavier guy up on the ledge.

The weight of the guy on the ledge is (mass) x (gravity) = (65 x 9.8) = 637 newtons.
That's his weight, pointing downward, against the ledge.
As his boots slip along the ledge, the friction force against his motion is
(his weight) x (the coefficient of kinetic friction between him and the ledge) =
(637 newtons) x (0.45) = 286.65 newtons.

The man on the ledge has the rope pulling him toward the edge with 539N of force,
and 286.65N of friction force holding him back. You can see that he's slipping toward
the edge, because the friction force isn't enough to hold him.

The net force on him is (539N forward) + (286.65N backward) = 252.35N forward.

Since the man on the ledge has a net force pulling him forward toward the edge,
he accelerates in that direction.

Force = (mass) x (acceleration)

Acceleration = (force) / (mass) = (252.35N) / (65kg) = 3.88 meters per second²

He's sliding toward the edge with an acceleration of about 0.396 G ... his speed is increasing about 39 or 40% as fast as it will after he falls over the edge, and the both of them proceed toward their ultimate and apparently unavoidable 'splut' below.

I've totally terrified myself answering this one.

after the box has left the rubber band but is still moving upward, what is the passenger's apparent weight?

Answers

If the upward acceleration is equal to the acceleration due to gravity ( $9.8 m/s^2$ ), then the apparent weight would be zero, indicating that the passenger is experiencing weightlessness or is in a state of free fall.

When the box has left the rubber band but is still moving upward, the passenger's apparent weight would be less than their actual weight.

The apparent weight of an object is the force exerted on it by a supporting surface, which in this case is the floor of the box. It is equal to the normal force acting on the object.

As the box accelerates upward, there are two main forces acting on the passenger: the gravitational force (weight) pulling them downward and the normal force pushing them upward. The normal force is responsible for providing the upward acceleration to the passenger.

Since the passenger is moving upward, the normal force exerted by the floor must be less than their actual weight to create the net upward force necessary for upward acceleration. Therefore, the passenger's apparent weight is reduced.

The exact value of the apparent weight depends on the magnitude of the upward acceleration and the mass of the passenger. The apparent weight can be calculated using the equation:

Apparent weight = Actual weight - (Mass of passenger * Magnitude of upward acceleration)

To know more about weight refer here

brainly.com/question/31659519#

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