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balandron [24]
3 years ago
10

Irene was investigating the behavior of gases at certain conditions. Which would most likely cause an increase in the rate of co

llisions between gas molecules in a container?
Physics
1 answer:
kipiarov [429]3 years ago
6 0

Answer:

Increase in temperature

Increase in pressure

Explanation:

For the rate of collisions to increase, It can happen in various ways

1. The gasses can be made to move faster: This can be done by increasing their temperature. An increase in the temperature of the gases directly relates to a corresponding increase in their kinetic energy. This means that the molecules of the gases can move with a greater velocity, which will increase collision rates.

2. The volume of the container can be reduced: Consider this; If the container is made smaller, the molecules of the gases will have less space to move about. This means that they will be bumping into the walls of the container more frequently. This reduction in volume leads to an increase in the pressure of the gas. This is exactly what happens in a piston-cylinder assembly

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Which electromagnetic waves have the shortest wavelength and the highest frequency?
ra1l [238]
It’s supposed to be gamma, what are your other options
5 0
3 years ago
A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangent
Julli [10]

Answer:

The kinetic energy of the merry-go-round is \bf{475.47~J}.

Explanation:

Given:

Weight of the merry-go-round, W_{g} = 826~N

Radius of the merry-go-round, r = 1.17~m

the force on the merry-go-round, F = 57.8~N

Acceleration due to gravity, g= 9.8~m.s^{-2}

Time given, t=3.47~s

Mass of the merry-go-round is given by

m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg

Moment of inertial of the merry-go-round is given by

I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}

Torque on the merry-go-round is given by

\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m

The angular acceleration is given by

\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}

The angular velocity is given by

\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}

The kinetic energy of the merry-go-round is given by

E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J

5 0
3 years ago
Can you ever absolutely prove that a hypothesis is correct? Explain.
Diano4ka-milaya [45]

Yes you can, with using scientific experiment.

Ask a question -- Do background Research -- Construct a Hypothesis --Test with an Experiment -- Procedure working? -- Yes or no? -- Analyze Data and Draw Conclusions

With an experiment you can discover if its correct or not.

Hope this helps ! <3

8 0
3 years ago
A motorcycle of mass 100 kilograms slowly rolls off the edge of a cliff and falls for three seconds before reaching the bottom o
Jet001 [13]

Answer:

C) 3,000 kg m/s

Explanation:

We can consider the horizontal velocity of the motorcycle to be zero, since it rolls off the edge of the cliff very slowly. So, we only need to find the vertical velocity at the time of the impact with the ground.

The vertical velocity of the motorcycle at time t is given by (free-fall motion):

v(t)=v_0 -gt

where

v_0=0 is the initial vertical velocity (zero, since the motorcycle is not moving)

g = 9.8 m/s^2 is the acceleration due to gravity

t is the time

Since the motorcycle hits the ground after t = 3 seconds, we have

v(3 s)=0-(9.8 m/s^2)(3 s)=-29.4 m/s

And since we know its mass, m=100 kg, we can find its momentum:

p=mv=(100 kg)(-29.4 m/s)=-2940 kg m/s \cdot -3000 kg m/s

and the negative sign simply means downward direction.

8 0
3 years ago
At which temperature does the motion of atoms and molecules stop?<br> 0°C<br> 0C<br> 0°K<br> 0K
anastassius [24]

Answer: 0K

Explanation:

Absolute 0 (0K) is the point where nothing could be colder and no heat energy remains in a substance.

7 0
3 years ago
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