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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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Answer:

792J

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Use the formula Force × Distance

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A maintenance worker wants to torque an engine bolt to 65.0 N m. If the torque wrench is 35cm in length, what is force applied t
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Answer:

Force = 186 N

Explanation:

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Torque = \vec{r} \times \vec{F}

Where \vec{r} is a vector that from the origin of the coordinate system to the point at which the force is applied (the position vector), \vec{F} is the applied force.

The easiest way of computing the force is by setting the origin of the coordinate system to the lowest point of the torque wrench.  By doing this we have that r (the magnitud of the position vector) is 35cm.

Before computing the force we need to set all our values to the international system of units (SI). The torque is already in SI. The one missing is the length of the torque wrench (it is in centimeters and we need it in meters). So :

35cm * \dfrac{1m}{100cm} = 0.35m

Now using the torque formula:

Torque = \vec{r} \times \vec{F}

Torque = rFsin(\theta)}

Where \theta is the smaller angle between the force and the position vector. Because the force is applied perpendiculary to the position vector  \theta = 90°, thus :

Torque = rFsin(\theta)}

65 N m = (0.35m)Fsin(90°)}

F = \dfrac{65N m} {(0.35m)sin(90°)}

F = \dfrac{65N m} {(0.35m)sin(90°)}

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3 years ago
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Answer:

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