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scoundrel [369]
3 years ago
10

What two characteristics are used to classify air masses

Physics
1 answer:
IgorLugansk [536]3 years ago
7 0

Answer: Temperature and humidity are the two characteristics used to classify air masses.

Explanation:

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As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arr
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Answer:

  • <u><em>1. Part A: 648N</em></u>
  • <u><em></em></u>
  • <u><em>2. Part B: 324J</em></u>
  • <u><em></em></u>
  • <u><em>3. Part C: 1,296N</em></u>

Explanation:

<em><u>1. Part A:</u></em>

The magnitude of the force is calculated using the Hook's law:

          |F|=k\Delta x

You know \Delta x=0.250m but you do not have k.

You can calculate it using the equation for the work-energy for a spring.

The work done to compress the springs a distance \Delta x is:

          Work=\Delta PE=(1/2)k(\Delta x)^2

Where \Delta PE is the change in the elastic potential energy of the "spring".

Here you have two springs, but you can work as if they were one spring.

You know the work (81.0J) and the length the "spring" was compressed (0.250m). Thus, just substitute and solve for k:

             81.0J=(1/2)k(0.250m)^2\\\\k=2,592N/m

In reallity, the constant of each spring is half of that, but it is not relevant for the calculations and you are safe by assuming that it is just one spring with that constant.

Now calculate the magnitude of the force:

         |F|=k\Delta x=2,592N/m\times 0.250m=648N

<u><em></em></u>

<u><em>2. Part B. How much additional work must you do to move the platform a distance 0.250 m farther?</em></u>

<u><em></em></u>

The additional work will be the extra elastic potential energy that the springs earn.

You already know the elastic potential energy when Δx = 0.250m; now you must calculate the elastic potential energy when  Δx = 0.250m + 0.250m = 0.500m.

          \Delta E=(1/2)2,592n/m\times(0.500m)^2=324J

Therefore, you must do 324J of additional work to move the plattarform a distance 0.250 m farther.

<em><u></u></em>

<em><u>3. Part C</u></em>

<u><em></em></u>

<u><em>What maximum force must you apply to move the platform to the position in Part B?</em></u>

The maximum force is when the springs are compressed the maximum and that is 0.500m

Therefore, use Hook's law again, but now the compression length is Δx = 0.500m

           |F|=k\Delta x=2,592N/m\times 0.500m =1,296N

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