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Bumek [7]
4 years ago
6

1) Frequency describes the rate in which waves move through a given location. In the simulation, the airplane emits waves as it

travels. The closer together the waves appear, the higher their frequency appears.
Imagine you’re standing at the center of the simulation, and the airplane flies directly overhead with an object speed of 50 meters per second. From your perspective, how does the frequency of the waves appear to change as the airplane approaches and then passes you?

2) Determine how visible light is affected by changes in frequency.
Physics
1 answer:
Mashcka [7]4 years ago
4 0

Explanation:

The Doppler effect describes the change in the observed frequency of a wave when there is relative motion between the wave source and the observer

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Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope
Kitty [74]

a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J   the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

       

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

  = -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

                     = 26,754 (0.816)

                     = 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c.  What is the work done by the gravitational force on the sled?

By using  the formula:

Gravity work = mgdsinθ

                    = 97.5 kg * 9.8 m/s² * 28 m * sin 60

                    = 23,170 J

23,170 J   the work, in joules done by the gravitational force on the sled .

       

D. What is the total work done?

By adding all the values

work done =  -1337.3 + 21,835 + 23,170

                 = 43,670 J

The net work done on the sled, in joules is 43,670 J.

Learn more about friction work here:

brainly.com/question/14619763

#SPJ1

4 0
2 years ago
Find the value of T1 if 1 = 30°, 2 = 60°, and the weight of the object is 139.3 newtons.
Lelechka [254]

Answer:

Option A (69.56 newtons) is the appropriate solution.

Explanation:

According to the question,

On the X-axis,

⇒ T_1Cos30^{\circ}-T_2Cos60^{\circ}=0

or,

    T_1Cos 30^{\circ}=T_2Cos60^{\circ}

On substituting the values, we get

      T_1\times \frac{\sqrt{3} }{2}=T_2\times \frac{1}{2}

      T_1\times \sqrt{3} =T_2....(equation 1)

On the Y-axis,

⇒ T_1Sin30^{\circ}+T_2Sin60^{\circ}=139.3 \ N

                        \frac{T_1}{2} +\frac{\sqrt{3} }{2} =139.2 \ N

                    T_1+\sqrt{3}T_2=139.2\times 2

From equation 1, we get

           T_1+\sqrt{3}\times \sqrt{3}T_1 =278.4 \ N

                        T_1+3T_1=278.4 \ N

                                4T_1=278.4 \ N

                                  T_1=\frac{278.4}{4}

                                       =69.6 \ N  

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