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harina [27]
3 years ago
10

Which change of state do particles in a material become farther apart?

Physics
1 answer:
crimeas [40]3 years ago
6 0
Particles become farther apart from each other in the change of state between liquid and gas.
Hope that helped =)
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Sam is playing football. She kicks the ball with an average force of 75 N.
damaskus [11]

Answer:

22.5J

Explanation:

Here the force is given. Also, the displacement is given as 30cm.

First we should check if all the values are in their standard form.

Here 30cm should be converted to metre by dividing it with 100.

Which would give us 0.3m

Now we use the equation W=force x displacement =75 x 0.3=22.5J

I hope this satisfies you. If u have any further questions please let me know.

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3 0
3 years ago
Two pipes of equal length are each open at one end. Each has a fundamental frequency of 470 Hz at 310 K. In one pipe the air tem
Tresset [83]

Answer:

Given:

Fundamental frequency: 470Hz

T1:310k,T2:315k

Calculating velocity

Recall v=(331m/s)✓[T1/273k)

V=331✓(310/273)

V1=331*(1.0656)=352.72m/s

V2=331✓(315/273)=355.5m/s

Fundamental frequency=4L

F2=F1(V2/V1)

F2=470(355.5/352.72)=474.4Hz

Beat=[F2-F1]=474.4-470=4.4Hz

Explanation:

7 0
3 years ago
The diagram shows a wave traveling through a medium.
m_a_m_a [10]
The answer is b. The crest.

Hope this helps!
6 0
3 years ago
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what will happen when two wave pulses that are the same size approach one another from opposite directions?
xenn [34]
When two or more waves meet, they interact with each other. The interaction of waves with other waves is called wave interference. Wave interference may occur when two waves that are traveling in opposite directions meet. The two waves pass through each other, and this affects their amplitude.
5 0
2 years ago
In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 1.00 mm away. part a
mihalych1998 [28]
<span>We can use Coulomb's law to find the force F acting on the proton that is released. F = k x Q1 x Q2 / r^2 k = 9 x 10^9 Q1 is the charge on one proton which is 1.6 x 10^{-19} C Q2 is the same charge on the other proton r is the distance between the protons F = (9x10^9) x (1.6 x 10^{-19} C) x (1.6 x 10^{-19} C) / (10^{-3})^2 F = 2.304 x 10^{-22} N We can use the force to find the acceleration. F = ma a = F / m a = (2.304 x 10^{-22} N) / (1.67 x 10^{-27} kg) a = 1.38 x 10^5 m/s^2 The initial acceleration of the proton is 1.38 x 10^5 m/s^2</span>
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3 years ago
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