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4 years ago

12

Complete the following sentences. a. The motion of the particles is a model of__________. . b. The movement of particles from on

e chamber to another is a model of________. . c. When the average motion of the particles is the same in both chambers, this is a model of _______.

1 answer:

Leno4ka [110]4 years ago

8 0

Effectiv was the morning time to go home

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A student observed a sample of water in three states of matter. The student should describe the liquid water as a state of matte

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B. Less volume than a solid state

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3 years ago

Question 1 (1 point)

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Answer:

The work done by the frictional force is 600J.

Explanation:

The work $W$ done by the frictional force is

$W= Fd$ .

Now, $F = 60N$ and $d =10m$ ; therefore,

$W= (60N)(10m)$

$\boxed{W = 600J.}$

Hence, the work done by friction is 660J.

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3 years ago

This diagram is a model of the nitrogen atom.

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C.
Nitrogen has the atomic no. of 7 so the proton and the electron no. is 7.
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3 years ago

You and your friend are going bungee jumping! You wait directly below them with a camera. When they leap from the bridge they be

Alex

Answer:

The amplitude is $A = 90.2 \ m$

Explanation:

From the question we are told that

The frequency of when sound is approaching observer is $f = 392 Hz$

The frequency as the move away from observer is $f_ a = 330 \ Hz$

The time between the pitch are $t = 10 \ s$

Here you are the observer and your friends are the source of the sound

The period is mathematically evaluated as

$T = 2 t$

as it is the time to complete one oscillation which from on highest pitch to the next highest pitch

Now T can also be mathematically represented as

$T = \frac{2 \pi}{w}$

Where $w$ is the angular velocity

=> $\frac{2 \pi}{w} = 2 * 10$

=> $w = 0.314 \ rad/sec$

Now using Doppler Effect,

The source of the sound is approaching the observer

The

$f = f_o (\frac{v}{v- wA} )$

$392 = f_o (\frac{v}{v- wA} )$

Where A is the amplitude

So when the source is moving away from the observer

$f_a = f_o (\frac{v}{v+ wA} )$

$330 = f_o (\frac{v}{v+ wA} )$

Here $f_o$ is the fundamental frequency

Dividing the both equation we have

$\frac{392}{330} = \frac{f_o(\frac{v}{v-wA} )}{f_o(\frac{v}{v+wA}}$

$1.1878 = \frac{v+wA}{v-wA}$

$1.1878 v - 1.1878 wA = v+wA$

$1.1878 v = 2.1878 wA$

=> $A = \frac{(0.1878 * (330))}{(2.1878)* (0.314)}$

$A = 90.2 \ m$

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3 years ago

If the number of crests that pass a point in a given time increases, what has increased?

zysi [14]

The frequency increasing makes the crests pass more quickly. Frequency is a count of how many times per second an event occurs. In waves, this event is the passing of an entire cycle. Once the cycle has passed, the wave repeats. The faster the wave repeats, the higher the frequency. For this reason, frequency has units of hertz, Hz. The unit of hertz is 1/s or "per second"

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3 years ago

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