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2 years ago

A force that pushes or pulls is known as

Physics

2 answers:

AlekseyPX2 years ago

8 0

Answer:contact force

Explanation:

Contact forces are forces that requires contact with the body to which they are applied. Examples are: push, pull, tension, friction

xeze [42]2 years ago

7 0

Answer:

A force that pushes or pulls is known as Newton's third law of Motion.

Explanation:

Newton's Third Law of Motion. Newton's Third Law of Motion states that for each action, there's an equal and opposite reaction. What this suggests is that pushing on an object causes that object to keep off against you, the precise same amount, but within the other way.

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Nuetrik [128]

Answer:

140m

70sec I cant show my work but I gave you a push

Explanation:

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2 years ago

One of the harmonic frequencies of tube A with two open ends is 576 Hz. The next-highest harmonic frequency is 648 Hz. (a) What

balu736 [363]

(a) 288 Hz

The difference between any two harmonics of an open-end tube is equal to the fundamental frequency, $f_1$ (first harmonic):

$f_{n+1}-f_n = f_1$ (1)

In this problem, we are told the frequencies of two successive harmonics:

$f_n = 576 Hz\\f_{n+1}=648 Hz$

So the fundamental frequency is:

$f_1 = 648 Hz-576 Hz=72 Hz$

Now we know that one of the the harmonics is $f_n=216 Hz$ , so its next highest harmonic will have a frequency of

$f_{n+1}=f_n+f_1 = 216 Hz+72 Hz=288 Hz$

(b) n=4

The frequency of the nth-harmonic is an integer multiple of the fundamental frequency:

$f_n=n f_1$ (2)

Since we know $f_n = 288 Hz$ , we can solve (2) to find the number n of this harmonic:

$n=\frac{f_n}{f_1}=\frac{288 Hz}{72 Hz}=4$

(c) 4445 Hz

For a closed pipe (only one end is open), the situation is a bit different, because only odd harmonics are allowed. This means that the frequency of the nth-harmonic is an odd-integer multiple of the fundamental frequency:

$f_n=(2n+1) f_1$ (2)

so, the difference between any two harmonics tube is equal to:

$f_{n+1}-f_n = (2(n+1)+1)f_1-(2n+1)f_1=(2n+3)f_1-(2n+1)f_1=2f_1$ (1)

In this problem, we are told the frequencies of two successive harmonics:

$f_n = 4699 Hz\\f_{n+1}=4953 Hz$

So, according to (1), the fundamental frequency is equal to half of this difference:

$f_1 = \frac{4953 Hz-4699 Hz}{2}=127 Hz$

Now we know that one of the harmonics is $f_n=4191 Hz$ , so its next highest harmonic will have a frequency of

$f_{n+1}=f_n+2f_1 = 4191 Hz+254 Hz=4445 Hz$

(d) n=17

We said that the frequency of the nth-harmonic is equal to an odd-integer multiple of the fundamental frequency:

$f_n=(2n+1) f_1$ (2)

Since we know $f_n = 4445 Hz$ , we can solve (2) to find the number n of this harmonic:

$n=\frac{1}{2}(\frac{f_n}{f_1}-1)=\frac{1}{2}(\frac{4445 Hz}{127 Hz}-1)=17$

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Reduced HDL cholesterol

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2 years ago

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Someone help please 1 You are cycling along a flat road. a Are you moving up or down? b What is the force that is pulling you

9966 [12]

Answer:

$\gamma \: 1a. \: i \: am \: moving \: downwards \: . \\ b. \: gravity \: is \: pulling \: me \: downwards \: \\ c. \: the \: force \: is \: acting \: downwards \: \\ the \: answer \: to \: the \: question \: c \: is \: downwars \: because \: its \: easy \: to \: pull \: the \: bike \: downwards \: but \: its \: hard \: to \: pull \: the \: bike \: upwards \: it \: would \: take \: more \: energy \: \gamma \\ \\$

Explanation:

<h3> $\infty \infty \: hope \: it \: helps \: you \: \\ have \: a \: good \: a \: day \: .$ </h3>

Slide my answer to see the full answer .

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3 years ago

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Answer: Euglena are photosynthetic

Explanation: USA test prep

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