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2 years ago

I need help on this and the first person who will answer this correctly gets BRANLIST

Physics

1 answer:

Nikolay [14]2 years ago

3 0

Answer:

i think it's b but i don't know for sure

Explanation:

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Stells [14]

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4 0

3 years ago

Read 2 more answers

E fundamental frequency of an open organ pipe corresponds to the middle c (261.6 hz on the chromatic musical scale). the third r

luda_lava [24]

The wavelength of the third resonance of the closed organ pipe is equal to the ratio between the speed of sound and the frequency of the 3rd harmonic:

$\lambda_3 = \frac{c}{f_3}=\frac{343 m/s}{261.6 Hz} =1.31 m$

The relationship between length of a closed pipe and wavelength of the standing waves inside is:

$L=\frac{n}{4}\lambda_n$

where n is the number of the harmonic. In this case, n=3, so the length of the pipe is

$L=\frac{3}{4}(1.31 m)=0.98 m$

8 0

3 years ago

When 2 forces are applied in opposite directions, how do you calculate the net force?

Effectus [21]

You find the net force by subtracting.

3 0

3 years ago

If F1 is the magnitude of the force exerted

aleksandrvk [35]

Answer: 3. F1 = F2

Explanation:

According to <u>Newton's law of Gravitation</u>, the force $F$ exerted <u>between two bodies</u> or objects of masses $M$ and $m$ and separated by a distance $r$ is equal to the product of their masses divided by the square of the distance:

$F=G\frac{Mm}{r^2}$ (1)

Where $G$ is the gravitational constant

Now, in the especific case of the Earth and the satellite, where the Earth has a mass $M$ and satellite a mass $m$ , being both separated a distance $r$ , the force exerted by the Earth on the satellite is:

$F1=G\frac{Mm}{r^2}$ (2)

And the force exerted by the satellite on the Earth is:

$F2=G\frac{Mm}{r^2}$ (3)

As we can see equations (2) and (3) are equal, hence the magnitude of the gravitational force is the same for both:

$F1=F2$

3 0

3 years ago

(a) If the maximum acceleration that is tolerable for passengers in a subway train is 1.34 m/s² and subway stations are located

andreev551 [17]

Answer:

The correct answer is 32.9 m/s

Explanation:

To solve this, we list out the known and the unknown variables as follows

Maximum allowable acceleration = 1.34 m/s²

Distance between sttions = 806 m

Therefore from the equation of motion

v = ut + 0.5·×at²

Where v = final velocity

u = initial velocity

S = distance covered

t = time

a = acceleration

Also v² = u² + 2·a·S

where u is the initial velocity, which we can take as u = 0, then

v² = 2·1.34·S = 2.68S m²/s² then

Also the train has to decelerate from maximum speed to stop at the next tran station wherev = 0, thus v² = u² -2·1.34·Z, so u² = 2.68Z

since u² = 2.68S from the previous calculation, then for v = 0

2.68S = 2.68Z thus S = Z which and to reach the next subway station S + Z must be = 806 m, then S = 806 m ÷ 2 = 403 m

and v² = 2.68S m²/s² = 1080.04 m²/s²

v = 32.9 m/s

The maximum speed a subway train can attain between stations is 32.9 m/s

3 0

3 years ago

I need help on this and the first person who will answer this correctly gets BRANLIST​

I need help on this and the first person who will answer this correctly gets BRANLIST