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Irina-Kira [14]
3 years ago
9

A rope vibrates every 0.5 s. what is the frequency of the waves?

Physics
2 answers:
julsineya [31]3 years ago
6 0

The rope vibrates every 0.5 s: it means that its period is

T=0.5 s


The frequency of an oscillation is equal to the reciprocal of the period:

f=\frac{1}{T}

Therefore, if we substitute T=0.5 s in the formula, we get:

f=\frac{1}{0.5 s}=2 Hz

umka2103 [35]3 years ago
4 0
F = 1/t
F = 1/0.5
F = 2Hz

Answer is 2Hz
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A 3.50 cm tall object is held 24.8 cm from a lens of focal length 16.0 cm. What is the image height?
Mandarinka [93]

Answer:

<h2>6.36 cm</h2>

Explanation:

Using the formula to first get the image distance

1/f = 1/u+1/v

f = focal length of the lens

u = object distance

v = image distance

Given f = 16.0 cm, u = 24.8 cm

1/v = 1/16 - 1/24.8

1/v = 0.0625-0.04032

1/v = 0.02218

v = 1/0.02218

v = 45.09 cm

To get the image height, we will us the magnification formula.

Mag = v/u = Hi/H

Hi = image height = ?

H = object height = 3.50 cm

45.09/24.8 = Hi/3.50

Hi = (45.09*3.50)/24.8

Hi = 6.36 cm

The image height is 6.36 cm

6 0
2 years ago
A bullet fired vertically at a velocity of 36m/s .after 45 the bullet hit the top of a bulid how height is a bulid?​
zzz [600]

Answer:

The height of the building is 8,302.5 m

Explanation:

Given;

velocity of the projectile, u = 36 m/s

time of motion, t = 45 s

Let the upward direction of the bullet be negative,

The height of the building is calculated as;

h = ut - \frac{1}{2} gt^2\\\\h = (36\times 45) - (\frac{1}{2} \times 9.8 \times 45^2)\\\\h = 1620 - 9922.5\\\\h = -8,302.5 \ m\\\\The \ height \ of \ the \ building \ is \ 8,302.5  \ m

3 0
2 years ago
calculate the amount of work done by a person while taking a bag of mass 100kg to the top of the building hight 10m. The mass of
vredina [299]

Explanation:

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Total weight=mass×gravitational field strength

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Work done=force×distance

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juin [17]

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Vesnalui [34]

Answer:

<em>The comoving distance and the proper distance scale</em>

<em></em>

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The comoving distance scale removes the effects of the expansion of the universe, which leaves us with a distance that does not change in time due to the expansion of space (since space is constantly expanding). The comoving distance and proper distance are defined to be equal at the present time; therefore, the ratio of proper distance to comoving distance now is 1. The scale factor is sometimes not equal to 1. The distance between masses in the universe may change due to other, local factors like the motion of a galaxy within a cluster.  Finally, we note that the expansion of the Universe results in the proper distance changing, but the comoving distance is unchanged by an expanding universe.

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