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3 years ago

Which variable stars have pulsation periods between 1.5 hours and 1.2 days?

2 answers:

7 0

D. Nebulas is a cloud of dust,
B. Protostars & A. Cepheid variables have longer pulsation periods.
Whereas, C. RR Lyrae variables have pulsation periods between 1.5 hours and 1.2 days. So the answer is C.

maria [59]3 years ago

3 0

Correct answer choice is :

C) RR Lyrae variables

Explanation:

RR Lyrae variables are periodical unsteady stars, usually discovered in globular groups. They are accepted as regular lights to measure galactic distances, as a part of the infinite distance ladder. This class of unsteady star is identified after the prototype and glorious example, RR Lyrae. R Lyraes are vibrating parallel section aging stars of phantom class A or F, with a mass of around half the Sun's.

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A wave travels at 295 m/s and has a wavelength of 2.50 m. What is the frequency of the wave?

posledela

Answer:

$118\; \rm Hz$ .

Explanation:

The frequency $f$ of a wave is equal to the number of wave cycles that go through a point on its path in unit time (where "unit time" is typically equal to one second.)

The wave in this question travels at a speed of $v= 295\; \rm m\cdot s^{-1}$ . In other words, the wave would have traveled $295\; \rm m$ in each second. Consider a point on the path of this wave. If a peak was initially at that point, in one second that peak would be

How many wave cycles can fit into that $295\; \rm m$ ? The wavelength of this wave $\lambda = 2.50\; \rm m$ gives the length of one wave cycle. Therefore:

$\displaystyle \frac{295\;\rm m}{2.50\; \rm m} = 118$ .

That is: there are $118$ wave cycles in $295\; \rm m$ of this wave.

On the other hand, Because that $295\; \rm m$ of this wave goes through that point in each second, that $118$ wave cycles will go through that point in the same amount of time. Hence, the frequency of this wave would be

Because one wave cycle per second is equivalent to one Hertz, the frequency of this wave can be written as:

$f = 118\; \rm s^{-1} = 118\; \rm Hz$ .

The calculations above can be expressed with the formula:

$\displaystyle f = \frac{v}{\lambda}$ ,

where

$v$ represents the speed of this wave, and
$\lambda$ represents the wavelength of this wave.

6 0

3 years ago

a gym consists of a rectangular region with a semi-circle on each end. if the perimeter of the room is to be a 200 m running tra

Nikolay [14]

The dimensions of the rectangle are:

l = 50 m

b = $100/\pi$ m

<h3>What is a perimeter in math?</h3>

The perimeter is the length of the outline of a shape. To find the perimeter of a rectangle or square you have to add the lengths of all the four sides.

<h3>How do we find a perimeter of a rectangle?</h3>

The perimeter of a rectangle,denoted by P is given by the formula, P=2l+2b, where l is the length and b is the breadth of the rectangle.

<h3>Given:</h3>

As per the question:

Perimeter of the room is given as P = 200 m

The region is rectangular having a semicircle at each end.

Now,

Let 'l' be the length of the rectangle, 'b' be its breadth and 'r' be the radius of the semi-circle at each end.

Then, Area of the given rectangle, A = lb

Perimeter of the room, P is = $\pi r+l+\pi r+l=2\pi r+2l=\pi b+2l$