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lesantik [10]
3 years ago
9

A pendulum is observed to complete 23 full cycles in 58 seconds. Use the definition of frequency to find the frequency.

Physics
1 answer:
Alex17521 [72]3 years ago
4 0

Answer: f = 0.397 Hz

Explanation:

f = cycles per second

f = 23 / 58 = 0.39655...

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Photons are particles of electromagnetic radiation.

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3 years ago
What is kepler's law??​
Elan Coil [88]
<h2>QUESTION:- </h2>

➜what is kepler's law??

\huge\red{\boxed{\huge\mathbb{\red A \pink{N}\purple{S} \blue{W}\orange{ER}}}}

Kepler gave the three laws or theorems of motion of the orbitals bodies

{\huge {\bold{ \red{ \star}}}}{ \blue{ \bold{FIRST \: \: \: LAW}}}

This law state that the celestial bodies revolves around the stars in elliptical orbit and star as a single focus.

Example :- Earth revolves around the Sun as assuming it as single focus

This also shows that earth revolves around the sun in elliptical orbit.

{\huge {\bold{ \blue{ \star}}}}{ \green{ \bold{SECOND \: \: \: LAW}}}

Area covered by the planet is equal in equal duration of time irrespective of the position of the planet.

It also states that Angular momentum is constant

As Angular momentum is constant it means areal velocity is also constant.

\frac{ \triangle \: A}{ \triangle \: T} = \frac{L}{2m}△T△A=2mL

where:-

A is the area.

T is the time.

L is the angular momentum.

M is the mass of the body.

{\huge {\bold{ \green{ \star}}}}{ \purple{ \bold{THIRD \: \: \: LAW}}}

square of the time of the revolution is directly proportional to the cube of the distance between the planet and star in Astronomical unit.

{T}^{2} = {a}^{3}T2=a3

where:-

T = time of revolution

a is the distance between the planet and star.

\purple\star \: {Thanks \: And \: Brainlist} \blue \star \\ {\orange{ \star}}{if \: U \: Like d \: My \: Ans} {\green{ \star }}

8 0
3 years ago
I need help if someone has coursehero plss some me the answer and I will give you brainliest plssss
Vinil7 [7]

Answer:

5. All of the answers are yes.

Explanation:

<h2><u><em>PLEASE MARK AS BRAINLIEST!!!!!</em></u></h2>
7 0
2 years ago
A wire of length 5mm and Diameter 2m extends by 0.25 when a force of 50N was use. calculate the
bazaltina [42]

Answer and Explanation:

Data provided in the question

Force = 50N

Length = 5mm

diameter = 2.0m = 2\times 10^{-3}

Extended by = 0.25mm = 0.25\times 10^{-3}

Based on the above information, the calculation is as follows

a. The Stress of the wire is

= \frac{force\ applied}{area\ of \ circle}

here area of circle = perpendicular to the are i.e cross-sectional  i.e

= \frac{\pi d^{2}}{4}

= \frac{\pi(2\times 10^{-3})^2}{4}

Now place these above values to the above formula

= \frac{4\times 50}{\pi\times 4 \times 10^{-6}} \\\\ = \frac{50}{\pi}

= 15.92 MPa

As 1Pa = 1 by N m^2

So,

MPa = 10^6 N m^2

b. Now the strain of the wire is

= \frac{Change\ in\ length}{initial\ length} \\\\ = \frac{0.25\times 10^{-3}}{5}

= 5 \times 10^{-5}

3 0
3 years ago
Kbhgyugybhhv vihiklmb
Ahat [919]

Answer:

?

Explanation:

Can u explain your question?

4 0
2 years ago
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