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erma4kov [3.2K]
3 years ago
14

Which statement is correct about waves?

Physics
1 answer:
grandymaker [24]3 years ago
7 0

Answer:

b is the correct answer

Explanation:

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PLEASE HELP ASAP ON TIMED QUIZ
Ganezh [65]

Answer:

It is C and 100 percent sure about it

4 0
3 years ago
An old lady sits on and pushes down on a park bench. What is the reaction force? A. The bench pushes up on her. B. The Earth pul
Vladimir [108]
B the earth pulls down because of gravity
3 0
4 years ago
Read 2 more answers
A planet is discovered orbiting around a star in the galaxy Andromeda at the same distance from the star as Earth is from the Su
aniked [119]

Answer:

The planet´s orbital period will be one-half Earth´s orbital period.

Explanation:

The planet in orbit, is subject to the attractive force from the sun, which is given by the Newton´s Universal Law of Gravitation.

At the same time, this force, is the same centripetal force, that keeps the planet in orbit (assuming to be circular), so we can put the following equation:

Fg = Fc ⇒ G*mp*ms / r² = mp*ω²*r

As we know to find out the orbital period, as it is the time needed to give a complete revolution around the sun, we can say this:

ω = 2*π / T (rad/sec), so replacing this in the expression above, we get:

Fg = Fc ⇒   G*mp*ms / r² = mp*(2*π/T)²*r

Solving for T²:

T² = (2*π)²*r³ / G*ms (1)

For the planet orbiting the sun in Andromeda, we have:

Ta² = (2*π)*r³ / G*4*ms (2)

As the radius of the orbit (distance to the sun) is the same for both planets, we can simplify it in the expression, so, if we divide both sides in (1) and (2), simplifying common terms, we finally get:

(Te / Ta)² =  4  ⇒ Te / Ta = 2 ⇒ Ta = Te/2

So, The planet's orbital period will be one-half Earth's orbital period.

7 0
3 years ago
What other theory does the author compare to the Big Bang to?
sveta [45]

Answer:

The two Russian authors based their exposition on what they called the Friedmann theory of a singular beginning of the universe, referring throughout to the “theory of the hot universe” as an alternative to the hot Big Bang theory.

7 0
3 years ago
Suppose that the strings on a violin are stretched with the same tension and each has the same length between its two fixed ends
lesya692 [45]

Answer:

0.00384 kg/m

Explanation:

The fundamental frequency of string waves is given by

f=\frac{1}{2L}\sqrt{\frac{F}{\mu}}

For some tension (F) and length (L)

f\propto\frac{1}{\mu}

Fundamental frequency of G string

f_G=196\ Hz

Fundamental frequency of E string

f_E=659.3\ Hz

Linear mass density of E string is

\mu_E=3.4\times 10^{-4}\ kg/m

So,

\frac{F_G}{F_E}=\sqrt{\frac{\mu_E}{\mu_G}}\\\Rightarrow \frac{F_G^2}{F_E^2}=\frac{\mu_E}{\mu_G}\\\Rightarrow \mu_G=3.4\times 10^{-4}\times \frac{659.3^2}{196^2}\\\Rightarrow \mu_G=0.00384\ kg/m

The linear density of the G string is 0.00384 kg/m

4 0
3 years ago
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