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3 years ago

What type of galaxy is shown?

2 answers:

8 0

Elliptical, because the shape of the galaxy isn’t like the others. It is unique to its own and doesn’t have another to compare to

Katena32 [7]3 years ago

7 0

Answer:

Elliptical

Explanation:

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If two is company and three is a crowd, what is 4 and 5?

LiRa [457]

4 and 5 is 9

(4 + 5)

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4 years ago

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Which of the following processes is not regarded as contributing to the creation of greenhouse gases? deforestation creating ele

DiKsa [7]

Creating electricity from wind is not regarded as a process contributing to the creation of greenhouse gases. Meanwhile, the processes such as deforestation, the creation of electricity from coal and the use of fertilizers are greatly contributing to the making of greenhouse gases.

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3 years ago

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A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit

Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

$P = \frac{1}{2} \mu \omega^2 A^2 v$

Here,

$\mu$ = Linear mass density of the string

$\omega =$ Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

$v = \sqrt{\frac{T}{\mu}} \rightarrow$ Here T is the Period

For the linear mass density we have that

$\mu = \frac{m}{l}$

And the angular frequency can be written as

$\omega = 2\pi f$

Replacing this terms and the first equation we have that

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})$

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})$

$P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})$

PART A ) Replacing our values here we have that

$P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.2320W$

PART B) The new amplitude A' that is half ot the wavelength of the wave is

$A' = \frac{1.8*10^{-3}}{2}$

$A' = 0.9*10^{-3}$

Replacing at the equation of power we have that

$P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.058W$

8 0

3 years ago

Ml(d^2θ/dt^2) =-mgθ

Nata [24]

The equation of motion of a pendulum is:

$\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\dfrac{g}{\ell}\sin\theta,$

where $\ell$ it its length and $g$ is the gravitational acceleration. Notice that the mass is absent from the equation! This is quite hard to solve, but for small angles ( $\theta \ll 1$ ), we can use:

$\sin\theta \simeq \theta.$

Additionally, let us define:

$\omega^2\equiv\dfrac{g}{\ell}.$

We can now write:

$\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\omega^2\theta.$

The solution to this differential equation is:

$\theta(t) = A\sin(\omega t + \phi),$

where $A$ and $\phi$ are constants to be determined using the initial conditions. Notice that they will not have any influence on the period, since it is given simply by:

$T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{g}{\ell}}.$

This justifies that the period depends only on the pendulum's length.

4 0

3 years ago

You and a friend are studying late at night. There are three 110 W light bulbs and a radio with an internal resistance of 56.0 Ω

Sliva [168]

Answer:

The total electrical power we are using is: 1316 W.

Explanation:

Using the ohm´s law $V=I*R$ and the formula for calculate the electrical power, we can find the total electrical power that we are using. First we need to find each electrical power that is using every single component, so the radio power is: $I=\frac{V}{R}=\frac{110 (v)}{56(ohms)}=1.96(A)$ , so the radio power is: $P=I*V=1.96(A)*110(v)=216(W)$ , then we find the pop-corn machine power as: $P=I*V=7(A)*110(v)=770(W)$ and finally there are three light bulbs of 110(W) so: P=3*110(W)=330(W) and the total electrical power is the adding up every single power so that: P=330(W)+770(W)+216(W)=1316(W).

8 0

3 years ago