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3 years ago

Which statement correctly describes variable stars?

Physics

1 answer:

Kryger [21]3 years ago

5 0

Answer:

Explanation:

The Period-Luminosity relationship tells us that luminosity increases with the period, and of course the more luminosity a star has the more far away they can be seen, so from this we know that:

A) False since lower luminosities can be observed when they are close.

B) False since longer periods means higher luminosities

C) False since lower luminosities can be observed when they are close.

D) True: Variable stars with shorter periods have lower luminosities, so they can only be observed when they are close.

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If vector C is added to vector D, the result is a third vector that is perpendicular to D and has a magnitude equal to 3D. What

Jet001 [13]

Answer:

(e) 3.2

Explanation:

We are given that vector C and D.

Let R be the magnitude of C+D.

According to question

R=3D

We have to find the ratio of the magnitude of C to that of D.

By using right triangle property

$C^2=R^2+D^2$

$C^2=(3D)^2+D^2$

$C^2=9D^2+D^2$

$C^2=10D^2$

$C=\sqrt{10D^2}=3.2D$

$\frac{C}{D}=3.2$

Hence, the ratio of the magnitude of C to that of D=3.2

(e) 3.2

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3 years ago

. The penalties for a person's second DUI conviction include completion of __________ hours of DUI school.

Ghella [55]

Answer:

Explanation:

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2 years ago

What is the difference in the speed of the generator with a small magnet and a generator with a large magnet?

Delvig [45]

With a small magnet with a generator it will be taken up quickly because how small it is while with a big generator it would take more force for it for the generator to attach because the larger the magnet that heavier it will be because it is attached to the North Pole magnet

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2 years ago

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What Energy source that produces wind

svetoff [14.1K]

Wind is usually caused by a difference in pressure that causes wind to be "Pushed" on the surface of earth. Air usually moves from high to low pressure and that is what we call wind. Please hit the thanks button if this helped :)

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2 years ago

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A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel

Dafna1 [17]

Answer:

$1.69\cdot 10^{10}J$

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

$E=-G\frac{mM}{2r}$

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

$M=5.98\cdot 10^{24}kg$ is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

$r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m$

So the initial total energy is

$E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J$

When the satellite hits the ground, it is now on Earth's surface, so

$r=R=6370 km=6.37\cdot 10^6 m$

so its gravitational potential energy is

$U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J$

And since it hits the ground with speed

$v=1.90 km/s = 1900 m/s$

it also has kinetic energy:

$K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J$

So the total energy when the satellite hits the ground is

$E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J$

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

$\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J$

8 0

3 years ago