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Varvara68 [4.7K]

3 years ago

13

Is the sun considered a star?

2 answers:

Sauron [17]3 years ago

8 0

Yes! The sun is a star, just like any star you can see in the sky at night. The sun is in the center of our solar system and the closest star to Earth.

wel3 years ago

6 0

Yes it is because it has all the characteristics of one
The sun is unremarkable
There are countless stars with the same attributes as the sun throughout the universe

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HELP ME ASAP!!!! PIC IS DOWN BELOW!

vagabundo [1.1K]

I think it is B Because it sounds like the best choice

7 0

3 years ago

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Anaya Goodloe decides to spike the football after a big touchdown (on Earth) and it

Llana [10]

Answer:

The football leaves with the velocity, u = 15.68 m/s

Explanation:

Given data,

The football bounces back up off the ground and is airborne for, t = 3.2 s

Let the football bounces back up off the ground in the vertical direction

The formula for time of flight is given by,

t = 2u /g

∴ u = gt / 2

Substituting the values,

u = 9.8 x 3.2 / 2

u = 15.68 m/s

Hence, the football leaves with the velocity, u = 15.68 m/s

6 0

4 years ago

An ultraviolet wave traveling through a vacuum has wavelength of 4.0 x 10^-7 m. The waves frequency, written in scientific notat

Naily [24]

Answer:

λ = c / f or f = c / λ

f = 3.0E8 / 4.0E-7 = .75E15 / sec = 7.5E14 / sec = 7.5 X 10^14 /sec

5 0

2 years ago

Force F acts between a pair of charges, q1 and q2, separated by a distance d. For each of the statements, use the drop-down menu

lora16 [44]

The initial force between the two charges is given by:

$F=k \frac{q_1 q_2}{d^2}$

where k is the Coulomb's constant, q1 and q2 the two charges, d their separation. Let's analyze now the other situations:

1. F

In this case, q1 is halved, q2 is doubled, but the distance between the charges remains d.

So, we have:

$q_1' = \frac{q_1}{2}\\q_2' = 2 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(\frac{q_1}{2})(2q_2)}{d^2}=k \frac{q_1 q_2}{d^2}=F$

So the force has not changed.

2. F/4

In this case, q1 and q2 are unchanged. The distance between the charges is doubled to 2d.

So, we have:

$q_1' = q_1\\q_2' = q_2\\d' = 2d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{q_1 q_2)}{(2d)^2}=\frac{1}{4} k \frac{q_1 q_2}{d^2}=\frac{F}{4}$

So the force has decreased by a factor 4.

3. 6F

In this case, q1 is doubled and q2 is tripled. The distance between the charges remains d.

So, we have:

$q_1' = 2 q_1\\q_2' = 3 q_2\\d' = d$

So, the new force is:

$F'=k \frac{q_1' q_2'}{d'^2}= k \frac{(2 q_1)(3 q_2)}{d^2}=6 k \frac{q_1 q_2}{d^2}=6F$

So the force has increased by a factor 6.

8 0

3 years ago

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Which of the following accurately describes the behavior of these two mechanical waves when they intersect?

Lerok [7]

D because I learned this 2 years ago

5 0

3 years ago