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Arte-miy333 [17]

3 years ago

12

The brightness of a star depends on its (color, composition of atmosphere, or distance from earth), and stars that are closer lo

ok (brighter, dimmer, or white).
(in parentheses is options)

2 answers:

larisa [96]3 years ago

5 0

The brightness of stars depends on its distance from earth. Stars that are closer look brighter. Star stuff is my favorite thing to learn about because I'm a sweaty nerd, so if you have anymore related questions I'd love to answer them.

Alex777 [14]3 years ago

4 0

The apparent brightness of a star depends on how bright it really IS, and also on its distance from Earth. Just like matches, flashlights, and fireflies, stars that are closer to us look brighter to us. (Think about the Sun.)

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Two charges that are separated by one meter exert 1-n forces on each other. if the magnitude of each charge is doubled, the forc

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The electrostatic force between the two charges is
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$F \sim q_1 q_2$
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Ivanshal [37]

If the wavelength increases (gets longer), then the frequency <em>decreases</em>.

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Water is traveling through a horizontal pipe with a speed of 1.7 m/s and at a pressure of 205 kPa. This pipe is reduced to a new

Arturiano [62]

Answer:

The velocity is $v_2 = 6.8 \ m/s$

The pressure is $P_2 = 204978 Pa$

Explanation:

From the question we are told that

The speed at which water is travelling through is $v = 1.7 \ m/s$

The pressure is $P_1 = 205 k Pa = 205 *10^{3} \ Pa$

The diameter of the new pipe is $d = \frac{D}{2}$

Where D is the diameter of first pipe

According to the principal of continuity we have that

$A_1 v_1 = A_2 v_2$

Now $A_1$ is the area of the first pipe which is mathematically represented as

$A_1 = \pi \frac{D^2}{4}$

and $A_2$ is the area of the second pipe which is mathematically represented as

$A_2 = \pi \frac{d^2}{4}$

Recall $d = \frac{D}{2}$

$A_2 = \pi \frac{[ D^2]}{4 *4}$

$A_2 = \frac{A_1}{4}$

So $A_1 v_1 = \frac{A_1}{4} v_2$

substituting value

$1.7 = \frac{1}{4} * v_2$

$v_2 = 4 * 1.7$

$v_2 = 6.8 \ m/s$

According to Bernoulli's equation we have that

$P_1 + \rho \frac{v_1 ^2}{2} = P_2 + \rho \frac{v_2 ^2}{2}$

substituting values

$205 *10^{3 }+ \frac{1.7 ^2}{2} = P_2 + \frac{6.8 ^2}{2}$

$P_2 = 204978 Pa$

4 0

3 years ago