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3 years ago

NEED HELP TODAY

Physics

2 answers:

larisa86 [58]3 years ago

7 0

Answer:

B- It is a constant when the light is traveling in a vacuum.

Explanation:

"Light in a vacuum always travels at the same speed." Those additional three words in a vacuum are very important. A vacuum is a region with no matter in it. So a vacuum would not contain any dust particles (unlike a vacuum cleaner, which is generally full of dust particles)

Free_Kalibri [48]3 years ago

6 0

Answer:

I am pretty sure it is B My friend hope you are well

Explanation:

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Three identical resistors are connected in parallel. The equivalent resistance increases by 630 when one resistor is removed and

strojnjashka [21]

Answer:

each resistor is 540 Ω

Explanation:

Let's assign the letter R to the resistance of the three resistors involved in this problem. So, to start with, the three resistors are placed in parallel, which results in an equivalent resistance $R_e$ defined by the formula:

$\frac{1}{R_e}=\frac{1}{R} } +\frac{1}{R} } +\frac{1}{R} \\\frac{1}{R_e}=\frac{3}{R} \\R_e=\frac{R}{3}$

Therefore, R/3 is the equivalent resistance of the initial circuit.

In the second circuit, two of the resistors are in parallel, so they are equivalent to:

$\frac{1}{R'_e}=\frac{1}{R} +\frac{1}{R}\\\frac{1}{R'_e}=\frac{2}{R} \\R'_e=\frac{R}{2} \\$

and when this is combined with the third resistor in series, the equivalent resistance ( $R''_e$ ) of this new circuit becomes the addition of the above calculated resistance plus the resistor R (because these are connected in series):

$R''_e=R'_e+R\\R''_e=\frac{R}{2} +R\\R''_e=\frac{3R}{2}$

The problem states that the difference between the equivalent resistances in both circuits is given by:

$R''_e=R_e+630 \,\Omega$

so, we can replace our found values for the equivalent resistors (which are both in terms of R) and solve for R in this last equation:

$\frac{3R}{2} =\frac{R}{3} +630\,\Omega\\\frac{3R}{2} -\frac{R}{3} = 630\,\Omega\\\frac{7R}{6} = 630\,\Omega\\\\R=\frac{6}{7} *630\,\Omega\\R=540\,\Omega$

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What percent of all bolts fire within a cloud

Leni [432]

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3 years ago

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PIT_PIT [208]

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1 year ago

a car whose mass is 1000kg is traveling at a constant speed of 10m/s. Neglecting any friction how much force will the engine hav

AURORKA [14]

This next statement is a big deal. It should be up on a board, surrounded
by flashing red and yellow lights, and hung on the wall of every Science
classroom. Although we never see it in our daily lives, it's fundamental to
the workings of the universe, and it's also Newton's first law of motion:

Without friction, it doesn't take ANY force to keep a moving object
moving. Force is only required to change the object's speed, or to
change the direction in which it's moving.

The answer to the question is: On a level road, and neglecting any friction,
the engine doesn't have to supply ANY force to keep the car going at the
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3 years ago

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