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agasfer [191]
1 year ago
10

Which part of the microscope is the circular area on the stage that light passes through?

Physics
1 answer:
patriot [66]1 year ago
6 0

Answer: The part of the microscope that is the circular area is the APERTURE

I hope this helped!

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ASAP ONLY CORRECT ANSWERS PLZZZZZZZZZZZZZ
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Answer b

Explanation:

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STATE THE HOOKE'S LAW
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The Hooke's law is a principal of physics that states that the force needed to extend or compress a spring by some distance x scales linearly with respect to that distance.
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Which statement correctly describes the current in a circuit that is made up of any two resistors connected in parallel with a b
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8 0
3 years ago
A certain circuit is composed of two series resistors. The total resistance is 10 Ohms. One of the resistors is 4 Ohms. The othe
Mariulka [41]
<h3><u>Given </u><u>:</u><u>-</u><u> </u></h3>

  • A certain circuit is composed of two series resistors
  • The total resistance is 10 ohms
  • One of the resistor is 4 ohms

<h3><u>To </u><u>Find </u><u>:</u><u>-</u></h3>

  • We have to find the value of other resistor?

<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u></h3>

We know that,

In series combination,

  • When a number of resistances are connected in series, the equivalent I.e resultant resistance is equal to the sum of the individual resistances and is greater than any individual resistance

<u>That </u><u>is</u><u>, </u>

Rn in series = R1 + R2 + R3.....So on

<u>Therefore</u><u>, </u>

<u>According </u><u>to </u><u>the </u><u>question</u><u>, </u>

We have,

R1 + R2 = 10 Ω

4 + R2 = 10Ω

R2 = 10 - 4

R2 = 6Ω

Hence, The value of R2 resistor in series is 6Ω

4 0
2 years ago
Star A and Star B have measured stellar parallax of 1.0 arc second and 0.75 arc second, respectively. Which star is closer? How
zhuklara [117]

Answer:

Star A is closer than Star B

Explanation:

As we know that in parallax method of distance measurement the angle subtended by the star when it covers a distance of one Parsec arc length, it is known as parallax angle

Here we can say

angle = \frac{1 Parsec}{distance}

so we have

distance = \frac{1 Parsec}{angle}

so here we have

angle subtended by Star A = 1 arc sec

angle subtended by star B = 0.75 arc sec

now we have

distance for star A is given as

d_a = \frac{1 Parsec}{1} = 1 Parsec

distance of star B is given as

d_b = \frac{1 Parsec}{0.75} = 1.33 Parsec

So star A is closer than star B

7 0
3 years ago
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