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Norma-Jean [14]
3 years ago
13

A light bulb will glow when electrons flow through it. As the electron flow increases, the brightness increases as well. A stude

nt hooks up two circuits containing three light bulbs in each circuit. In one circuit the lights are connected in series and in the other circuit the lights are hooked up in parallel.
If you could only see the lights in the circuit and the wires were covered from view, how could you determine the type of circuit the lights are arranged in?

A) If all the lights are equal brightness, it is a series circuit.
B) If all the lights are equal brightness, it is a parallel circuit.
C) Unscrew one light, if the others remain on it is a series circuit.
D) Unscrew one light, if the others remain on it is a parallel circuit.
Physics
2 answers:
Inga [223]3 years ago
6 0
The correct answer is: <span>Unscrew one light, if the others remain on it is a parallel circuit.</span>
Juliette [100K]3 years ago
3 0

The answer on USATESTPREP is D

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The objective lens of a microscope has a focal length of 5.5mm. Part A What eyepiece focal length will give the microscope an ov
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Complete Question

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .

What eyepiece focal length will give the microscope an overall angular magnification of 300?

Answer:

The  eyepiece focal length is  f_e  = 0.027 \ m

Explanation:

From the question we are told that

    The focal length is  f_o =  5.5 \ mm =  -0.0055 \ m

This negative sign shows the the microscope is diverging light

     The  angular magnification is m = 300

     The  distance between the objective and the eyepieces lenses is  Z =  19 \ cm  = 0.19 \ m

Generally the magnification is mathematically represented as

        m  =  [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ]

Where f_e is the eyepiece focal length of the microscope

  Now  making f_e the subject  of the formula

         f_e  = \frac{Z}{1 - [\frac{M  *  f_o }{0.25}] }

substituting values

        f_e  = \frac{ 0.19 }{1 - [\frac{300  *  -0.0055 }{0.25}] }

         f_e  = 0.027 \ m

     

5 0
2 years ago
What is the (magnitude of the) centripetal acceleration (as a multiple of g=9.8~\mathrm{m/s^2}g=9.8 m/s ​2 ​​ ) towards the Eart
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Answer:

The centripetal acceleration as a multiple of g=9.8 m/s^{2} is 1.020x10^{-3}m/s^{2}

Explanation:

The centripetal acceleration is defined as:

a = \frac{v^{2}}{r}  (1)

Where v is the velocity and r is the radius

Since the person is standing in the Earth surfaces, their velocity will be the same of the Earth. That one can be determined by means of the orbital velocity:

v = \frac{2 \pi r}{T}  (2)

Where r is the radius and T is the period.

For this case the person is standing at a latitude 71.9^{\circ}. Remember that the latitude is given from the equator. The configuration of this system is shown in the image below.

It is necessary to use the radius at the latitude given. That radius can be found by means of trigonometric.

\cos \theta = \frac{adjacent}{hypotenuse}

\cos \theta = \frac{r_{71.9^{\circ}}}{r_{e}} (3)

Where r_{71.9^{\circ}} is the radius at the latitude of 71.9^{\circ} and r_{e} is the radius at the equator (6.37x10^{6}m).

r_{71.9^{\circ}} can be isolated from equation 3:

r_{71.9^{\circ}} = r_{e} \cos \theta  (4)

r_{71.9^{\circ}} = (6.37x10^{6}m) \cos (71.9^{\circ})

r_{71.9^{\circ}} = 1.97x10^{6} m

Then, equation 2 can be used

v = \frac{2 \pi (1.97x10^{6} m)}{24h}

Notice that the period is the time that the Earth takes to give a complete revolution (24 hours), this period will be expressed in seconds for a better representation of the velocity.

T = 24h . \frac{3600s}{1h} ⇒ 84600s

v = \frac{2 \pi (1.97x10^{6} m)}{84600s}

v = 146.31m/s

Finally, equation 1 can be used:

a = \frac{(146.31m/s)^{2}}{(1.97x10^{6}m)}

a = 0.010m/s^{2}

Hence, the centripetal acceleration is 0.010m/s^{2}

To given the centripetal acceleration as a multiple of g=9.8 m/s^{2}​ it is gotten:

\frac{0.010m/s^{2}}{9.8 m/s^{2}} = 1.020x10^{-3}m/s^{2}

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Answer:

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Newton’s first law states that if an objects velocity is changing a _____ must be acting on it.
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<h3>Answer: </h3>

(a)

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