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3 years ago

Compared to a virtual image, a real image

Physics

1 answer:

Rom4ik [11]3 years ago

5 0

Compared to a virtual image, a real image can be observed on a screen.

Explanation

When a beam of light meets at a specific point after the reflection from the mirror, the image formed at the point is said to be a real image.

Real images are generally inverted.

When a beam of light seems to appear to meet at a certain point after the reflection from the mirror, the image formed at the point is said to be a virtual image.

Virtual images are generally erect.

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3 years ago

Two light bulbs have resistances of 400 Ω and 800 ΩThe two light bulbs are connected in series across a 120- V line. Find the cu

Natasha2012 [34]

1) Current in each bulb: 0.1 A

The two light bulbs are connected in series, this means that their equivalent resistance is just the sum of the two resistances:

$R_{eq}=R_1 + R_2 = 400 \Omega + 800 \Omega=1200 \Omega$

And so, the current through the circuit is (using Ohm's law):

$I=\frac{V}{R_{eq}}=\frac{120 V}{1200 \Omega}=0.1 A$

And since the two bulbs are connected in series, the current through each bulb is the same.

2) 4 W and 8 W

The power dissipated by each bulb is given by the formula:

$P=I^2 R$

where I is the current and R is the resistance.

For the first bulb:

$P_1 = (0.1 A)^2 (400 \Omega)=4 W$

For the second bulb:

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2 years ago

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2 years ago

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A truck is traveling at 27 m/s down the interstate highway where you are changing a flat tire. frequency of 185 Hz.

Reil [10]

Answer:

(a) the observed frequency is 200 Hz

(b) the observed frequency is 188 Hz.

Explanation:

speed of the truck, Vs = 27 m/s

frequency of the truck as it approaches, Fs = 185 Hz

(a) Apply Doppler effect to determine the frequency you will hear.

As the truck approaches you, the observed frequency will be higher than the source frequency because of decrease in distance.

$F_s = F_o [\frac{V}{V_S + V} ]$

Where;

Fo is the observed frequency which is the frequency you will hear.

V is speed of sound in air

$F_s = F_o [\frac{V}{V_S + V} ]\\\\185 = F_o [\frac{340}{27 + 340} ]\\\\185 = F_o (0.926)\\\\F_o = \frac{185}{0.926}\\\\F_o = 199.78 \ Hz$

$F_o = 200 \ Hz$

(b) Apply the following formula for a moving observer and a moving source;

$F_o = F_s[\frac{V-V_o}{V} ](\frac{V}{V-V_S} )$

The observed frequency is negative since you are driving away from the truck and the source frequency is also negative since it is driving towards you.

$F_o = F_s[\frac{V-V_o}{V} ](\frac{V}{V-V_S} )\\\\F_o = 185[\frac{340-22}{340} ](\frac{340}{340-27} )\\\\F_o = 185(0.9353)(1.0863)\\\\F_o = 188 \ Hz$

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3 years ago