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Naddika [18.5K]
3 years ago
7

5. When an object is placed 8 millimeters from a concave spherical mirror, a clear image can be projected on a screen 16 millime

ters in front of the mirror. If the object has a height of 4 millimeters, the height of the image is _________ millimeters.
A. 8
B. 4
C. 12
D. 2
Physics
2 answers:
Cloud [144]3 years ago
6 0

A...............................................................

















....

Sophie [7]3 years ago
6 0

The correct answer to the question is A). 8 mm.

CALCULATION:

As per the question, the object distance u = 8 mm.

The image distance v = 16 mm.

The height of the object h_{0}=\ 4\ mm.

As per the rule of sign convention, the distance measured along the direction of light is taken as positive and opposite to the direction of light is taken as negative.

All the transverse measurements above the principal axis are taken as positive, and the transverse measurements below the principal axis are negative.

Hence, the object distance  u = -8 mm

          The image distance v = -16 mm.

          The object height h_{o}=\ +4\ mm.

We are asked to calculate the image height h_{i}.

The transverse magnification of the spherical concave mirror is given as -

                                  m=\ \frac{h_{i}} {h_{o}}=\ \frac{-v}{u}

                               ⇒  \frac{h_{i}} {h_{0}}=\ \frac{-v}{u}

                                ⇒   h_{i} =\ h_{0}\times \frac{-v}{u}

                                        =\ 4\times \frac{-(-16)}{(-8)}\ mm

                                        =\ -8\ mm    

The negative sign signifies that the image is inverted in nature.

Hence, the image height will be 8 mm.

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Oksana_A [137]

Answer:

When the temperature of the coffee is 50 °C, the time will be 20.68 mins

Explanation:

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According to Newton's law of cooling;

\frac{dT}{dt} \alpha (T-21)\\\\\frac{dT}{dt} = k (T-21)\\\\\frac{dT}{T-21} = kdt\\\\\int\limits {\frac{dT}{T-21}}  =  \int\limits kdt\\\\Log(T-21) =kt +  Logc \\\\Log (\frac{T-21}{c} ) = kt\\\\T -21 = ce^{kt}\\\\At \ t = 0, T = 95\\\\95-21 = ce^0\\\\74 = c\\\\New, equation: T -21 = 74e^{kt}\\\\Again; when \ t= 5\ min, T = 80\\\\80 -21 = 74e^{5k}\\\\59 = 74e^{5k}\\\\e^{5k} = \frac{59}{74}\\\\ 5k = ln(\frac{59}{74})\\\\5k = -0.2265\\\\k = -0.0453

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The driver of a train moving at 23m/s applies the breaks when it pases an amber signal. The next signal is 1km down the track an
pochemuha

Answer:

3.2 m/s

Explanation:

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Δx = 1000 m

v₀ = 23 m/s

a = -0.26 m/s²

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Find: v

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v = at + v₀

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Δx = ½ (v + v₀) t

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v² = v₀² + 2aΔx

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Or:

Δx = vt − ½ at²

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As you can see, you get slightly different answers depending on which variables you use.  Since 1000 m has 1 significant figure, compared to the other variables which have 2 significant figures, I recommend using the first equation.

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