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3 years ago

5

A concave makeup mirror is designed so the virtual image it produces is four times the size of the object when the distance betw

een the object and the mirror is 18.0 cm. What is the radius of curvature of the mirror?

1 answer:

kipiarov [429]3 years ago

7 0

Answer:

Radius of curvature = 48 cm.

Explanation:

Given that

Magnification,m=4

Distance between object and mirror,u=18 cm

We know that magnification of concave mirror given as

m= -v/u

So

4 = -v/18

v= - 72 cm

As we know that

$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$

$\dfrac{1}{f}=\dfrac{1}{18}-\dfrac{1}{72}$

f=24 cm

So r= 2 x f

r=48 cm

Radius of curvature = 48 cm.

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Answer:

585×10⁸ m

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d = (2.998×10⁸ m/s) (3.25 min) (60 s/min)

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An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s 1045 rad/s ). If a particular disk is

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Explanation:

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The wavelength of a wave on a string is 1.2 meters. If the speed of the wave is 60 m/s, what is the wave frequency? 5 points 0.2

vagabundo [1.1K]

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Read 2 more answers

Lewiston and Vernonville are 208 miles apart. A car leaves Lewiston traveling towards Vernonville, and another car leaves Verno

iragen [17]

Answer:

Average speed of the car A = 70 miles per hour

Average speed of the car B = 60 miles per hour

Explanation:

Average speed of the car A is $v_{A} =\frac{x_{A} }{t_{A} }$ (Equation A) and Average speed of the car B is $v_{B} =\frac{x_{B} }{t_{B} }$ (Equation B), where $x_{A}$ and $x_{B}$ are the distances and $t_{A}$ and $t_{B}$ are the times at which are travelling the cars A and B respectively.

We have to convert the time to the correct units:

1 hour and 36 minutes = 96 minutes

$96 minutes . \frac{1 hour}{60 minutes} = 1.6 h$

From the diagram (Please see the attachment), we can see that at the time they meet, we have:

$v_{A} = \frac{208-x}{1.6h} + 10\frac{miles}{h}$ (Equation C)

$v_{B} = \frac{208-x}{1.6h}$ (Equation D)

From Equation A and C, we have:

$\frac{208-x}{1.6}+10 = \frac{x}{1.6}$

208-x+16 = x

208 + 16 = 2x

$x = \frac{224}{2}$

x = 112 miles

Replacing x in Equation A:

$v_{A} = \frac{112miles}{1.6h}$

$v_{A} = 70 miles per hour$

Replacing x in Equation B:

$v_{B} = \frac{208miles-112miles}{1.6h}$

$v_{B} = \frac{96miles}{1.6h}$

$v_{B} = 60 miles per hour$

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