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Lana71 [14]
3 years ago
9

PLEASE HELP SOMEONE!!! An object is located 5.0 cm from a convex lens. The lens focuses light at a distance of 10.0 cm. What is

the image distance? Use the equation attached.
A.-10.0 Cm

B.10.0

C.3.33Cm

D.-3.33Cm

Physics
1 answer:
OverLord2011 [107]3 years ago
8 0

Answer:

-10.0 cm

Explanation:

OK so in the equation they're having you use the variables are:

d_o = 5.0 cm\\\\f = 10.0 cm\\\\d_i = ?

So we simply plug in the variables:

d_i = \frac{d_of}{d_o-f} \\\\d_i = \frac{5.0 * 10.0}{5.0 - 10.0}\\\\d_i = \frac{50}{-5}\\\\d_i = -10.0 cm

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To find the final velocity, we would use the first equation of motion;

V = U + at

Substituting into the equation, we have

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3 years ago
A sailboat took 25 hours to cover 1/4 of a journey. Then, it
ozzi

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A mass with mass 4 is attached to a spring with spring constant 24 and a dashpot giving a damping 20. The mass is set in motion
aleksandrvk [35]

Answer: x(t) = 14e^{-2t} - 10e^{-3t}

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x'(0) = -2C_{1} -3C_{2} = 2\\-2(4-C_{2}) -3C_{2} = 2\\C_{2} = -10\\C_{1} = 4 - C_{2}\\C_{1} = 14

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The position function for this system is: x(t) = 14e^{-2t} - 10e^{-3t}

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3 years ago
How far must you be from a 110 W speaker to have an intensity of 0.0439 W/m^2? (Treat the speaker as a point source.)
Lina20 [59]

Answer:

d=14.12m

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Assume a point sound source that emits a sound power<em> P </em>(in W) evenly in all directions of space. Let us also assume that the medium does not absorb this sound power when it passes through it. At a distance<em> d </em>from the source this power will have been evenly distributed over the surface of a sphere of radius <em>d</em>. Therefore, the acoustic intensity I at distance d will be worth:

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3 0
3 years ago
How many protons are in one kg of lead? b) How many electrons are in one kg of lead? c) What is the total negative charge of the
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207.2 g of lead contains 82\times 6.023\times 10^{23} protons

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Also, for neutral atom, number of protons = number of electrons.

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<u>Number of electrons = 2.3836\times 10^{26}</u>

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