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uranmaximum [27]
4 years ago
6

You place a light bulb 8 cm in front of a concave mirror. You then move a sheet of paper back and forth in front of the mirror.

The image of the light bulb focuses on the paper when the paper is 12 cm in front of the mirror. What is the focal length of the mirror? Please show work too.
Physics
1 answer:
Alika [10]4 years ago
5 0

sorry - late reply...just stumbled across tis...hope u can still use it :)


By the mirror equation: 1/di + 1/do = 1/f

<span>
</span>

<span>where di = distance to image = +12cm (+ for real image)</span>


and do = distance to object = +8cm


Substitute and solve for f, the focal length

<span><span>
1/12 + 1/8 = 1/f
</span><span>
1/f = (8 + 12) / 12 * 8 = 20/96
</span><span>
so f = 96/20 = 4.8 cm</span>
</span>
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Where are metamorphic rocks formed? And how are they formed
den301095 [7]

Answer:

Metamorphic rocks form when rocks are subjected to high heat, high pressure, hot mineral-rich fluids or, more commonly, some combination of these factors. Conditions like these are found deep within the Earth or where tectonic plates meet.

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Which object has the least amount of kinetic energy? a car driving down a road a soccer ball rolling down a hill a bicycle locke
mars1129 [50]
<span> a bicycle locked to a bike rack im pretty sure</span>
7 0
3 years ago
Read 2 more answers
A rocket with a mass of 62,000 kg (including fuel) is burning fuel at the rate of 150 kg/s and the speed of the exhaust gases is
mezya [45]

Answer:

h≅ 58 m

Explanation:

GIVEN:

mass of rocket M= 62,000 kg

fuel consumption rate =  150 kg/s

velocity of exhaust gases v= 6000 m/s

Now thrust = rate of fuel consumption×velocity of exhaust gases

=6000 × 150 = 900000 N

now to need calculate time t = amount of fuel consumed÷ rate

= 744/150= 4.96 sec

applying newton's law

M×a= thrust - Mg

62000 a=900000- 62000×9.8

acceleration a= 4.71 m/s^2

its height after 744 kg of its total fuel load has been consumed

h= \frac{1}{2}at^2

h= \frac{1}{2}4.71\times4.96^2

h= 58.012 m

h≅ 58 m

4 0
3 years ago
Below is an oscilloscope trace from an AC supply. Calculate the frequency of the supply if each horizontal division represents 0
GuDViN [60]

Answer:

20 hertz of frequency produced.

Explanation:

\boxed{frequency = \frac{1}{period} }

Here we will find frequency and period should be in second, here given: 0.05 seconds

using the formula:

\boxed{frequency = \frac{1}{0.05} }

\boxed{frequency =20}

8 0
3 years ago
A parallel RC circuit has a capacitive reactance of 962 ohms and a resistance of 1,200 ohms what is the impedance of this circui
Alex777 [14]

Answer:

<em>A. 751 ohm</em>

Explanation:

Impedance: <em>This is the total opposition to the flow of current in an a.c circuit by any or all of the three circuit elements ( R, L, C). The unit of impedance is Ohms (Ω). The impedance in a parallel circuit is gives a s</em>

<em>Z = RXₐ/√(Xₐ² + R²)............................... Equation 1</em>

<em>Where Z = The impedance of the a.c circuit, Xₐ = capacitive reactance, R = resistance.</em>

<em>Given: Xₐ = 962 Ω, R = 1200 Ω</em>

<em>Substituting these values into equation 1,</em>

<em>Z = 962×1200/√(962² + 1200²)</em>

<em>Z = 1154400/√(925444 + 1440000)</em>

<em>Z = 1154400/√(925444+1440000</em>

<em>Z = 1154400/1538</em>

<em>Z = 750.59 Ω</em>

<em>Z≈ 751 Ω</em>

<em>Therefore the impedance of the circuit = 751 Ω</em>

<em>The right option is A. 751 ohm</em>

6 0
4 years ago
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