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3 years ago

The image of an object placed 30cm from a diverging lens is formed 10cm in front of the lens.

Physics

1 answer:

luda_lava [24]3 years ago

6 0

Answer:

15cm

Explanation:

Since the lens is a diverging lens, the image distance is negative (virtual)

v = -30cm

u = 10cm

Required

focal length f

Using the lens formula;

1/u + 1/v = 1/u

1/10 - 1/30 = 1/f

(3-1)/30 = 1/f

2/30 = 1/f

f = 30/2

f = 15cm

Hence the focal length of the lens is 15cm

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You are trying to overhear a juicy conversation, but from your distance of 20.0 m , it sounds like only an average whisper of 30

12345 [234]

Answer:

r₂ = 0.2 m

Explanation:

given,

distance = 20 m

sound of average whisper = 30 dB

distance moved closer = ?

new frequency = 80 dB

using formula

$\beta = 10 log(\dfrac{I_1}{I_0})$

I₀ = 10⁻¹² W/m²

now,

$30 = 10 log(\dfrac{I_1}{10^{-12}})$

$\dfrac{I_1}{10^{-12}}= 10^3$

$I_1= 10^{-8}\ W/m^2$

to hear the whisper sound = 80 dB

$80 = 10 log(\dfrac{I_2}{10^{-12}})$

$\dfrac{I_2}{10^{-12}}= 10^8$

$I_2= 10^{-4}\ W/m^2$

we know intensity of sound is inversely proportional to square of distances

$\dfrac{I_1}{I_2}=\dfrac{r_2^2}{r_1^2}$

$\dfrac{10^{-8}}{10^{-4}}=\dfrac{r_2^2}{20^2}$

$10^{-4}=\dfrac{r_2^2}{20^2}$

r₂ = 0.2 m

6 0

3 years ago

You might say that this experiment was an attempt to build a scale, and then calibrate it against a scale that we trust (the ele

Allushta [10]

No.

Since repeated measurements are taken and the average and 95% confidence interval are calculated, the possibility of the lack of agreement being a random error has been minimized or even eliminated.

<h3>What is a random error?</h3>

Random error is defined as the deviation of the total error from its mean value due to chance.

Random errors can result from the instrument not being precise or from mistakes by the researcher.

Random errors can be minimized by taking multiple readings and averaging the results.

Since repeated measurements are taken and the average and 95% confidence interval are calculated, the possibility of the lack of agreement being a ransom error has been minimized.

Learn more about random errors at: brainly.com/question/22041172

3 0

3 years ago

Read 2 more answers

A flute player hears four beats per second when she compares her note to a 523 HzHz tuning fork (the note C). She can match the

laiz [17]

Answer:

527 Hz

Solution:

As per the question:

Beat frequency of the player, $\Delta f = 4\ beats/s$

Frequency of the tuning fork, f = 523 Hz

Now,

The initial frequency can be calculated as:

$\Delta f = f - f_{i}$

$f_{i} = f \pm \Delta f$

when

$f_{i} = f + \Delta f = 523 + 4 = 527 Hz$

when

$f_{i} = f - \Delta f = 523 - 4 = 519 Hz$

But we know that as the length of the flute increases the frequency decreases

Hence, the initial frequency must be 527 Hz

7 0

3 years ago

Read 2 more answers

The loudness of a sound is the waves amplitude <br><br><br><br> True or false

kakasveta [241]

Answer:

true i think

Explanation:

The amplitude of a sound wave determines its loudness or volume. A larger amplitude means a louder sound, and a smaller amplitude means a softer sound. In Figure 10.2 sound C is louder than sound B. The vibration of a source sets the amplitude of a wave.

4 0

3 years ago

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When numbers are very small or very large, it is convenient to either express the value in scientific notation and/or by using a

Oxana [17]

Answer:

5 mg, $5\cdot 10^{-3}g$

Explanation:

First of all, let's rewrite the mass in grams using scientific notation.

we have:

m = 0.005 g

To rewrite it in scientific notation, we must count by how many digits we have to move the dot on the right - in this case three. So in scientific notation is

$m=5\cdot 10^{-3}g$

If we want to convert into milligrams, we must remind that

1 g = 1000 mg

So we can use the proportion

$1 g : 1000 mg = 0.005 g : x$

and we find

$x=\frac{(1000 mg)(0.005 g)}{1 g}=5 mg$

4 0

4 years ago