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andreev551 [17]

3 years ago

8

Discuss the correlation or connection between stars with a higher mass and the amount of fuel they have to work with

1 answer:

dimulka [17.4K]3 years ago

3 0

Larger stars have a higher amount of fuel in order to keep the process of nuclear fusion going.

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The speaker at the concert has the sound intensity level of 100 dB if we listen from the distance 5 m.How far from the speaker d

Mademuasel [1]

Answer:

$28.11m$ far from the speaker the intensity drops to 85 dB.

Explanation:

In the equation for the Decibel scale

$(1). \: \:\beta =10 log(\dfrac{I}{I_0})$

The ratio of the intensities can be written as

$\frac{I}{I_0} = \dfrac{\frac{P}{A} }{\frac{P}{A_0} }$

$\dfrac{I}{I_0} = \dfrac{A_0}{A}.$

And since

$A = 4\pi r^2$

and

$A_0 = 4 \pi r_0^2$ ,

$\dfrac{A_0}{A} = \dfrac{4 \pi r_0^2}{4 \pi r^2} = \dfrac{r_0^2}{r^2}$

meaning

$\dfrac{I}{I_0} = \dfrac{r_0^2}{r^2}.$

Putting this into equation (1), we get:

$\boxed{ (2).\: \: \beta = 10log(\dfrac{r_0^2}{r^2})}$

Now, if the intensity is 100 dB when the distance is 5 meters, we have:

$100dB=10 log(\dfrac{r_02}{(5m)^2})$

$10= log(\dfrac{r_0^2}{25})$

by taking both sides to the exponent:

$10^{10}= \dfrac{r_0^2}{25}$

$r^2 = 25 *10^{10}\\r = 5 *10^5$

Now equation (2) becomes

$\beta = 10log(\dfrac{25*10^{10}}{r^2})$

when the intensity level is 85 dB we have

$85 = 10log(\dfrac{25*10^{10}}{r^2})$

$8.5 = log(\dfrac{25*10^{10}}{r^2})$

take both sides to exponents and we get:

$10^{8.5} =10^{ log(\dfrac{25*10^{10}}{r^2})}$

$10^{8.5} =\dfrac{25*10^{10}}{r^2}$

$r^2 = \dfrac{25*10^{10}}{10^{8.5}}$

$\boxed{r = 28.11m}$

Thus, $28.11m$ far from the speaker the intensity drops to 85 dB.

7 0

3 years ago

The key difference between the binomial and hypergeometric distribution is that

Ivenika [448]

Explanation:

Both distributions describe the number of times an event occurs in a givn number of trials. In the binomial distribution, the probability is the same for each trial. While in the hypergeometric distribution, each trial changes the probability of each subsequent trial, since there is no replacement.

5 0

3 years ago

A diffraction pattern is formed on a screen 130 cm away from a 0.420-mm-wide slit. Monochromatic 546.1-nm light is used. Calcula

notka56 [123]

Answer:

The fractional Intensity $\frac{I}{I_{max}}$ = 0.0146

Given:

wavelength of the light, $\lambda = 546.1 nm = 546.1\times 10^{-9} m$

slit and screen separation difference, D = 130 cm = 1.3 m

distance of the point from the center of the principal maximum, y = 4.10 mm = 0.041 m

slit width, d = 0.420 mm = $0.420\times 10^{-3}$

Solution:

To calculate the fractional intensity, we use the given formula:

$\frac{I}{I_{max}} = \frac{sin^{2}\delta }{\frac({\delta}{2})^{2}}$ (1)

$\delta = \frac{\pi }{\lambda}dsin\theta$

For very small angle:

$\delta = \frac{\pi dy}{\lambda D}$ (2)

where

$\delta = total phase angle$

$\theta = angle of deviation$

Using eqn (2):

$\delta = \frac{\pi \times 0.42\times 10^{-3}\times 4.1\times 10^{-3} }{546.1\times 10^{-9}\times 1.3} = 7.6202 radians$

Now, using eqn (1):

$\frac{I}{I_{max}} = \frac{sin^{2}(7.6202) }{(\frac{7.6202}{2})^{2}} = 0.0146$

4 0

4 years ago

What’s the total system momentum before collision?

Soloha48 [4]

The momentum would be 9 kg

3 0

3 years ago

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Chemistry<br> determine the pressure

sashaice [31]

I think it's 1.03412969 or 1.03

8 0

4 years ago