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liubo4ka [24]
3 years ago
10

PLEASE HELP ME IM TIMED

Physics
1 answer:
Novosadov [1.4K]3 years ago
6 0
Yeah it’s the mantle
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Find the magnitude of the force at point x = 15cm if k=10500 N/m when the work done by this force is WD= 1/2k x2
Ann [662]

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sqdqk3

Explanation:

qdjqưbdkq ưqdhud j

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3 years ago
A rescue team is searching for Andrew, a geologist who was stranded while conducting research in the mountains of Colorado. The
Sindrei [870]

Answer:

I = 9.82 10⁻⁷ W / m²

Explanation:

The intensity of the sound wave is the energy of the wave between the order per unit area of ​​the same

      I = P / A = E / T A

the energy is calculated by integrating the mechanical energy in a period, where the mass is changed by the density and ‘s’ is the amplitude of the sound wave

     I = ½ ρ v (w s)²

   

     I = ½ 1.35 328 (2π 530 2.00 10⁻⁸)²

     I = 221.4 (4.435 10⁻⁹)

     I = 9.82 10⁻⁷ W / m²

4 0
3 years ago
Which is purely physical and which is both physical and chemical
algol [13]
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3 0
3 years ago
A client with a head injury is being monitored for increased intracranial pressure (ICP). The client's blood pressure is 90/60 m
artcher [175]

Answer:

CPP=52mm \ Hg

Explanation:

<em>Cerebral Perfusion Pressure</em><em> </em>is obtained by subtracting <em>IntraCranial Pressure</em>(ICP) from the <em>Mean Arterial Pressure</em>(MAP). Adequate cerebral perfusion requires a minimum goal of 70mm \ Hg. MAP is obtained using the formula:-

MAP=\frac{(diastolic \ blood \ pressure\times2)+\ systolic \ blood \ pressure)}{3}\\\\MAP=\frac{2\times60+90}{3}\\\\MAP=70mm \ Hg\\CPP=MAP-ICP\\CPP=70-18\\CPP=52mm \ Hg

5 0
3 years ago
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
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