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d1i1m1o1n [39]
3 years ago
8

About once every 30 minutes, a geyser known as Old Faceful projects water 11.0 m straight up into the air. Use g = 9.80 m/s^2, a

nd take atmospheric pressure to be 101.3 kPa. The density of water is 1000 kg/m^3. What is the speed of the water when it emerges from the ground?
Physics
1 answer:
Maurinko [17]3 years ago
7 0

Answer:

The speed of the water is 14.68 m/s.

Explanation:

Given that,

Time = 30 minutes

Distance = 11.0 m

Pressure = 101.3 kPa

Density of water = 1000 kg/m³

We need to calculate the speed of the water

Using equation of motion

v^2=u^2+2gs

Where, u = speed of water

g = acceleration due to gravity

h = height

Put the value into the formula

0=u^2-2\times9.8\times11.0

u=\sqrt{2\times9.8\times11.0}

u=14.68\ m/s

Hence, The speed of the water is 14.68 m/s.

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A vacuum cleaner produces sound with a measured sound level of 75.0 dB. (a) What is the intensity of this sound in W/m2? W/m2 (b
jonny [76]

Answer:

Part a)

I = 3.16 \times 10^{-5} W/m^2

Part b)

P_o = 0.162 Pa

Explanation:

Part a)

Level of sound = 75 dB

now we know that

L = 10 Log\frac{I}{I_0}

here we know that

I_0 = 10^{-12} W/m^2

now we have

75 = 10 Log(\frac{I}{10^{-12}})

I = 3.16 \times 10^{-5} W/m^2

Part b)

Intensity of sound wave is given as

I = \frac{1}{2}\rho A^2\omega^2 c

here we know that

A = \frac{P_o}{Bk}

so we have

I = \frac{1}{2}\rho(\frac{P_o}{Bk})^2\omega^2 c

I = \frac{1}{2}\rho P_o^2 \frac{c^3}{B^2}

now we know

\rho = 1.2 kg/m^3

c = 340 m/s

B = 1.4 \times 10^5 Pa

now we have

3.16 \times 10^{-5} = \frac{1}{2}(1.2)P_o^2\frac{340^3}{(1.4\times 10^5)^2}

P_o = 0.162 Pa

5 0
3 years ago
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