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Deffense [45]
3 years ago
8

Water from a fire hose is directed horizontally against at a rate of 50.0 kg/s and a speed of 42.0 m/s.

Physics
1 answer:
ss7ja [257]3 years ago
8 0

Answer:

Force, |F| = 2100 N

Explanation:

It is given that,

Water from a fire hose is directed horizontally against at a rate of 50.0 kg/s, \dfrac{m}{t}=50\ kg/s

Initial speed, v = 42 m/s

The momentum is reduced to zero, final speed, v = 0

The relation between the force and the momentum is given by :

F=\dfrac{p}{t}

F=\dfrac{mv}{t}

F=50\ kg/s\times 42\ m/s

|F| = 2100 N

So, the magnitude of the force exerted on the wall is 2100 N. Hence, this is the required solution.

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An insulator can do which of the following? Conduct charge through it. Become positively or negatively charged. Transfer protons
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The correct answer is:
<span>Become positively or negatively charged

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What type of metal are conducting wires most often made of?<br><br> #16
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2 years ago
A sound is recorded at 19 decibels. What is the intensity of the sound?
sp2606 [1]

1 \times 10^{-10.1} \mathrm{Wm}^{-2} is the intensity of the sound.

Answer: Option B

<u>Explanation:</u>

The range of sound intensity that people can recognize is so large (including 13 magnitude levels). The intensity of the weakest audible noise is called the hearing threshold. (intensity about 1 \times 10^{-12} \mathrm{Wm}^{-2}). Because it is difficult to imagine numbers in such a large range, it is advisable to use a scale from 0 to 100.

This is the goal of the decibel scale (dB).  Because logarithm has the property of recording a large number and returning a small number, the dB scale is based on a logarithmic scale. The scale is defined so that the hearing threshold has intensity level of sound as 0.

                     \text { Intensity }(d B)=(10 d B) \times \log _{10}\left(\frac{I}{I_{0}}\right)

Where,

I = Intensity of the sound produced

I_{0} = Standard Intensity of sound of 60 decibels = 1 \times 10^{-12} \mathrm{Wm}^{-2}

So for 19 decibels, determine I as follows,

                   19 d B=(10 d B) \times \log _{10}\left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=\frac{19}{10}

                  \log _{10}\left(\frac{1}{1 \times 10^{-12} \mathrm{Wm}^{-2}}\right)=1.9

When log goes to other side, express in 10 to the power of that side value,

                  \left(\frac{I}{1 \times 10^{-12} W m^{-2}}\right)=10^{1.9}

                  I=1 \times 10^{-12} \mathrm{Wm}^{-2} \times 10^{1.9}=1 \times 10^{-12-1.9}=1 \times 10^{-10.1} \mathrm{Wm}^{-2}

5 0
3 years ago
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