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Levart [38]
3 years ago
11

In every trench over 4 feet (1.2 m) deep, there must be an exit every​

Physics
1 answer:
ss7ja [257]3 years ago
5 0

Explanation:

25 ft I believe that this is the best answer

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A rock band playing an outdoor concert produces sound at 120 dB 5.0 m away from their single working loudspeaker. What is the so
joja [24]

ANSWER:100dB

f<em>rom the  sound intensity level,the sound intensity is calculated as:</em>

<em />\beta<em>₁=</em>\beta<em>=(10dB)㏒₁₀(l₁/l₀)</em>

<em>inserting numbers:</em>

<em>120dB=(10dB)㏒₁₀[l₁/10⁻¹²W/m⁸] or 12=㏒₁₀[l₁/(10⁻¹²)W/m²</em>

<em>Getting the antilog of both sides and obtain 10¹²=l₁(10⁻¹²W/m²)which </em>

<em>can be used to solve for l₁ and get</em>

<em>          l₁=(10⁻¹²W/m²)(10¹²)=1 W/m²</em>

<em>since the sound  intensity is related to the power and that the power does not change,the sound intensity at any other point can be solved.Plugging-in</em>

<em>ᵃ = 4πr²,into P=l₁ₐ₁=l₂ₐ₂ and get:</em>

<em>l₂=l₁(r₁/r₂)² =(1W/m²)(5/35) =2.04×10⁻²W/m²</em>

<em>since we know the sound intensity at the sound point 2r,the sound intensity level at the point can be solved.We have:</em>

<em>     </em>\beta<em>₂=</em>\beta<em>=(10dB)㏒₁₀(l₂/l₀)=(10dB)㏒₁₀(2.04×10⁻²/1×10⁻²)</em>

<em>    </em>\beta<em>₂=(10dB)㏒₁₀(2.04×10¹⁰)</em>

<em />\beta<em> =(10dB)[㏒₁₀(2.04)+㏒₁₀(10¹⁰)]=10dB[0.32+10]=103dB=100dB</em>

<em />

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4 years ago
A physical property that describes how something feels is called
Norma-Jean [14]
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4 years ago
Which technology is used to service space vehicles and other equipment?
Crank
You’re correct hope this helps :)
6 0
3 years ago
At one instant a bicyclist is 21.0 m due east of a park's flagpole, going due south with a speed of 13.0 m/s. Then 21.0 s later,
andre [41]

Answer:

Part a)

d = 21\sqrt2 = 29.7 m

Part b)

Direction is 45 degree North of West

Part c)

v_{avg} = 1.41 m/s

Part d)

direction of velocity will be 45 Degree North of West

Part e)

a = 0.875 m/s^2

Part f)

\theta = 45 degree North of East

Explanation:

Initial position of the cyclist is given as

r_1 = 21.0 m due East

final position of the cyclist after t = 21.0 s

r_2 = 21.0 m due North

Part a)

for displacement we can find the change in the position of the cyclist

so we have

d = r_2 - r_1

d = 21\hat j - 21\hat i

so magnitude of the displacement is given as

d = 21\sqrt2 = 29.7 m

Part b)

direction of the displacement is given as

\theta = tan^{-1}\frac{y}{x}

\theta = tan^{-1}\frac{21}{-21}

so it is 45 degree North of West

Part c)

For average velocity we know that it is defined as the ratio of displacement and time

so here the magnitude of average velocity is defined as

v_{avg} = \frac{\Delta x}{t}

v_{avg} = \frac{29.7}{21}

v_{avg} = 1.41 m/s

Part d)

As we know that average velocity direction is always same as that of average displacement direction

so here direction of displacement will be 45 Degree North of West

Part e)

Here we also know that initial velocity of the cyclist is 13 m/s due South while after t = 21 s its velocity is 13 m/s due East

So we have

change in velocity of the cyclist is given as

\Delta v = v_f - v_i

\Delta v = 13\hat i - (-13\hat j)

now average acceleration is given as

a = \frac{\Delta v}{\Delta t}

a = \frac{13\hat i + 13\hat j}{21}

so the magnitude of acceleration is given as

a = \frac{13\sqrt2}{21} = 0.875 m/s^2

Part f)

direction of acceleration is given as

\theta = tan^{-1}\frac{y}{x}

\theta = tan^{-1}\frac{13}{13}

\theta = 45 degree North of East

4 0
3 years ago
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