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user100 [1]
3 years ago
13

Suppose a certain jet plane creates an intensity level of 124 dB at a distance of 5.01 m. What intensity level does it create on

the ground directly underneath it when flying at an altitude of 2.25 km?
Physics
1 answer:
Zolol [24]3 years ago
6 0

Answer:71 dB

Explanation:

Given

sound Level \beta _1=124 dB

distance r_1=5.01 m

From sound Intensity

\beta =10dB\log (\frac{I_1}{I_0})

124=10dB\log (\frac{I_1}{I_0})

12.4=\log (\frac{I_1}{I_0})

I_1=(1\times 10^{-12})\times 10^{12.4}

I_1=2.51 W/m^2

we know Intensity I\propto ^\frac{1}{r^2}

I_1r_1^2=I_2r_2^2

I_2=I_1(\frac{r_1}{r_2})^2

I_2=2.51\cdot (\frac{5.01}{2.25\times 10^3})^2

I_2=1.24\times 10^{-5} W/m^2

Sound level corresponding to I_2

\beta _2=10\log (\frac{I_2}{I_0})

\beta _2=10\log (\frac{1.24\times 10^{-5}}{1\times 10^{-12}})

\beta _2=70.93\approx 71 dB

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inn [45]

Answer:

1.8 × 10⁻⁸ Hm

Explanation:

Given that:

The refractive index of the film = 19

The wavelength of the light = 136.8 μ m

The thickness can be calculated by using the formula shown below as:

Thickness=\frac {\lambda}{4\times n}

Where, n is the refractive index of the film

{\lambda} is the wavelength

So, thickness is:

Thickness=\frac {136.8\ \mu\ m}{4\times 19}

Thickness = 1.8 μ m

Since,

1 μ m = 10⁻⁸ Hm

So,

Thickness = 1.8 × 10⁻⁸ Hm

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2 years ago
Anyone know the answer ?
dlinn [17]
Momentum = mass x velocity, so 500kg x 2m/s = 1000 kg m/s
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3 years ago
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Describe a situation where you can be traveling at a low speed but have an extremely high velocity
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Answer:

Cruising at 35,000 feet in an airliner, straight toward the east,

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2 years ago
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A person travels by car from one city to another with different constant speeds between pairs of cities. She drives for 35.0 min
Lubov Fominskaja [6]

Explanation:

Solution:

Let the time be

t1=35min = 0.58min

t2=10min=0.166min

t3=45min= 0.75min

t4=35min= 0.58min

let the velocities be

v1=100km/h

v2=55km/h

v3=35km/h

a. Determine the average speed for the trip. km/h

first we have to solve for the distance

S=s1+s2+s3

S= v1t1+v2t2+v3t3

S= 100*0.58+55*0.166+35*0.75

S=58+9.13+26.25

S=93.38km

V=S/t1+t2+t3+t4

V=93.38/0.58+0.166+0.75+0.58

V=93.38/2.076

V=44.98km/h

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3 years ago
Evaluate the gravitational potential energy between two 5.00-kg spherical steel balls separated by a center-tocenter distance of
navik [9.2K]

Answer:

U = 8.30×10-⁹J

Explanation:

m1 = m2 = 5.00kg masses of the spheres

d = 15.0cm = 15×10-²m

r = 5.10cm = 5.10×10-²m

R = d + r = 15×10-² + 5.10×10-²

R = 20.10 ×10-²m = 0.201m

G = 6.67×10-¹¹Nm²/kg²

U = Gm1×m2/R = potential energybetween the spheres

U = 6.67×10-¹¹×5.00×5.00/0.201

U = 8.30×10-⁹J

7 0
3 years ago
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