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Elden [556K]
2 years ago
5

To practice Problem-Solving Strategy 16.1 Sound Intensity. A rescue team is searching for Andrew, a geologist who was stranded w

hile conducting research in the mountains of Colorado. The team uses electronic listening devices in order to detect any shouts for help. The sound waves from his shouts reaching the base camp can be approximated by a sinusoidal wave with a frequency f = "520" Hz and displacement amplitude A = 4.50×10−8 m , where the sound wave properties are valid at the base camp where the measurements are being made. What sound intensity level will the rescue team measure from the frightened researcher? Assume the speed of sound is v=333m/s and the density of air rho = 1.28 kg/m3 .
Physics
1 answer:
scoundrel [369]2 years ago
3 0

Answer:

Explanation:

Intensity of wave is

2 π² f²A²v p

= 2 π² (520)² (4.5 ×10∧-8)² (333) (1.28)

I= 4.607 × 10∧-6

Intensity level is

β = 10 dB ㏒ (I÷Io)

    =10 dB ㏒ (4.607 × 10∧-6 ÷ 10∧-12)

β = 66.6 dB

Bulk modulus is

B = v² p

 = (333)² (1.28)

 = 1.419 ×10∧5 Pa

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A brass lid screws tightly onto a glass jar at 20 degrees C. To help open the jar, it can be placed into a bath of hot water. Af
omeli [17]

Answer:

0.0016 cm

Explanation:

\alpha_b = Thermal coefficient of expansion of brass = 19\times 10^{-6}\ /^{\circ}C

\alpha_g = Thermal coefficient of expansion of glass = 9\times 10^{-6}\ /^{\circ}C

\Delta T = Change in temperature = (60-20)^{\circ}C

R_0 = Initial radius = 4 cm

Change in radius of material is given by

R=R_0(1+\alpha\Delta T)

Difference in radii of the lid and jar

\Delta R=R_b-R_g\\\Rightarrow \Delta R=R_0(1+\alpha_b\Delta T)-R_0(1+\alpha_g\Delta T)\\\Rightarrow \Delta R=R_0(\alpha_b-\alpha_g)\Delta T\\\Rightarrow \Delta R=4\times (19\times 10^{-6}-9\times 10^{-6})\times (60-20)\\\Rightarrow \Delta R=0.0016\ cm

The size of the gap is 0.0016 cm or 0.000016 m

8 0
2 years ago
You are using a lightweight rope to pull a sled along level ground. The sled weighs 485 N, the coefficient of kinetic friction b
Bogdan [553]

Answer:

N=459.01N

Explanation:

According to Newton's first law:

N+F_y-W=0

The component of the force on the y-axis can be obtained through the Pythagorean Theorem. This is because the components are the cathetus of a right triangle and its hypotenuse is the magnitude of the force:

sin12^\circ=\frac{F_y}{F}\\F_y=Fsin12^\circ

Replacing and solving for N:

N=W-Fsin12^\circ\\N=485N-(125N)sin12^\circ\\N=459.01N

5 0
2 years ago
Alex goes cruising on his dirt bike. He rides 700m north, 300m east, 400 m north, 600m west, 1200m south, 300m east and finally
Bess [88]

Find Displacement and Distance

displacement ...

north is 700+400+100 =1200m n

south=1200m

1200-1200=0


east is 300+300=600m

west is 600m

600-600=0

back at dtart. displ zero


distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m  = 3600m


3 0
2 years ago
A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes
OlgaM077 [116]

Answer:

Explanation:

mass of car, m = 1000 kg

initial velocity, u = 20 m/s

final velocity, v = 0 m/s

distance, s = 120 m

Let a be the acceleration of motion

use third equation of motion

v² = u² + 2 as

0 = 20 x 20 + 2 x a x 120

a = - 1.67 m/s²

Let F be the force

Force, F  mass x acceleration

F = - 1000 x 1.67

F = - 1666.67 N

The direction of force is towards south and the magnitude of force is 1666.67 N.

8 0
2 years ago
To practice Problem-Solving Strategy 29.1: Faraday's Law. A metal detector uses a changing magnetic field to detect metallic obj
ExtremeBDS [4]

Answer:

1.138\times 10^{-3}V

Explanation:

Apply Faraday's Newmann Lenz law to determine the induced emf in the loop:

\epsilon=\frac{d\phi}{dt}

where:

d\Phi-variation of the magnetic flux

dt-is the variation of time

#The magnetic flux through the coil is expressed as:

\Phi=NBA \ Cos \theta

Where:

N- number of circular loops

A-is the Area of each loop(A=\pi r^2=\pi \times 5^2=78.5398)

B-is the magnetic strength of the field.

\theta=15\textdegree- is the angle between the direction of the magnetic field and the normal to the area of the coil.

\epsilon=-\frac{d(78.5398\times 10^{-3}NB \ Cos \theta)}{dt}\\\\=-(78.5398\times 10^{-3}N\ Cos \theta)}{\frac{dB}{dt}

\frac{dB}{dt}-=0.0250T/s is given as rate at which the magnetic field increases.

#Substitute in the emf equation:

=-(78.5398\times 10^{-3} m^2 \times 6\ Cos 15 \textdegree)\times 0.0250T/s\\\\=1.138\times 10^{-3}V

Hence, the induced emf is 1.138\times 10^{-3}V

4 0
2 years ago
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