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icang [17]
3 years ago
15

Q 17.10: Receiver A detects a sound level of 75 dB at a distance of 25 m from a speaker, which emits sound equally in all direct

ions. Receiver B detects a sound level of 55 dB from the same speaker. How far is receiver B from the speaker
Physics
1 answer:
Ronch [10]3 years ago
4 0

Answer:

250 m

Explanation:

We expect the distance to be more because the sound level has decreased.

Sound level (in decibels) is related to distance by

r_2 = r_1 \times  10^{\frac{|L_1-L_2|}{20}}

where L_1 is the sound level at a distance of r_1 and

L_2 is the sound level at a distance of r_2.

Using the values in the question,

r_{2} = 25\times10^{\frac{|75-55|}{20}} \text{ m}

r_2 = 25\times10^1 \text{ m}= 250 \text{ m}

As a note, when distance increases by 10, the sound level drops by 20 dB, which is what we have in the question.

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The answer is C, because they moved from a stand still to down the hill
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Compare the properties of Titan’s atmosphere with those of Earth’s atmosphere.
Aloiza [94]

Answer and Explanation:

Comparison between the Titan's atmosphere and earth atmosphere

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3 years ago
May you help me answer this​
Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is 2.2 m/s^2

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

1)

There are three Kepler's law of planetary motion:

  1. 1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
  2. 2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
  3. 3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, T^2 \propto r^3, where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

M g sin \theta - T = Ma (1)

where

Mg sin \theta is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

g=9.8 m/s^2 (acceleration of gravity)

\theta=35^{\circ}

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

T-mg sin \theta = ma (2)

where

m = 3.5 kg (mass of the block)

\theta=35^{\circ}

From (2) we get

T=mg sin \theta + ma

Substituting into (1),

M g sin \theta - mg sin \theta - ma = Ma

Solving for a,

a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2

2b)

The tension in the string can be calculated using the equation

T=mg sin \theta + ma

where

m = 3.5 kg (mass of lighter block)

g=9.8 m/s^2

\theta=35^{\circ}

a=2.2 m/s^2 (acceleration found in part 2)

Substituting,

T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

U=mgh

where

m is the mass of the object

g is the strength of the gravitational field

h is the heigth of the object relative to the ground

3b)

The kinetic energy of an object is given by

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

For the object in this problem,

m = 500 g = 0.5 kg

v = 3 m/s

Substituting, we find its kinetic energy:

K=\frac{1}{2}(0.5)(3)^2=2.25 J

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

And about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

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A filled water tower sits on the top of the highest hill near a city. The cylindrical tower has a height of 55.0 m
Reika [66]

Answer:

539 kPa

Explanation:

Pressure equals density times acceleration of gravity times depth.

P = ρgh

Water has a density of 1000 kg/m³, and acceleration of gravity is 9.8 m/s².

P = (1000 kg/m³) (9.8 m/s²) (55.0 m)

P = 539,000 Pa

P = 539 kPa

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3 years ago
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