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mel-nik [20]
3 years ago
8

The sound intensity at a distance of 16 m from a noisy generator is measured to be 0.25 W/m2. What is the sound intensity at a d

istance of 28 m from the generator?
Physics
1 answer:
Margarita [4]3 years ago
4 0

Answer:

0.1111 W/m²

Explanation:

If all other parameters are constant, sound intensity is inversely proportional to the square of the distance of the sound. That is,

I ∝ (1/r²)

I = k/r²

Since k can be the constant of proportionality. k = Ir²

We can write this relation as

I₁ × r₁² = I₂ × r₂²

I₁ = 0.25 W/m²

r₁ = 16 m

I₂ = ?

r₂ = 24 m

0.25 × 16² = I₂ × 24²

I₂ = (0.25 × 16²)/24²

I₂ = 0.1111 W/m²

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Which transformation of energy occurs in a hydroelectric power plant? A. Potential to chemical energy B. Chemical to potential e
motikmotik

The correct answer to the question is D). Kinetic to electrical.

EXPLANATION:

Hydro electric power plants are the electricity generation plants which capture the energy of falling water to produce current.

The water present at the top of a dam has potential energy. When the water is allowed to fall from certain height, the potential energy of the water is converted into kinetic energy.

The falling water moving with high speed is allowed to fall on a turbine. The kinetic energy of the water will rotate the turbine. The turbine is attached to a generator which will produce electricity due to the electromagnetic induction.

Hence, we see that the kinetic energy of the water is converted into electric energy.

7 0
3 years ago
Read 2 more answers
Caleb rides his motorcycle with a constant speed of 72km/h. How far can he Travel in 3 hours 40 minutes
lina2011 [118]
264 km

72 x 3 hr = 216 km
72/60 (min in an hr) =  1.2(km per min) x 40 = 48 km
216 + 48 = 264

5 0
3 years ago
How long will it take you to travel 20 miles on a bus that drives 60 miles/h?
ValentinkaMS [17]

Answer:

<u>20 Minutes</u>

<u></u>

Explanation:

Well we know Mph (Miles per hour) is distance over time : \frac{distance}{time} \\

R (rate) = 60

d (distance) = 20

t (time) = Unknown

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

 R = \frac{d}{t}  

   ↓

60 = \frac{20}{t}

   ↓

 t = \frac{20}{60}

    ↓

 t = \frac{1}{3}  or 0.3333

<em>So basically it would take one third of an hour. Lets change these units to minutes.</em>

60 * 0.333333 = 20

<em>So it would take you </em><u><em>20 minutes</em></u><em> to drive 20 miles on a bus that drives 60 mph</em>

<em />

Hope that helps

<em>~Siascon~</em>

8 0
2 years ago
b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul
Nat2105 [25]

Answer:

b) Betelgeuse would be \approx 1.43 \cdot 10^{6} times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is \approx 1.37 \cdot 10^{-5} } the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

B = \displaystyle{\frac{L}{4\pi d^2}} (1)

where B is the brightness, L is the star luminosity and d, the distance from the star to the point where the brightness is calculated (measured). Thus:

b) B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}} and B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}} where L_{Sun} is the Sun luminosity (3.9 x 10^{26} W) but we don't need to know this value for solving the problem. ly is light years.

Finding the ratio between the two brightness we get:

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is 1.581 \cdot 10^{-5}\ ly. Then

\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

4 0
3 years ago
A car is traveling at 120 km/h (75 mph). When applied the braking system can stop the car with a deceleration rate of 9.0 m/s2.
Bumek [7]

Answer:

the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

Explanation:

Given that;

speed of car V  = 120 km/h = 33.3333 m/s

Reaction time of an alert driver = 0.8 sec

Reaction time of an alert driver = 3 sec

extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec

now, extra distance that car will travel in case of sleepy driver  will be'

S_d = V × 2.2 sec

S_d = 33.3333 m/s × 2.2 sec

S_d = 73.3333 m

hence, number of car of additional car length  n will be;

n = S_n / car length

n = 73.3333 m / 5m

n = 14.666 ≈ 15

Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

8 0
3 years ago
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