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andrew11 [14]
3 years ago
7

At a maximum level of loudness, the power output of a 75 piece orchestra radiated as sound is 70.0 w. What is the intensity of t

hese sound waves to a listener who is sitting 25.0 m from the orchestra?

Physics
1 answer:
Bogdan [553]3 years ago
6 0

Answer:

0.0089 W/m^2

Explanation:

The intensity of a sound wave is given by

I=\frac{P}{A}

where

P is the power of the wave at the source

A is the area over which the sound wave propagates

In this problem, we have

P = 70.0 W is the power

We want to know the intensity at a distance of r = 25.0 m from the orchestra, so since the sound waves propagates radially, we must consider a spherical surface of radius r, so the area is

A=4\pi r^2 = 4\pi (25.0 m)^2=7850 m^2

Therefore, the intensity of the wave is

I=\frac{70 W}{7850 m^2}=0.0089 W/m^2

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Olivia wants to find out whether a substance will fluoresce. She says she should put it in a microwave oven. Do you agree with h
andreyandreev [35.5K]
Disagree.
Fluoresce objects will only glow when put under actual Ultraviolet light. This is due to the molecules becoming excited by the ultraviolet radiation.


Microwaves give micro-waves that are present in another spectrum of wave length and will not be able to fluoresce the molecules. If it’s not “ultra violet “.... it’s not going to glow.
4 0
2 years ago
Read 2 more answers
(a) what will an object weigh on the moon's surface if it weighs 100 n on earth's surface
juin [17]
We know the equation

weight = mass × gravity

To work out the weight on the moon, we will need its mass, and the gravitational field strength of the moon.
Remember that your weight can change, but mass stays constant.

So using the information given about the earth weight, we can find the mass by substituting 100N for weight, and we know the gravity on earth is 10Nm*2 (Use the gravitational field strength provided by your school, I am assuming yours in 10Nm*2)

Therefore,

100N = mass × 10
mass= 100N/10
mass= 10 kg

Now, all we need are the moon's gravitational field strength and to apply this to the equation

weight = 10kg × (gravity on moon)
4 0
3 years ago
Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical
lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}} (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}

\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2 (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio \frac{T^{2}}{a^{3}} is constant for all the planets orbiting the same sun, so we can said that:

\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}

\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83

6 0
3 years ago
Read 2 more answers
A 06-C charge and a .07-C charge are apart at 3 m apart. What force attracts them?
Andru [333]

Answer:

F = 37.8 × 10^(6) N

Explanation:

The charges are 0.06 C and 0.07 C.

Thus;

Charge 1; q1 = 0.06 C

Charge 2; q2 = 0.07 C

Distance between them; r = 3 m

Formula for the force in between them is;

F = kq1•q2/r²

Where k is a constant = 9 × 10^(9) N.m²/C²

Thus;

F = (9 × 10^(9) × 0.06 × 0.07)/3²

F = 37.8 × 10^(6) N

3 0
3 years ago
The parking brake on a 1200kg automobile has broken, and the vehicle has reached a momentum of 7800kg.M/s. What is the velocity
AysviL [449]
Ok so the equation for momentum is:
v=p/m

So you would do:
7800/1200=6.5

So the answer is:
6.5 m/s

Hope this helps :)
6 0
3 years ago
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