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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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Answer:

if you stretch a spring with k = 2, with a force of 4N, the extension will be 2m. the work done by us here is 4x2=8J. in other words, the energy transferred to the spring is 8J. but, the stored energy in the spring equals 1/2x2x2^2=4J (which is half of the work done by us in stretching it).

8 0
3 years ago
The revolution of the earth around the sun demonstrate what motion?​
olga2289 [7]

Answer:

Anticlockwise directions

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3 0
2 years ago
What effort force will be required to lift the 20 N object using the pulley above?
Umnica [9.8K]

Answer:

I think it is 5N.

Explanation:

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3 years ago
A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 3.0 km/h
Yuki888 [10]

Answer:

The magnitude of F₁ is 3.7 times of F₂

Explanation:

Given that,

Time = 10 sec

Speed = 3.0 km/h

Speed of second tugboat = 11 km/h

We need to calculate the speed

v_{1}=\dfrac{3.0\times10^{3}}{3600}

v_{1}=0.833\ m/s

The force F₁is constant acceleration is also a constant.

F_{1}=ma_{1}

We need to calculate the acceleration

Using formula of acceleration

a_{1}=\dfrac{v}{t}

a_{1}=\dfrac{0.833}{10}

a_{1}=0.083\ m/s^2

Similarly,

F_{2}=ma_{2}

For total force,

F_{3}=F_{2}+F_{1}

ma_{3}=ma_{2}+ma_{1}

The speed of second tugboat is

v=\dfrac{11\times10^{3}}{3600}

v=3.05\ m/s

We need to calculate total acceleration

a_{3}=\dfrac{v}{t}

a_{3}=\dfrac{3.05}{10}

a_{3}=0.305\ m/s^2

We need to calculate the acceleration a₂

0.305=a_{2}+0.083

a_{2}=0.305-0.083

a_{2}=0.222\ m/s^2

We need to calculate the factor of F₁ and F₂

Dividing force F₁ by F₂

\dfrac{F_{1}}{F_{2}}=\dfrac{m\times0.83}{m\times0.22}

\dfrac{F_{1}}{F_{2}}=3.7

F_{1}=3.7F_{2}

Hence, The magnitude of F₁ is 3.7 times of F₂

3 0
3 years ago
When are you most aware of your motion in a moving vehicle: when it is moving steadily in a straight line or when it is accelera
Ghella [55]

Answer:

Acceleration is percieved, not constant velocity.

Explanation:

You are most aware when the vehicle is accelerating. At constant velocity you would not be aware of the motion. Only if the system is accelerated the dynamics must be solved considering a pseudo-force (of inertial origin) acting.

It's because of this that:

(A) False. The acceleration can be detected from the inside of a closed car.

(B) False. You would be aware of the motion, but not because humans can sense speed but acceleration.

(C) False. Constant velocity cannot be felt in a closed car.

(D) False. Again, you can't feel constant speed.

6 0
3 years ago
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