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Annette [7]
4 years ago
9

A person shouting at the top of his lungs emits aboue 1.0 w of energy as sound waves. What is the sound intensity 1.0 m from suc

h a person
Physics
1 answer:
umka21 [38]4 years ago
3 0

Answer:

I=0.0795\ W/m^2

Explanation:

When a person shouts it emits 1 W of energy as sound waves, P = 1 W

Let I is the sound intensity 1 meters from such a person. We know that the power per unit area is called the sound intensity of the person. Its formula is given by :

I=\dfrac{P}{A}

I=\dfrac{P}{4\pi r^2}

I=\dfrac{1\ W}{4\pi (1\ m)^2}  

I=0.0795\ W/m^2

So, the sound intensity of such a person is 0.0795\ W/m^2. Hence, this is the required solution.

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What is a “plate” in geological terms
Tju [1.3M]

Answer:a plate is a rigid segment Of The earth's lithosphere that moves horizontally and adjoins other plates along then zone of seismic activity.

Explanation:

A plate is a rigid segment Of The earth's lithosphere that moves horizontally and adjoins other plates along then zone of seismic activity.plates may include portion of both continents and ocean basin.

3 0
4 years ago
7. A bicycle accelerates at 2.0 m/s2. If the mass of the bicycle and rider together is
kkurt [141]

Answer:

170 N

Explanation:

Since Force F = ma were m = mass = 85 kg and a = acceleration = 2.0 m/s².

So the net force on the bicycle is

F = ma = 85 kg × 2.0 m/s² = 170 N

4 0
3 years ago
An ideal monatomic gas initially has a temperature of 300 K and a pressure of 5.79 atm. It is to expand from volume 420 cm3 to v
maxonik [38]

Answer:

a) The final pressure is 1.68 atm.

b) The work done by the gas is 305.3 J.

Explanation:

a) The final pressure of an isothermal expansion is given by:

T = \frac{PV}{nR}

T_{i} = T_{f}

\frac{P_{i}V_{i}}{nR} = \frac{P_{f}V_{f}}{nR}

Where:

P_{i}: is the initial pressure = 5.79 atm

P_{f}: is the final pressure =?

V_{i}: is the initial volume = 420 cm³

V_{f}: is the final volume = 1450 cm³

n: is the number of moles of the gas

R: is the gas constant

P_{f} = \frac{P_{i}V_{i}}{V_{f}} = \frac{5.79 atm*420 cm^{3}}{1450 cm^{3}} = 1.68 atm

Hence, the final pressure is 1.68 atm.

b) The work done by the isothermal expansion is:

W = P_{i}V_{i}ln(\frac{V_{f}}{V_{i}}) = 5.79 atm*\frac{101325 Pa}{1 atm}*420 cm^{3}*\frac{1 m^{3}}{(100 cm)^{3}}ln(\frac{1450 cm^{3}}{420 cm^{3}}) = 305.3 J

Therefore, the work done by the gas is 305.3 J.

I hope it helps you!        

3 0
3 years ago
• How much work is<br>required to lift a 2kg<br>object 2m high?<br>​
pychu [463]

Answer You need to consider that the gravity on earth is 9.8 m/s/s. This means any object you let go on the earths surface will gain 9.8 m/s of speed every second. You need to apply a force on the object in the opposite direction to avoid this acceleration. If you are pushing something up at a constant speed, you are just resisting earths acceleration. The more massive and object is, the greater force is needed to accelerate it. The equation is Force = mass*acceleration. So for a 2kg object in a 9.8 m/s/s gravity you need 2kg*9.8m/s/s = 19.6 Newtons to counteract gravity. Work or energy = force * distance. So to push with 19.6 N over a distance of 2 meters = 19.6 N*2 m = 39.2 Joules of energy. There is an equation that puts together those two equations I just used and it is E = mgh

The amount of Energy to lift an object is (mass) * (acceleration due to gravity) * (height)

:Hence, the Work done to life the mass of 2 kg to a height of 10 m is 196 J. Hope it helps❤️❤️❤️

Explanation:

7 0
3 years ago
Imagine that Kevin can instantly transport himself between Planet X and Planet Y. Which statement could be said about Kevin in t
Over [174]
What are the choices ? 

Without some directed choices, I'm, free to make up any
reasonable statement that could be said about Kevin in this
situation.  A few of them might be . . .

-- Kevin will have no trouble getting back in time for dinner.

-- Kevin will have no time to enjoy the scenery along the way.

-- Some simple Physics shows us that Kevin is out of his mind.
He can't really do that.

           -- Speed = (distance covered) / (time to cover the distance) .

If time to cover the distance is zero, then speed is huge (infinite).

           -- Kinetic energy = (1/2) (mass) (speed)² .

If speed is huge (infinite), then kinetic energy is huge squared (even more).
There is not enough energy in the galaxy to push Kevin to that kind of speed.

         -- Mass = (Kevin's rest-mass) / √(1 - v²/c²)

-- As soon as Kevin reaches light-speed, his mass becomes infinite.
-- It takes an infinite amount of energy to push him any faster.
-- If he succeeds somehow, his mass becomes imaginary.
-- At that point, he might as well turn around and go home ...
     if he ever reached Planet-Y, nobody could see him anyway.
8 0
3 years ago
Read 2 more answers
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