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3 years ago

What is the loudness, in decibels, of a sound 1 trillion times as loud as the softest audible sound?

Physics

1 answer:

Rufina [12.5K]3 years ago

5 0

Answer:

What is the loudness, in decibels, of a sound 100 trillion times as loud as the softest audible sound? The loudness of this sound in decibels is dB. (Type an integer or a decimal.)

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A blackbody curve relates _____. intensity of radiation to energy intensity of radiation to wavelength wavelength to energy surf

neonofarm [45]

A blackbody curve represents the relation between <u>intensity of radiation with wavelength.</u>

Here in this curve we can see that all ideal blackbody radiates almost all wavelength of radiations and these radiations are of different intensity.

here intensity will be maximum for a given wavelength of radiation and the relation of this wavelength with the temperature of the object is given by Wein's law

It is given by

$\lambda = \frac{b}{T}$

now if we increase the temperature the maximum intensity for which wavelength is given will shift to the left.

Using this all we can also compare the temperature of two blackbody for which radiation graph is given to us.

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3 years ago

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What is v^2=0.05-4.9 please i need this asap

Margaret [11]

Answer:

v =2.02

Explanation:

v^2=0.05-4.9

v^2=-4.85

square root both side

v=2.02

^^^^this is a not a perfect square

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3 years ago

When buying a car, you tell the salesman that you want to make sure the car can go from 0 miles per hour to 60 miles per hour in

MaRussiya [10]

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3 years ago

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A projectile is fired with a velocity of 22 m/s at an angle of 25°. What is the vertical component of the velocity?

7nadin3 [17]

Answer:

Vertical component of velocity is 9.29 m/s

Explanation:

Given that,

Velocity of projection of a projectile, v = 22 m/s

It is fired at an angle of 22°

The horizontal component of velocity is v cosθ

The vertical component of velocity is v sinθ

So, vertical component is given by :

$v_y=v\ sin(25)$

$v_y=22\ m/s\times\ sin(25)$

$v_y=9.29\ m/s$

Hence, the vertical component of the velocity is 9.29 m/s

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3 years ago

A swimming pool has the shape of a right circular cylinder with radius 21 feet and height 10 feet. Suppose that the pool is full

AysviL [449]

Answer:

The water required to pump all the water to a platform 2 feet above the top of the pool is is 6061310.32 foot-pound.

Explanation:

Given that,

Radius = 21 feet

Height = 10 feet

Weighing = 62.5 pounds/cubic

Work = 4329507.37572

Height = 2 feet

Let's look at a horizontal slice of water at a height of h from bottom of pool

We need to calculate the area of slice

Using formula of area

$A=\pi r^2$

Put the value into the formula

$A=\pi\times21^2$

$A=441\pi\ feet^2$

Thickness of slice $t=\Delta h\ ft$

The volume is,

$V=(441\pi\times\Delta h)\ ft^3$

We need to calculate the force

Using formula of force

$F=W\times V$

Where, W = water weight

V = volume

Put the value into the formula

$F=62.5\times(441\pi\times\Delta h)$

$F=27562.5\pi\times\Delta h\ lbs$

We need to calculate the work done

Using formula of work done

$W=F\times d$

Put the value into the formula

$W=27562.5\pi\times\Delta h\times(10-h)\ ft\ lbs$

We do this by integrating from h = 0 to h = 10

We need to find the total work,

Using formula of work done

$W=\int_{0}^{h}{W}$

Put the value into the formula

$W=\int_{0}^{10}{27562.5\pi\\times(10-h)}dh$

$W=27562.5\pi(10h-\dfrac{h^2}{2})_{0}^{10}$

$W=27562.5\pi(10\times10-\dfrac{100}{2}-0)$

$W=4329507.37572$

To pump 2 feet above platform, then each slice has to be lifted an extra 2 feet,

So, the total distance to lift slice is (12-h) instead of of 10-h

We need to calculate the water required to pump all the water to a platform 2 feet above the top of the pool

Using formula of work done

$W=\int_{0}^{h}{W}$

Put the value into the formula

$W=\int_{0}^{10}{27562.5\pi\\times(12-h)}dh$

$W=27562.5\pi(12h-\dfrac{h^2}{2})_{0}^{10}$

$W=27562.5\pi(12\times10-\dfrac{100}{2}-0)$

$W=1929375\pi$

$W=6061310.32\ foot- pound$

Hence, The water required to pump all the water to a platform 2 feet above the top of the pool is is 6061310.32 foot-pound.

8 0

3 years ago