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3 years ago

A photon has a wavelength of 500nm what is the energy associated

1 answer:

8 0

Given: Plank's constant h = 6.626 x 10⁻³⁴ J.s

Speed of light c = 2.998 x 10⁸ m/s

Wavelength λ = 500 nm convert to meter = 5 x 10⁻⁷ m

Required: Energy E =?

Formula: E = hc/λ

E = (6.626 X 10⁻³⁴ J.s)(2.998 X 10⁸ m/s)/5 x 10⁻⁷ m

E = 3.97 X 10⁻¹⁹ J or in termS of electron volt

E = 2.48 eV

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Three noise sources produce volume (loudness) levels of 70, 73, and 80 dB when acting separately. When the sources act together,

Mashutka [201]

Sound at 70 dB is 70 dB louder than the human reference level. That's 10⁷ times as much as the reference sound power.

Sound at 73 dB is 73 dB louder than the human reference level. That's 10⁷.³ or 2 x 10⁷ times as much as the reference sound power.

Sound at 80 dB is 80 dB louder than the human reference level. That's 10⁸ or 10 x 10⁷ times as much as the reference sound power.

Now we can adumup:

Intensity of all 3 sources = (10⁷) + (2 x 10⁷) + (10 x 10⁷)

Intensity = (13 x 10⁷) times the sound power reference intensity.

Intensity in dB = 10 log (13 x 10⁷) = 10 (7 + log(13)

Intensity = 70 + 10 log(13)

Intensity = 70 + 10 (1.114)

Intensity = 70 + 11.14

Intensity = <em>81.14 dB</em>

Looking at the questioner's profile, I seriously wonder whether I'll ever get a comment in return from this creature, and how I'll ever find out if my solution is correct. For that matter, I'm also seriously questioning how and whether my solution will ever be used for anything.

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3 years ago

Look in the comments. I d a r e y o u

Alika [10]

Answer:

Where is the comments??

Explanation:

3 0

3 years ago

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What do we measure energy in

Goshia [24]

Energy is measured by joule(j)

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3 years ago

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A stuffed toy with a mass of 0.900 kilograms sits on the edge of a bed at a height of 0.830 if the toy falls off the bed what wi

Olenka [21]

Mechanical energy (ME) is the sum of potential energy (PE) and kinetic energy (KE). When the toy falls, energy is converted from PE to KE, but by conservation of energy, ME (and therefore PE+KE) will remain the same.

Therefore, ME at 0.500 m is the same as ME at 0.830 m (the starting point). It's easier to calculate ME at the starting point because its just PE we need to worry about (but if we wanted to we could calculate the instantaneous PE and KE at 0.500 m too and add them to get the same answer).

At the start:

ME = PE = mgh
ME = 0.900 (9.8) (0.830)
ME = 7.32 J

8 0

3 years ago

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km fr

mestny [16]

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

What is the tension between her ears?

Would the astronaut find it difficult to keep from being torn apart by the gravitational forces?

Answer:

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

Explanation:

Given that:

Orbital radius of the spacecraft (R) = 120 Km = 120 × 10³ m

Mass of the black hole (m) = $5 \ * (M \ _{sun})$

where : $M_{sun} = 1.99*10^{33} \ kg$

Then; we have:

$m = 5*(1.99*10^{30} \ kg ) \\ = 9.95*10^{30} kg$

Schwarzchild radius of the black hole

r - 15.0 km

Mass of each ear $m_{ear} = 0.030 \ kg$

Farther distance between one ear and the black hole (d) = 6.0 cm

= 0.06 m

Closer distance between the other ear and the black home is (d) 6.0 cm

= 0.6 cm

NOW, If we assume that the tension force should be T; then definitely the two ears will posses the same angular velocity .

The net force on the ear closer to the black hole will be:

$\frac{GMm_{ear} }{(R-d)}- T = m_{ear} (R - d) \omega^2$

$\frac{GMm_{ear} }{(R-d)^2}- \frac{T}{(R-d)} = m_{ear} \omega^2 \ ----> \ (1)$

The net force on the ear farther to the black hole is :

$\frac{GMm_{ear} }{(R+d)}- T = m_{ear} (R + d) \omega^2$

$\frac{GMm_{ear} }{(R+d)^2}- \frac{T}{(R+d)} = m_{ear} \omega^2 \ ----> \ (2)$

Equating equation (1) and (2) & therefore making (T) the subject of the formula; we have:

$T = \frac{3GMm_{ear}d}{R^3}$

$T = \frac{3(6.67*10^{-11}N.m^2/kg^2)(1.95*10^{30}kg)(0.03kg)(0.06m)}{(120*10^3m)^3}$

$T = 2073.9 N\\T = 2.07 KN$

The tension between the ears = 2.07 KN

The astronaut will find it difficult to keep and will eventually be in trouble because the tension is now greater compared to the tension in the human tissues.

3 0

3 years ago