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3 years ago

The visible light spectrum ranges between

Physics

2 answers:

natulia [17]3 years ago

8 0

Answer: 380 nm (violet-blue) and 780 nm (red)

Explanation:

The portion visible by the human eye of the electromagnetic spectrum is between 380 nm (violet-blue) and 780 nm (red) approximately.

This part of the spectrum is located between ultraviolet light and infrared light.

It should be noted that the fact only part of the whole electromagnetic spectrum is visible to humans is because the receptors in our eyes are only sensitive to these wavelengths.

aleksandr82 [10.1K]3 years ago

4 0

Red goes in the first blank and violet goes in the second blank if you're using plato

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Select the major problems associated with the production of nuclear energy.

liberstina [14]

Hazardous solid wastes

3 0

3 years ago

Read 2 more answers

The velocity of the transverse waves produced by an earthquake is 8.9 km/s, and that of the longitudinal waves is 5.1 km/s. A se

Brrunno [24]

Answer: The distance is 723.4km

Explanation:

The velocity of the transverse waves is 8.9km/s

The velocity of the longitudinal wave is 5.1 km/s

The transverse one reaches 68 seconds before the longitudinal.

if the distance is X, we know that:

X/(9.8km/s) = T1

X/(5.1km/s) = T2

T2 = T1 + 68s

Where T1 and T2 are the time that each wave needs to reach the sesmograph.

We replace the third equation into the second and get:

X/(9.8km/s) = T1

X/(5.1km/s) = T1 + 68s

Now, we can replace T1 from the first equation into the second one:

X/(5.1km/s) = X/(9.8km/s) + 68s

Now we can solve it for X and find the distance.

X/(5.1km/s) - X/(9.8km/s) = 68s

X(1/(5.1km/s) - 1/(9.8km/s)) = X*0.094s/km= 68s

X = 68s/0.094s/km = 723.4 km

6 0

4 years ago

IBM has a fast computer that it calls the Blue Gene/L that can do '136.8

dmitriy555 [2]

Answer:

138.6 megacalculations

Explanation:

This is a pretty straightforward one.

All it needs is to convert the degree of measurement.

Dimensions in physics are attributed names, which state the power to which they're are raised. Just as how

Kilo and Mega means the numbers are raised to the power of 3 and 6 respectively. There also exists the ones that indicates how small, such as milli and micro, which are to the powers of -3 & -6.

The question says the IBM computer calculates at an astonishing 136.8 teracalculations.

Tera in physics means it's raised to the power of 12. Thus, the IBM calculates at an astonishing rate of

136.8*10^12 calculations per second.

We're then asked how many calculations it does in 1 micro second. Like I had highlighted earlier, 1 micro second is 1 raised to the power of -6. Or succinctly put,

1 micro second = 1*10^-6.

If the IBM does

138.6*10^12 = 1 second,

Then it does

x = 1*10^-6 second.

When we cross multiply, we have

138.6*10^12 * 1*10^-6, and that is

138.6*10^6 calculations, or say, 138.6 megacalculations.

The IBM does 138.6 megacalculations in 1 micro second, which is still astonishing, by the way

6 0

3 years ago

A 5 kg mass is oscillating on a spring with a time period of 2.8 seconds. What is the spring constant k of the spring?