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11Alexandr11 [23.1K]

3 years ago

12

A light source emits light with dominant wavelengths in the range of 650 to 690 nm. what is the principle color of light emitted

by the source?

1 answer:

dsp733 years ago

6 0

Answer:

Please see below as the answer is self-explanatory.

Explanation:

The visible range extends roughly from 400 nm (violet) to 700 nm (red).
Below the violet is the ultra-violet spectrum (with higher energy) and above red, we have the infra-red spectrum.
The wavelengths in the range of 650 to 690 nm have red as the dominant color.

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What is the potential difference between the ends of the wire??

malfutka [58]

The potential difference between the ends of the wire is (V - IR)

7 0

3 years ago

A hollow cylinder with an inner radius of and an outer radius of conducts a 3.0-A current flowing parallel to the axis of the cy

Artemon [7]

Complete Question:

A hollow cylinder with an inner radius of 4.0mm and an outer radius of 30mm conducts a 3.0-A current flowing parallel to the axis of the cylinder. If the current density is uniform throughout the wire, what is the magnitude of the magnetic field at a point 12mm from its center?

Answer:

The magnitude of the magnetic field = 7.24 μT

Explanation:

Inner radius, a = 4.0 mm = 0.004 m

Outer radius, b = 30 mm = 0.03 m

Radius, r = 12 mm = 0.012 m

let h² = b² - a²

h² = 0.03² - 0.004²

h² = 0.000884

Let d² = r² - a²

d² = 0.012² - 0.004²

d² = 0.000128

Current I = 3A

μ = 4π * 10⁻⁷

The magnitude of the magnetic field is given by:

$B = \frac{\mu I d^{2} }{2\pi r h^{2} } \\B = \frac{4\pi * 10^{-7} * 3* 0.000128^{2} }{2\pi *0.012* 0.000884^{2} }$

B = 7.24 * 10⁻⁶T

B = 7.24 μT

7 0

3 years ago

Vary the sled’s height and mass. Observe the effect of each change on the potential energy of the sled.

qaws [65]

Answer:

a. Potential energy of the sled is increased when height of sled is increased.

b. Potential energy of the sled is increased when height of sled is increased.

c. P.E₂₀ = 2 P.E₁₀

d. P.E₂₀₀ = 2 P.E₁₀₀

Explanation:

The potential energy of the sled can be given by the following:

$Potential\ Energy = P.E = mgh\\$

where,

m = mass of sled

g = acceleration due to gravity

h = height of sled

a.

It is clear from the formula that potential energy of sled is directly proportional to the height of sled.

<u>Therefore, potential energy of the sled is increased when height of sled is increased.</u>

b.

It is clear from the formula that potential energy of sled is directly proportional to the mass of sled.

<u>Therefore, potential energy of the sled is increased when mass of sled is increased.</u>

<u></u>

c.

$P.E\ at\ 10\ m:\\P.E_{10} = 10mg\\P.E\ at\ 20\ m:\\P.E_{20} = 20mg\\\frac{P.E_{20}}{P.E_{10}} = \frac{20mg}{10mg}$

<u>P.E₂₀ = 2 P.E₁₀</u>

<u></u>

d.

$P.E\ at\ 100\ kg:\\P.E_{100} = 100gh\\P.E\ at\ 200\ m:\\P.E_{200} = 200gh\\\frac{P.E_{200}}{P.E_{100}} = \frac{200gh}{100gh}$

<u>P.E₂₀₀ = 2 P.E₁₀₀</u>

7 0

3 years ago

If a 20kg mass hangs from a spring, whose elastic constant is 1800 N / m, the value of the spring elongation is

almond37 [142]

Explanation:

F = kx

mg = kx

(20 kg) (10 m/s²) = (1800 N/m) x

x = 0.11 m

4 0

3 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