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What is the basic difference between ultraviolet, visible, and infrared radiation?

N76 [4]

To me the basic difference is their frequency and wavelength.

For frequency :
Infrared < visible light < ultraviolet

For wavelength :
Infrared > visible light > ultraviolet

3 0

4 years ago

10. If the mass of the Earth is... increased by a factor of 2, then the Fgrav is ______________ by a factor of _______. ... incr

12345 [234]

Answer:

If the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3, then Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4

Explanation:

In order to solve this question, we must take into account that the force of gravity is given by the following formula:

$F_{g0}=G \frac{mM_{E0}}{r^{2}}$

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=2M_{E0}$

so:

$F_{gf}=G \frac{2mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{2mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=2$

so if the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3

So if the mass of the earth is increased by a factor of 2, this means that:

$M_{Ef}=3M_{E0}$

so:

$F_{gf}=G \frac{3mM_{E0}}{r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{3mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=3$

so if the mass of the Earth is increased by a factor of 3, then the Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4

So if the mass of the earth is decreased by a factor of 4, this means that:

$M_{Ef}=\frac{M_{E0}}{4}$

so:

$F_{gf}=G \frac{mM_{E0}}{4r^{2}}$

Therefore:

$\frac{F_{gf}}{F_{g0}}=\frac{G \frac{mM_{E0}}{4r^{2}}}{G \frac{mM_{E0}}{r^{2}}}$

When simplifying we end up with:

$\frac{F_{gf}}{F_{g0}}=\frac{1}{4}$

so if the mass of the Earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4.

4 0

3 years ago

Simplified. The absorption spectra of ions have been used to identify the presence of the elements in the atmospheres of the sun

Nookie1986 [14]

Answer:

The answer is " $3.83 \times 10^9 \ m$ "

Explanation:

Z=2, so the equation is $E= \frac{-4B}{n^2}$

Calculate the value for E when:

n=2 and n=9

The energy is the difference in transformation, name the energy delta E Deduct these two energies

In this transition, the wavelength of the photon emitted is:

$\Delta E=2.18 \times 10^{-18} ( \frac{1}{4}- \frac{1}{81})$

$\lambda = \frac{h c}{\Delta E}$

$h ( Planck's\ constant) = 6.62 \times 10^{(-34)} \ Js \\\\ speed \ of \ light = 3 \times 10^{8} \ \frac{m}{s}\\\\= \frac{6.62 \times 10^{(-34)} \times 3 \times 10^ {8}}{2.18 \times 10^{-18}} (\frac{1}{4}- \frac{1}{81}) \\\\=3.83 \times 10^9 \ m\\\\$

6 0

3 years ago

A skateboarder is accelerating forward. How would his acceleration change if he had more mass?

exis [7]

Answer:

If the force remains the same, the acceleration would decrease

Explanation:

According to Newton's second law, the acceleration of an object is given by

$a=\frac{F}{m}$

where

F is the force applied to the object

m is the mass of the object

As we see from the formula, the acceleration a is inversely proportional to the mass, m. Therefore, if the force F remains constant, this means that if the mass of the skateboarder increases, then the acceleration will decrease.

4 0

4 years ago

A package is pushed across the floor a distance of 90 feet by exerting a force of 32 lbs downward at an angle of 24 ∘ with the h

Crazy boy [7]

Answer:

2632 foot-pound

Explanation:

Work done: Work is said to be done when ever a force moves a body through a given distance. The S.I unit of force is Newton (N).

From the question,

The expression for work done is given as,

W = Fdcos∅......................... Equation 1

Where W = work done, F = force, d = distance, ∅ = angle between the force and the horizontal.

Given: F = 32 lbs, d = 90 feet, ∅ = 24°

substitute into equation 1

W = 32×90×cos24

W = 2880(0.914)

W = 2632.32

W = 2632 foot-pound

4 0

3 years ago