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3 years ago

How as the energy of infrared, visible light and ultraviolet rays related?and why

2 answers:

8 0

A) Infrared contains the least energy and ultraviolet contains the most energy. The energy of a wave is proportional to its wavelength.

B) Infrared contains the most energy and ultraviolet contains the least energy. The energy of a wave is proportional to its wavelength.

C) Infrared contains the least energy and ultraviolet contains the most energy. The energy of a wave is inversely proportional to its wavelength.

D) Infrared contains the most energy and ultraviolet contains the least energy. The energy of a wave is inversely proportional to its wavelength.

Eliminate

sleet_krkn [62]3 years ago

5 0

All are forms of electromagnetic radiation. hope this helps!

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You drive a race car around a circular track of radius 300 m at a constant spe 108 km/h. Your acceleration will be: A) 39.9 m/s

morpeh [17]

Answer:

Explanation:

Centripetal acceleration is:

a = v² / r

First, convert km/h to m/s:

108 km/h × (1000 m / km) × (1 h / 3600 s) = 30 m/s

Therefore, the acceleration is:

a = (30 m/s)² / 300 m

a = 3 m/s²

8 0

4 years ago

How does the today's model of the atom DIFFER from the Rutherford’s model? 25 Points

Yuliya22 [10]

Answer:

The Rutherford Model shows an atom with electrons orbiting a fixed, positively charged nucleus in set, predictable paths. The Bohr model shows electrons travel in defined circular orbits around the nucleus.

Explanation:

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3 years ago

Read 2 more answers

A student was asked to explain how a prism separates white light into its separate colors. The student explanation is below but

Nonamiya [84]

Answer:

pp i hate khan academy pls help me

3 0

3 years ago

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A 100-year flood is the flooding elevation that has a one percent chance of happening each year. Why is the name misleading

fenix001 [56]

One percent annual chance of flooding does not mean it will happen with absolute certainty once every 100 years. Just that the expected value of a flood in 100 years is one.

As an example, heads has a 50% chance of happening every coin flip. With 2 coin flips, the expected value is one. However, it would be a misnomer to call "heads" a 2-flip outcome.

5 0

4 years ago

Water is flowing through a 45° reducing pipe bend at a rate of 200 gpm and exits into the atmosphere (P2 = 0 psig). The inlet to

Nataliya [291]

Answer:

F1=177.88 Newtons

Explanation:

Let's start with the Bernoulli's equation:

$P_{1} + \frac{1}{2}\beta V_{1} ^{2} + \beta gh_{1} =P_{2} + \frac{1}{2}\beta V_{2} ^{2} + \beta gh_{2}$

Where:

P is pressure, V is Velocity, g is gravity, h is height and β is density (for water β=1000 kg/m3); at the points 1 and 2 respectively.

From the Bernoulli's equation and assuming that h is constant and P2 is zero (from the data), we have:

$P_{1} + \frac{1}{2}\beta V_{1} ^{2} = \frac{1}{2}\beta V_{2} ^{2}$

As we know, P1 must be equal to $\frac{F_{1} }{A_{1}}$ , so, replacing P1 in the equation, we have:

$P_{1} = \frac{F_{1}}{A_{1}} = \frac{1}{2}\beta(V_{2} ^{2} - V_{1} ^{2})$

And

$F_{1} = {A_{1}} ( \frac{1}{2}\beta(V_{2} ^{2} - V_{1} ^{2}))$

Now, let's find the velocity to replace the values on the expression:

We can express the flow in function of velocity and area as $Q = V A$ , where Q is flow, V is velocity and A is area. As the same, we can write this: $Q_{1} = V_{1} A_{1}\\Q_{2} = V_{2} A_{2}$ . In the last two equations, let's clear Velocities.

$V_{1} = \frac{Q_{1}}{A_{1}}\\V_{2} = \frac{Q_{2}}{A_{2}}$

and replacing V1 and V2 on the last equation resulting from Bernoulli's (the one that has the force on it):

$F_{1} = {A_{1}} ( \frac{1}{2}\beta((\frac{Q_{2}}{A_{2}})^{2} - (\frac{Q_{1}}{A_{1}})^{2}))$

First, we have to consider that from a mass balance, the flow is the same, so Q1=Q2, what changes, is the velocity. Knowing this, let's write the areas, diameters, density and flow on International Units System (S.I.), because the exercise is asking us the answer in Newtons.

$D_{1}=1.5 inches=0.0381 mts\\D_{2}=1 inches=0.0254 mts\\A_{1}=\frac{\pi D_{1}^{2} }{4}=0.00114mts^{2}\\A_{2}=\frac{\pi D_{2}^{2} }{4}=0.000507mts^{2}\\\beta=1000 kgs/m^{3}\\Q=200gpm=0.01mts^{3}/seg$

Replacing the respective values in this last expression, we obtain:

F1 = 177.88 N

3 0

3 years ago