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d1i1m1o1n [39]
3 years ago
11

What is the approximate range of wavelengths for visible light?

Physics
1 answer:
Korvikt [17]3 years ago
7 0
We can answer the question by looking at the spectrum of the visible light, for example in this figure:
https://en.wikipedia.org/wiki/Light#/media/File:EM_spectrum.svg

We see from the figure that the lowest part of the spectrum (violet) corresponds approximately to a wavelength of 380 nm, while the highest part of the spectrum (red light) corresponds approximately to a wavelength of 750 nm.
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HELLLP PLEASE || the graph below shows a conversion of energy for a skydive jumping out of a plane and landing safely on the gro
fenix001 [56]

Answer: I maybe wrong but i'm pretty sure its C) Kinetic energy

5 0
4 years ago
Gravity and magnetism are both
ella [17]

Answer:

Explanation:

Comment

You have to read this carefully enough that you don't mix up energy and forces.

Gravity is a force. If you don't believe me try jumping off a building. Which way are you going to go and why? Down because gravity attracts your mass.

So Magnetism must be a force as well. It acts in one direction, but not a specific one the way gravity acts). It also either attracts or repulses (pushes an object away)

Answer A

3 0
2 years ago
Read 2 more answers
Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is
Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of c_p is determined via the expression:

c_p = \frac{KR}{K_1}

where ;

R = universal gas constant = \frac{8.314 \ J}{28.97 \ kg.K}

k = constant = 1.4

c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}

c_p= 1.004 \ kJ/kg.K

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}     ------ equation(1)

we can rewrite the above equation as :

0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}

T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}

where:

T_1  = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K

T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}

T_2 = 402.36 \ K

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation:\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}

P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}

P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}

P_2 = 17.79 \ bar

Therefore, the exit pressure is 17.79 bar

7 0
3 years ago
The objects listed are placed at the top of a ramp and roll down to the bottom without slipping. Assuming that there is no air r
Thepotemich [5.8K]

Explanation:

For each object, the initial potential energy is converted to rotational energy and translational energy:

PE = RE + KE

mgh = ½ Iω² + ½ mv²

For the marble (a solid sphere), I = ⅖ mr².

For the basketball (a hollow sphere), I = ⅔ mr².

For the manhole cover (a solid cylinder), I = ½ mr².

For the wedding ring (a hollow cylinder), I = mr².

If we say k is the coefficient in each case:

mgh = ½ (kmr²) ω² + ½ mv²

For rolling without slipping, ωr = v:

mgh = ½ kmv² + ½ mv²

gh = ½ kv² + ½ v²

2gh = (k + 1) v²

v² = 2gh / (k + 1)

The smaller the value of k, the higher the velocity.  Therefore:

marble > manhole cover > basketball > wedding ring

7 0
4 years ago
Which statement is true of earths poles?
djverab [1.8K]
Answer choice d is correct
7 0
3 years ago
Read 2 more answers
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