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3 years ago

¿Cuál es el rango de frecuencias comprendido entre las longitudes de onda de 220 nm, 350 nm,

Physics

1 answer:

olchik [2.2K]3 years ago

3 0

Answer:

f1 = 1.36*10^15 Hz

f2 = 8.57*10^14 Hz

f3 = 3.75*10^14 Hz

f4 = 8.57*10^13 Hz

Explanation:

To know which are the frequency ranges of the wavelengths you use the following formula:

$c=\lambda f\\\\f=\frac{c}{\lambda}$

c: speed of light = 3*10^8m/s

λ: wavelength = 220nm, 350nm, 800nm, 3500nm

f: frequency of the wave

You calculate f for each wavelength, furthermore, you add the kind of electromagnetic wave:

λ1 = 220nm = 220*10^-9 m

$f=\frac{3*10^8m/s}{220*10^{-9}m}=1.36*10^{15}Hz$ ULTRAVIOLET

λ2 = 350nm = 350*10^-9 m

$f=\frac{3*10^8m/s}{350*10^{-9}m}=8.57*10^{14}Hz$ VISIBLE LIGHT

λ3 = 800nm = 350*10^-9 m

$f=\frac{3*10^8m/s}{800*10^{-9}m}=3.75*10^{14}Hz$ INFRARED

λ4 = 3500nm = 350*10^-9 m

$f=\frac{3*10^8m/s}{3500*10^{-9}m}=8.57*10^{13}Hz$ MICROWAVE - INFRARED

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Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of

emmainna [20.7K]

Answer:

$5,2$

Explanation:

From the question we are told that:

Speed of light $C=3.0×10^8 m/s.$

Generally the equation for Average Speed is mathematically given by

$V_{avg}=\frac{d}{t}$

Where

d=Distance between the Earth and the sun

$d=1.5*10^11m$

Therefore

$t=\frac{d}{V_{avg}}$

$t=\frac{1.5*10^11m}{3.0×10^8 m/s.}$

$t=5*10^2s$

Since m and n is given in the form of

$m*10^n$

Therefore

$m=5 & n=2$

$5,2$

3 0

3 years ago

Jack and Jill are on two different floors of their high rise office building and looking out of their respective windows. Jack s

Maru [420]

Answer:

a) speed when Jack sees the pot : 12.92 meters per second

b) height difference 163.115 meters

Explanation:

First to calculate te initial speed we use the acceleration formula:

a= v1-v0/t

Acceleration being gravity's acceleration (9.8 m/s^2)

v1 being the speed when Jill sees the pot

v0 when Jack sees it

and t the time between

Solving for v0 it would be

v1 - a*t = v0

replacing

$58 m/s - 9.8 m/s^2 *4.6 s = v0 = 12.92 m/s$

For the second question we use the position formula setting y0 and t0 as the position and time when jack sees the pot. (and setting the positive axis downward I.E. one meter below jack would be 1m not -1m)

The formula is

$y0 + v0*t + 1/2 g *t^2 = yt$

replacing

$0m + 12.92m/s* 4.6 s + 1/2 * 9.8 m/s^2 * (4.6 s)^2 = 163.115 m$

5 0

3 years ago

7. A mother pushes her 9.5 kg baby in her 5kg baby carriage over the grass with a force of 110N @ an angle

jasenka [17]

Weight of the carriage $=(m+M)g =142.1\ N$

Normal force $=Fsin(\theta) + W = 197.1\ N$

Frictional force $=\mu N=27.59\ N$

Acceleration $=4.66\ m\ s^{-2}$

Explanation:

We have to look into the FBD of the carriage.

Horizontal forces and Vertical forces separately.

To calculate Weight we know that both the mass of the baby and the carriage will be added.

So Weight(W) $=(m+M)\times g =(9.5+5)\ kg \times 9.8 =142.1\ Newton\ (N)$

To calculate normal force we have to look upon the vertical component of forces, as Normal force is acting vertically.We have weight which is a downward force along with $F_x$ , force of $110\ N$ acting vertically downward.Both are downward and Normal is upward so Normal force $=Summation\ of\ both\ forces$