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nevsk [136]
3 years ago
6

Light with a frequency of 7.30 x 1014 hz lies in the violet region of the visible spectrum. What is the wavelength of this frequ

ency of light?
Physics
1 answer:
Leona [35]3 years ago
6 0

From the theory we know that:

c = λ / T

f = 1 / T

Where:

c = 3.10^{8} / m   (the speed of light)

λ is the wavelengh (in meters)

T is the period (in seconds)

f is the frequency (in Hz)

We were told that:

f = 7.30 . 10^{14}

And we want to find out the value of λ.

c = λ / T  

c = λ . 1/T

Swaping 1/T = f

c = λ . f

λ = c / f

λ = 3 . 10^{8} /  7.30 . 10^{14}

λ = 4.12 10^{-7} m

Response: 4.12 10^{-7} m = 412 nm

:-)

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A horizontal force of 300.0 N is used to push a 145-kg mass 30.0 m horizontally in 3.00 s. Calculate the power developed.
Svetradugi [14.3K]

Answer:

3 * 10³J/s

Explanation:

Given :

Force applied, F = 300 N

Distance, d = 30 m

Time, t = 3 seconds

Power, P = Workdone / time

Recall :

Workdone = Force * distance

Workdone = 300 N * 30 m = 9000 Nm

Workdone = 9 * 10³ J

Power = (9 * 10³ J) / 3s

Power = 3 * 10³J/s

4 0
3 years ago
Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.
Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

Explanation:

Given that,

Positive charge = 24.00  μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})

\theta=\sin^{-1}(\dfrac{1}{4})

\theta=14.47^{\circ}

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

E=\dfrac{2k\lambda}{r}\times2\cos\theat

Put the value into the formula

E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}

E=408094.00\ N/C

E=4.08\times10^{5}\ N/C

Hence, The magnitude of the electric field at a point equidistant from the lines is 4.08\times10^{5}\ N/C

6 0
3 years ago
8. A sprinter on a school track team is running north at a velocity of 6.0 m/s. After 5.0 s, she
Marysya12 [62]

Answer:

acc. = 4-(-6) /5= 10/5=2 m/s^2

6 0
3 years ago
Which of the following occurs with both a cold front and a mountain breeze?
BigorU [14]

Answer:

b warm air rises

Explanation:

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SSSSS [86.1K]
You said that she's losing 1.9 m/s of her speed every second.

So it'll take

             (6 m/s) / (1.9 m/s²)  =  3.158 seconds  (rounded)

to lose all of her initial speed, and stop.
6 0
3 years ago
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