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vladimir1956 [14]
3 years ago
15

A wave of infrared light has a speed of 6m/s and a wave length of 12 m. What is the frequency of this wave?

Physics
1 answer:
erastovalidia [21]3 years ago
6 0
Frequency = speed / wavelength

(6 m/s) / (12 m) = 0.5 Hz.

That's not infrared light.
Infrared light waves move about 50 million times faster than that, and they're only about 0.00000007 as long as that.
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A 62.00-cm guitar string under a tension of 70.000 N has a mass per unit length of 0.10000 g/cm. What is the highest resonant fr
quester [9]

Answer:

17,947.02 Hz

Explanation:

length (L) = 62 cm = 0.62 m

tension (T) = 70 N

mass per unit length (μ) = 0.10000 g/cm = 0.010000 kg/m

maximum frequency = 18,000 Hz

f = \frac{n}{2L} x \sqrt{\frac{T}{μ}}

f = \frac{n}{2 x 0.62} x \sqrt{\frac{70}{0.01} }

f = n x 67.47

18,000 = n x 67.47

n = 266.8≈ 266

the 267th overtone is the highest overtone that can be heard by this person, and its frequency would be 26 x 67.47 = 17,947.02 Hz

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2 years ago
How much electrical energy is used by a 400 W toaster that is operating for 5
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The answer is C. 120,000 J.

Explanation:

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For questions 1 - 4 complete the representations for the four patterns below. Provide the mathematical
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Suppose an event is measured to be at a = (0,-2, 3, 5) in one reference frame. Find the components of this event in another refe
LenaWriter [7]

Answer:

The components of the moving frame is (8.07c, -2, 3, 9.493)

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As per the question:

Velocity of moving frame w.r.t original frame v_{m} 0.85c

Point 'a' of an event in one reference frame corresponds to the (x, y, z, t) coordinates of the plane

a = (0, - 2, 3, 5)

Now, according the the question, the coordinates of moving frame, say (X, Y, Z, t'):

New coordinates are given by:

X = \frac{x - v_{m}t}{\sqrt{1 - \frac{v_{m}^{2}}{c^{2}}}}

X = \frac{0 - 0.85c\times 5}{\sqrt{1 - \frac{(0.85c)^{2}}{c^{2}}}}

X = 8.07 c

Now,

Y = y = - 2

Z = z = 3

Now,

t' = \frac{t - \frac{vx}{c}^{2}}{\sqrt{1 - (\frac{v}{c})^{2}}}

t' = \frac{5 - 0}{\sqrt{1 - (\frac{0.85c}{c})^{2}}} = 9.493 s

4 0
3 years ago
A train that has wheels with a diameter of 91.44 cm (36 inches used for 100 ton capacity cars) slows down from 82.5 km/h to 32.5
podryga [215]

Answer:

The answer is below

Explanation:

The initial velocity = u = 82.5 km/h = 22.92 m/s, the final velocity = 32.5 km/h = 9.03 m/s, diameter = 91.55 cm = 0.9144 cm

radius (r) = diameter / 2 = 0.9144 / 2= 0.4572 m

a) Initial angular velocity (\omega_o) = u /r = 22.92 / 0.4572 = 50.13 rad/s, final velocity  (ω) = v / r = 9.03 / 0.4592 = 19.67 rad / s

θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad

angular acceleration (α) is:

\omega^2=\omega_o^2+2\alpha \theta\\\\19.67^2-50.13^2=2\alpha(272.9)\\\\19.67^2=50.13^2+2\alpha(272.9)\\\\2\alpha(272.9)=-2126.108\\\\\alpha=-3.89\ rad/s^2\\\\

b)

\omega=\omega_o+\alpha t\\\\19.67=50.13+(-3.89t)\\\\3.89t=50.13-19..67\\\\3.89t=30.46\\\\t=7.83\ s

c) θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad

a) When it stops, the final angular velocity is 0. Hence:

\omega^2=\omega_o^2+2\alpha \theta\\\\0=50.13^2+2(-3.89)\theta\\\\2(3.89)\theta=50.13^2\\\\2(3.89)\theta=2513\\\\\theta=323\ rad\\\\revolutions=\frac{\theta}{2\pi r}=\frac{323}{2\pi(0.4572)}  =112.4\ rev

θ = 323 rad

4 0
2 years ago
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