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Ulleksa [173]
3 years ago
13

A spaceship from a friendly, extragalactic planet flies toward Earth at 0.203 0.203 times the speed of light and shines a powerf

ul laser beam toward Earth to signal its approach. The emitted wavelength of the laser light is 691 nm . 691 nm. Find the light's observed wavelength on Earth.

Physics
2 answers:
Serjik [45]3 years ago
7 0

Answer:

567.321nm

Explanation:

See attached handwritten document for more details

Rudik [331]3 years ago
7 0

Answer:

The observed wavelenght on earth will be 5.51x10^-7 m

Explanation:

Speed of light c = 3x10^8 m/s

Ship speed u = 0.203 x 3x10^8 = 60900000 = 6.09x10^7 m/s

Wavelenght of laser w = 691x10^-9 m

Observed wavelenght w' = w(1 - u/c)

u/c = 0.203

w' = 691x10^-9(1 - 0.203)

w' = 5.51x10^-7 m

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3 years ago
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If you are driving 95 km????h along a straight road and you look to the side for 2.0 s, how far do you travel during this inatte
Gala2k [10]

Answer:

52.7 m

Explanation:

Given that

speed of the vehicle, v = 95 km/h

time of inattentiveness, t = 2 s

distance travelled, s = ?

Since we have the speed in km/h and the time in s, it would be best if we converted one of them to make sure we have all units in the same rank.

95 km/h = 95 * 1000/3600 m/s

95 km/h = 95000/3600 m/s

95 km/h = 26.38 m/s

Now, we use our derived speed in m/s

Speed of a moving vehicle is given by,

v = s/t, where

v = speed in m/s

s = distance travelled, in m

t = time spent, in s

if we make d the subject of formula by rearranging the equation, we have

s = v * t

distance travelled, s = 26.38 * 2

distance travelled, s = 52.7 m

therefore, during this inattentive period, 52.7 m was travelled.

3 0
3 years ago
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates
34kurt

The given question is incomplete. The complete question is as follows.

A parallel-plate capacitor has capacitance C_{0} = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 \times 10^{4} V/m?

Explanation:

It is known that relation between electric field and the voltage is as follows.

             V = Ed

Now,  

              Q = CV

or,           Q = C \times Ed

Therefore, substitute the values into the above formula as follows.

              Q = C \times Ed

                  = 8.50 pF \times (\frac{10^{-12} F}{1 pF})(3 \times 10^{4} m/s)(1 mm)(\frac{10^{-3} m}{1 mm})

                  = 2.55 \times 10^{-10} C

Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is 2.55 \times 10^{-10} C.

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Answer:

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Work = (force) x (distance) =

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6 0
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