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3 years ago

How many years is a light year

Physics

2 answers:

n200080 [17]3 years ago

6 0

10,108 earth years equals 1 light year

Sholpan [36]3 years ago

5 0

A "light year" is not an amount of time.

It's an amount of distance ... the distance that light travels
through space in one year.

1 light year = 5,878,625,372,000 miles
(rounded to the nearest thousand miles)

1 light-year = 9,460,730,473,000 kilometers
(rounded to the nearest thousand kilometers)

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dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

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3 years ago

A bar magnet was placed underneath a sheet of paper where a pile of iron filings sits. In the presence of the energy stored in t

Naya [18.7K]

Answer: 1. The field energy will increase

2. The energy increases, and the lines of force are denser

3. It points toward the field of earths magnetic poles

4. 1 and 2 only

5. 2, 4, 1, 3

Explanation: just took it

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3 years ago

A 1500 kg truck travelling north at 60 km/hr collides with a 1200 kg car moving east at 15km/hr. If the two cars remain locked t

Alenkinab [10]

Answer:

33.33j+6.67i km/hr

Explanation:

From the law of conservation of momentum,

Applying,

mu+m'u' = V(m+m')............... Equation 1

Where m = mass of the truck, m' = mass of the car, u = initial velocity of the truck, u' = initial velocity of the car, V = Final velocity.

Note: let j represent the north, and i represent the east

From the question,

Given: m = 1500 kg, u = 60j, m' = 1200 kg, u' = 15i

Substitute these values into equation 1

1500*60j+1200*15i = V(1500+1200)

90000j+18000i = 2700V

V = (90000j+18000i)/2700

V = 33.33j+6.67i km/hr

6 0

3 years ago

A drop

exis [7]

Answer:

$27.5\ m$

Explanation:

As we know that volume of cylinder is

$v=\pi r^{2} *h$

Where v=volume , h= height or thickness and r= radius

Here,

$v= 10 m ,\ diameter= 10, \ r=\frac{diameter}{2} \ r=\frac{10}{2}\\ r=5$

Putting these values in the previous equation , we get

$10\ = \frac{22}{7} *5 *5*h\\ 14\ =\ 110*h\\h=\frac{110}{14} \\h=\frac{55}{2} \\\\h=27.5\ m$

Therefore thickness is 27.5 m

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icang [17]

Answer:

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4 years ago