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pantera1 [17]
3 years ago
5

What is the diameter of the sun

Physics
2 answers:
Contact [7]3 years ago
3 0
Diameter of sun = 865,370 miles
lozanna [386]3 years ago
3 0

Answer: 865,370 miles

Explanation: I did the quiz:] Have a nice day!

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Reasons why inductors opposes charges passing through it<br>​
saul85 [17]

Answer:

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Explanation:

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8 0
2 years ago
Which of the following quantities provide enough information to calculate the tension in a string of mass per unit length μ that
Bad White [126]

Answer:

A. the wave speed v and Wavelength

Explanation:

Given that

Mass density per unit length=μ

Frequency = f

The velocity V given as

\mu=\dfrac{T}{V^2}\ kg/m

V=\sqrt{\dfrac{T}{\mu}}

T=Tension

V=Velocity

V= f λ

λ=Wavelength

Therefore to find the tension ,only wavelength and speed is required.

The answer is A.

8 0
3 years ago
Two trains travel toward each other on the same track, beginning 100 miles apart. One train travels at 40 miles per hour; the ot
Paladinen [302]

D = distance between th two trains at the start of the motion = 100 miles

V = speed of the faster train towards slower train = 60 mph

v = speed of the slower train towards faster train = 40 mph

t = time taken by the two trains to collide = ?

time taken by the two trains to collide is given as

t = D/(V + v)

t = 100/(60 + 40) = 1 h

v' = speed of the bird = 90 mph

d = distance traveled by the bird

distance traveled by the bird is given as

d = v' t

d = 90 x 1

d = 90 miles

7 0
3 years ago
Read 2 more answers
When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin
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

5 0
3 years ago
The distance from the Earth to the Sun equals 1 AU. Neptune is 30 AU from the Sun. How far is Neptune from the Earth?AU
Ipatiy [6.2K]

Answer:

Depending on the relative position of the Earth the Sun and Neptune in the Earths orbit the distances are;

The closest (minimum) distance of Neptune from the Earth is 29 AU

The farthest (maximum) distance of Neptune fro the Earth is 31 AU

Explanation:

The following parameters are given;

The distance from the Earth to the Sun = 1 AU

The distance of Neptune from the Earth = 30 AU

We have;

When the Sun is between the Earth and Neptune, the distance is found by the relation;

Distance from the Earth to Neptune = 30 + 1 = 31 AU

When the Earth is between the Sun  and Neptune, the distance is found by the relation;

Distance from the Earth to Neptune = 30 - 1 = 29 AU

Therefore, the closest distance from Neptune to the Earth in the Earth's Orbit is 29 AU

The farthest distance from Neptune to the Earth in the Earth's orbit is 31 AU.

8 0
3 years ago
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