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Butoxors [25]
3 years ago
6

This distance from the Earth to the Sun given above is a standard for measuring other distances in the solar system and is calle

d an Astronomical Unit or AU. 1AU=_________miles. Write this as a ratio:_________
Physics
1 answer:
sineoko [7]3 years ago
5 0

Answer:

The Astronomical Unit (AU) is approximately 92.96\times 10^{6}\,mi, the ratio of Astronomical Units per mile is r = 1\,AU\,:\,92.96\times 10^{6}\,mi.

Explanation:

The distance from the Earth to the Sun is approximately 92.96\times 10^{6}\,mi, then the Astronomical Unit (AU) is approximately 92.96\times 10^{6}\,mi. The ratio can be described by the quantity of miles per Astronomical Unit, then we define:

r = 1\,AU\,:\,92.96\times 10^{6}\,mi.

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Explanation:

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A wave emitted from a source has a frequency of 10 Hz and wavelength 2.5 m. How much time will it take to reach a person located
const2013 [10]

Answer:

time taken by the wave to reach the person is 0.2 s

Explanation:

As we know that the speed of the wave is given as

v = \lambda f

here we know that the wavelength of the wave is

\lambda = 2.5 m

f = 10 Hz

now speed of the wave is given as

v = 10(2.5)

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Now time taken by the wave to reach 5 m distance is

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4 0
3 years ago
Two equally charged tiny spheres of mass 1.0 g are placed 2.0 cm apart. When released, they begin to accelerate away from each o
zhannawk [14.2K]

Answer:

The magnitude of the charge on each sphere is 0.135 μC

Explanation:

Given that,

Mass = 1.0

Distance = 2.0 cm

Acceleration = 414 m/s²

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Using newton's second law

F= ma

a=\dfrac{F}{m}

Put the value of F

a=\dfrac{kq^2}{mr^2}

Put the value into the formula

414=\dfrac{9\times10^{9}\times q^2}{1.0\times10^{-3}\times(2.0\times10^{-2})^2}

q^2=\dfrac{414\times1.0\times10^{-3}\times(2.0\times10^{-2})^2}{9\times10^{9}}

q^2=1.84\times10^{-14}

q=0.135\times10^{-6}\ C

q=0.135\ \mu C

Hence, The magnitude of the charge on each sphere is 0.135μC.

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3 years ago
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Answer:

it b

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bc A water droplet falling in the atmosphere is spherical

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2 years ago
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