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Lemur [1.5K]
3 years ago
11

The speed of light is 3.00×108m/s. How long does it take for light to travel from Earth to the Moon and back again? Express your

answer using two significant figures.
Physics
1 answer:
bija089 [108]3 years ago
7 0

Answer:

v = 3×10^8 m/s

s= 384,400 km= 3.84×10^8 m/s

t = ?

v = s/t = 2s/t

t = 2s/v

t = (2×3.84×10^8) ÷ 3×10^8

t = 2.56 seconds

Explanation:

Earth's moon is the brightest object in our

night sky and the closest celestial body. Its

presence and proximity play a huge role in

making life possible here on Earth. The moon's gravitational pull stabilizes Earth's wobble on its axis, leading to a stable climate.

The moon's orbit around Earth is elliptical. At perigee — its closest approach — the moon comes as close as 225,623 miles (363,104 kilometers). At apogee — the farthest away it gets — the moon is 252,088 miles (405,696

km) from Earth. On average, the distance fromEarth to the moon is about 238,855 miles (384,400 km). According to NASA , "That means 30 Earth-sized planets could fit in between Earth and the moon."

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A) A mass spectrometer has a velocity selector that allows ions traveling at only one speed to pass with no deflection through s
KengaRu [80]

Answer:

(A). The speed of the ions is 1.2\times10^{6}\ m/s

(B). The radius of curvature of a singly charged lithium ion is 2.0\times10^{6}\ m

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Given that,

Electric field = 60000 N/C

Magnetic field = 0.0500 T

(A). We need to calculate the velocity

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F_{E}=F_{B}

Eq=Bqv

v = \dfrac{E}{B}

v=\dfrac{60000}{0.0500}

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(B). We need to calculate the radius

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Bqv=\dfrac{mv^2}{r}

r=\dfrac{mv^2}{Bqv}

Put the value into the formula

r=\dfrac{1.16\times10^{-26}\times(1.2\times10^{6})^2}{0.0500\times1.6\times10^{-19}}

r=2.0\times10^{6}\ m

Hence, (A). The speed of the ions is 1.2\times10^{6}\ m/s

(B). The radius of curvature of a singly charged lithium ion is 2.0\times10^{6}\ m

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3 years ago
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