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taurus [48]
3 years ago
9

The nearest star to the Earth is the red dwarf star Proxima Centauri, at a distance of 4.218 light-years. Convert this distance

to barleycorns. (1barleycorn=1/3 inch.)
Physics
1 answer:
alukav5142 [94]3 years ago
3 0

Explanation:

The nearest star to the Earth is the red dwarf star Proxima Centauri, at a distance of 4.218 light-years.  

Light year is the unit of distance covered by the heavenly bodies. 1 light year is equal to :

1\ light\ year=3.72\times 10^{17}\ inches

So, 4.218\ light\ year=4.218\times 3.72\times 10^{17}\ inches

4.218\ light\ year=1.56\times 10^{18}\ inches

We need to convert 4.218 light-years barley corns.  

Since, 1 barleycorn = 1/3 inch  

1\ inch=3\ barleycorn

1.56\times 10^{18}\ inches=3\times 1.56\times 10^{18}=4.68\times 10^{18}\ barleycorn

So, the nearest star to the Earth is at a distance of 4.68\times 10^{18}\ barleycorn. Hence, this is the required solution.

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In the formula GPE = mgh, which symbols represent the constants, and which symbols represent the variables?​
lorasvet [3.4K]

Answer:

Explanation:

m = mass of the body

g = acceleration of gravity

h = height above surface of earth

 

m and g are constants while h can vary as the body is moving.  Unless h is extremely large, you can take g to be a constant 9.81 m/s2

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3 years ago
What are bio photographers how does this work contribute to the health sciences
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Explanation:

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4 years ago
Why is an air mass unlikely to form over the rocky mountains of north america?
Shkiper50 [21]
The mountains can and will block airflow from higher pressure systems that come in from a coast and won't combine to nake storms
3 0
3 years ago
A copper wire has a square cross section 2.0 mm on a side. The wire is 5.0 m long and carries a current of 2.0 A. The density of
kondor19780726 [428]

Answer:

30.22 hours

Explanation:

Given data:

A= l² = (2 x 10^{-3})² = 4 x 10^{-6} m²

Length 'L' = 5m

current 'I' = 2 A

density of free electrons 'n'= 8.5 x 10^{28} /m³

Current Density 'J' = I/ A

J= 2/4 x 10^{-6}

J= 5 x 10^{5} A/m²

We can determine the  time required for an electron to travel the length of the wire by

T= L/ Vd

Where,

L is length and Vd is drift velocity.

Vd can be defined by J/ n|q|

where,

n is the charge-carrier number density

|q| is is the charge carried by each charge carrier =>1.6 x 10^{-19}C

T= L/ Vd

Therefore,

T= L . n|q| / J

T= (4 x 8.5 x 10^{28} x |1.6 x 10^{-19}|)/5 x 10^{5}

T= 108800 seconds =>1813.33 minutes

Converting minute into hours:

T= 30.22 hours

Thus, time that is required for an electron to travel the length of the wire is 30.22 hours

4 0
3 years ago
A large pendulum with a 200-lb gold-plated bob 12 inches in diameter is on display in the lobby of the United Nations building.
Tpy6a [65]

Answer:

2.4s

Explanation:

The length of the pendulum = 75ft

Diameter d = 12 inches

The time period of the pendulum is given as

T = 2pi(L/g)^1/2

Then the time it takes to move from displacement to equilibrium is given as:

t = T/4

= (Pi/2)*(L/g)^1/2

= pi/2 x [(75x0.3048)/9.81]^0.5

= 1.57x[22.86/9.81)^0.5

= 2.4s

2.4 seconds is the least amount of time that it would take.

5 0
3 years ago
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