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Neko [114]
3 years ago
7

On the 1-to-10-billion scale, about how far is it to the nearest stars besides the sun? on the 1-to-10-billion scale, about how

far is it to the nearest stars besides the sun? 400 kilometers 4 kilometers 1,000 kilometers 10,000 kilometers 4,400 kilometers
Physics
1 answer:
aliina [53]3 years ago
8 0
Not sure what your question means but the nearest star is Alpha Centauri which is about 4.2 light years (ly) away.  This is roughly 4x10¹³ km away.  A billion is 10⁹ so this is 4x10⁴ larger than a billion.  I'd say the last one then...
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If the battery of your phone can provide 2 mA of current to your phone and holds a charge of 130 C, how long will it take a full
Naya [18.7K]

Answer: 65000 seconds

Explanation:

Given that,

Current (I) = 2 mA

(Since 1 mA = 1 x 10^-3A

2 mA = 2 x 10^-3A)

Charge (Q) = 130 C

Time taken for a fully charged phone to die (T) = ?

Recall that the charge is the product of current and time taken.

i.e Q = I x T

130C = 2 x 10^-3A x T

T = 130C / (2 x 10^-3A)

T = 65000 seconds (time will be in seconds because seconds is the unit of time)

Thus, it will take a fully charged phone 65000 seconds to die

5 0
3 years ago
A 2.0kg mass is attached to a horizontal spring having a spring constant of 0.05Nm.
Alex Ar [27]

Good.  You can do some very interesting experiments with that equipment.

3 0
3 years ago
A 5.0-μC charge is placed at the 0 cm mark of a meter stick and a -4.0 μC charge is placed at the 50 cm mark. At what point on a
Maksim231197 [3]

Answer:

The distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

Explanation:

Given that

q₁ = 5 μ C

q₂ = - 4 μ C

The distance between charges = 50 cm

d= 50 cm

Lets take at distance x from the charge μ C ,the electrical field is zero.

That is why the distance from the charge - 4 μ C =  50 - x cm

We know that ,electric field is given as

E=K\dfrac{q}{r^2}

K\dfrac{5\ \mu}{x^2}=K\dfrac{4\mu }{(50-x)^2}\\\\\dfrac{5}{x^2}=\dfrac{4 }{(50-x)^2}\\\\\\5(50-x)^2=4x^2\\(50-x)^2=0.8x^2\\\\50-x =0.89x\\\ x=\dfrac{50}{1.89}\ cm\\\\\\x=26.45\ cm\\

Therefore the distance from charge 5 μ C = 26.45 cm and the distance from - 4 μ C is 23.55 cm.

3 0
3 years ago
Merry-go-rounds are a common ride in park playgrounds. The ride is a horizontal disk that rotates about a vertical axis at their
Nataliya [291]

Answer:

Though the question is not specified here, but this information can determine the following quantity: period T= 6 secs, Frequency F=1/6 Hz, speed of rotation V= 2 pi ft/sec and wave length =pi/3 ft

Explanation:

7 0
3 years ago
The amplitude of a paricular wave is 4.0 m. The crest to trough distance
kozerog [31]

Answer:

The crest to trough distance = 8 m

Explanation:

Given that,

The amplitude of a particular wave is 4.0 m.

We need to find the crest to trough distance.

We know that,

Amplitude = The distance from the base line to the crest or the the distance from the baseline to the trough.

It means,

Distance from crest to trough = 2(Amplitude)

= 2(4)

= 8 m

Hence, the crest to trough distance is equal to 8 m.

6 0
3 years ago
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