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Gnesinka [82]
3 years ago
6

The nearest star to our planet other than the sun , is 4.4 light year away.One light-year is the distance light travels in a yea

r.
* HOW LONG DOES IT TAKE FOR TO REACH US?
* SUGGEST WHY LIGHT-YEAR IS USED AS THE UNIT FOR MEASURING THE DISTANCE BETWEEN DISTANT STARS AND OUR PLANET, RATHER THAN THE UNIT 'KILOMETER,
Physics
1 answer:
lozanna [386]3 years ago
3 0

Answer:

1606

The main reason for using light years, however, is because the distances we deal with in space are immense. If we stick to miles or kilometers we quickly run into unwieldy numbers just measuring the distance to the nearest star: a dim red dwarf called Proxima Centauri that sits a mere 24,000,000,000,000 miles away!

Explanation:

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A solenoid of length 18 cm consists of closely spaced coils of wire wrapped tightly around a wooden core. The magnetic field str
Kisachek [45]

Answer:

B_2 = 1.71 mT

Explanation:

As we know that the magnetic field near the center of solenoid is given as

B = \frac{\mu_0 N i}{L}

now we know that initially the length of the solenoid is L = 18 cm and N number of turns are wounded on it

So the magnetic field at the center of the solenoid is 2 mT

now we pulled the coils apart and the length of solenoid is increased as L = 21 cm

so we have

\frac{B_1}{B_2} = \frac{L_2}{L_1}

now plug in all values in it

\frac{2.0 mT}{B_2} = \frac{21}{18}

B_2 = 1.71 mT

3 0
3 years ago
1 6. Which of the following astronomical units is c/osest to the distance of the earth to the sun?
Ganezh [65]

Answer:

astronomical unit

Explanation:

https://en.wikipedia.org/wiki/Astronomical_unit

6 0
3 years ago
U need help so can some one help me
GrogVix [38]

Answer:

are sure this is a question

7 0
3 years ago
What, roughly, is the percent uncertainty in the volume of a spherical beach ball whose radius is 5.66 0.09 m?
iren2701 [21]

Answer:

  • 4.77 %

Explanation:

We know that the volume V for a sphere of radius r is

V(r) = \frac{4}{3} \ \pi \ r^3

If we got an uncertainty \Delta r the formula for the uncertainty of V is:

\Delta V(r) = \sqrt{  (\frac{dV}{dr} \Delta r)^2  }

We can calculate this uncertainty, first we obtain the derivative:

\frac{dV}{dr}  = 3 * \frac{4}{3} \ \pi \ r^2

\frac{dV}{dr}  = 4 \ \pi \ r^2

And using it in the formula:

\Delta V(r) = \sqrt{  (4 \ \pi \ r^2\Delta r)^2  }

\Delta V(r) = \sqrt{  4^2 \ \pi^2 \ r^4 \Delta r^2  }

\Delta V(r) =  4 \  \pi \ r^2 \Delta r

The relative uncertainty is:

\frac{\Delta V(r)}{V(r)}

\frac{ 4 \  \pi \ r^2 \Delta r  }{ \frac{4}{3} \ \pi \ r^3}

\frac{ 3  \Delta r  }{  r}

Using the values for the problem:

\frac{ 3 * 0.09 m  }{  5.66 m} = 0.0477

This is, a percent uncertainty of 4.77 %

4 0
3 years ago
If the pendulum took longer to complete one oscillation, how would the graph change?
pickupchik [31]

We don't know what kind of graph it is.

For example, it might be a graph of the pendulum's distance from center,

angle from center, speed, acceleration, total distance swung since it was

started, mass, weight, temperature, etc.


If the graph shows the pendulum's distance from center, angle from center,

speed, or acceleration, then the graph will look like a wave, with the period

of the wave being the period of the pendulum's oscillation. If the pendulum

took longer to complete one oscillation, that means its PERIOD increased,

and the distance between the peaks of the graph would be longer.


If it was a graph of total distance the pendulum swung since it was started,

the graph wouldn't look like a wave, just a steadily rising wiggle line. If the

pendulum took longer to complete one oscillation, the wiggles in the line

would be farther apart, and the average slope of any large section of the

line would be less.


If it was a graph of the pendulum's mass, weight, temperature, cost, etc.,

then the graph would be a horizontal line, and nothing that might change

the period of oscillation would have any effect on the graph.

7 0
3 years ago
Read 2 more answers
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