Suppose we wanted to represent all of these objects on the 1-to-10-billion scale from Chapter 1, on which the Sun is about the s
1 answer:
Answer:
6.62 m
About 44.14 grapefruits lined in a single line would be the length of the diameter of star Aldebaran
Explanation:
First we know,
Diameter of the Sun - 1.39 million km (Ds)
Diameter of Aldbaran - 61.40 million km (Da)
Thus,
=
≈ 44.14
We know that in this scale the size of the sun is that of a grapefruit.
The average diameter of a grapefruit is about 15cm
Thus Ds in this scale is 15cm
Thus Da in this scale,
Da = Ds *44.14
Da = 0.15 * 44.14 ≈ 6.62 m
Thus the diameter of Aldebaran in this scale is 6.62 m
You might be interested in
Answer:
R = 9.85 ohm , r = 0.85 ohm
Explanation:
Let the two resistances by r and R.
when they are connected in series:
V = 12 V
i = 1.12 A
The equivalent resistance when they are connected in series is
Rs = r + R
So, By using Ohm's law
V = i Rs
Rs = V / i = 12 / 1.12 = 10.7 ohm
R + r = 10.7 ohm .... (1)
When they are connected in parallel:
V = 12 V
i = 9.39 A
The equivalent resistance when they are connected in parallel
So, By using Ohm's law
V = i Rp
Rp = V / i = 12 / 9.39 = 1.28 ohm
.... (2)
by substituting the value of R + r from equation (1) in equation (2), we get
r R = 8.36 ..... (3)
..... (4)
By solvng equation (1) and (4), we get
R = 9.85 ohm , r = 0.85 ohm