Answer:
a

b

Explanation:
From the question we are told that
The speed of the spaceship is 
Here c is the speed of light with value 
The length is 
The distance of the star for earth is 
The speed is 
Generally the from the length contraction equation we have that
![l = l_o \sqrt{1 -[\frac{v}{c } ]}](https://tex.z-dn.net/?f=l%20%20%3D%20%20l_o%20%20%5Csqrt%7B1%20-%5B%5Cfrac%7Bv%7D%7Bc%20%7D%20%5D%7D)
Now the when at rest the length is 
So



Considering b
Applying above equation
![l =l_o \sqrt{1 - [\frac{v}{c } ]}](https://tex.z-dn.net/?f=l%20%20%3Dl_o%20%5Csqrt%7B1%20-%20%20%5B%5Cfrac%7Bv%7D%7Bc%20%7D%20%5D%7D)
Here 
So



Hello My Dear Friend!Do you mean
Sun or actually Moon?If Moon is what you're looking for...then that is
absolutely IMPOSSIBLE, due to that the Earth is WAY bigger than the Moon, therefore, not even HALF of the Earth would fit into the Moon since it's too small XD.But if it's
Sun...well then, t<span>he Answer is that it would take </span>
1.3 million Earths<span>
to fill up the Sun. That's a lot of Earths LOL XD</span>
I Hope my answer has come to your Help. Thank you for posting your question here in

We hope to answer more of your questions and inquiries soon.
Have a nice day ahead! :)

The magnification of the ornament is 0.25
To calculate the magnification of the ornament, first, we need to find the image distance.
Formula:
- 1/f = u⁻¹+v⁻¹.................... Equation 1
Where:
- f = Focal length of the ornament
- u = image distance
- v = object distance.
make u the subject of the equation
- u = fv/(f+v)................ Equation 2
From the question,
Given:
Substitute these values into equation 2
- u = (12×4)/(12+4)
- u = 48/16
- u = 3 cm.
Finally, to get the magnification of the ornament, we use the formula below.
- M = u/v.................. Equation 3
Where
- M = magnification of the ornament.
Substitute these values above into equation 3
Hence, The magnification of the ornament is 0.25
Answer:
205 V
V
= 2.05 V
Explanation:
L = Inductance in Henries, (H) = 0.500 H
resistor is of 93 Ω so R = 93 Ω
The voltage across the inductor is

w = 500 rad/s
IwL = 11.0 V
Current:
I = 11.0 V / wL
= 11.0 V / 500 rad/s (0.500 H)
= 11.0 / 250
I = 0.044 A
Now
V
= IR
= (0.044 A) (93 Ω)
V
= 4.092 V
Deriving formula for voltage across the resistor
The derivative of sin is cos
V
= V
cos (wt)
Putting V
= 4.092 V and w = 500 rad/s
V
= V
cos (wt)
= (4.092 V) (cos(500 rad/s )t)
So the voltage across the resistor at 2.09 x 10-3 s is which means
t = 2.09 x 10⁻³
V
= (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))
= (4.092 V) (cos (500 rads/s)(0.00209))
= (4.092 V) (cos(1.045))
= (4.092 V)(0.501902)
= 2.053783
V
= 2.05 V