A. an accelerating charged charged particle or changing magnetic fields
The correct answer for the question that is being presented above is this one: "a. only from an instructor or supervisor." Ideally, rewards should be given immediately and frequently but <span>only from an instructor or supervisor to show authority. </span>
Answer;
=15855.40 kg/m^3
Explanation;
Volume (V) of the cylinder = pi x r^2 x h
V = 3.14 x (44/2 x 10^-3)^2 x 41.5 x 10^-3
V = 6.307 x 10^-5 m^3
By density = m/V
mass = 1 kg
density = 1/(6.307 x 10^-5) = 15855.40 kg/m^3
We would have to search at least 5,000,000,000 (5 billion) stars before we would expect to hear a signal.
To find out the number of stars that we will need to search to find a signal, we need to use the following formula:
- total of stars/civilizations
- 500,000,000,000 (500 billion) stars / 100 civilization = 5,000,000,000 (5 billion)
This shows it is expected to find a civilization every 5 billion stars, and therefore it is necessary to search at least 5 billion stars before hearing a signal from any civilization.
Note: This question is incomplete; here is the complete question.
On average, how many stars would we have to search before we would expect to hear a signal? Assume there are 500 billion stars in the galaxy.
Assuming 100 civilizations existed.
Learn more about stars in: brainly.com/question/2166533
Answer:
The answer to the question is
The ladybug begins to slide
Explanation:
To solve the question we assume that the frictional force of the ladybug and the gentleman bug are the same
Where the frictional force equals 
= μ×N = m×g×μ
and the centripetal force is given by m·ω²·r
If we denote the properties of the ladybug as 1 and that of the gentleman bug as 2, we have
m₁×g×μ = m₁·ω²·r₁ ⇒ g×μ = ω²·r₁
and for the gentleman bug we have
m₂×g×μ = m₂·ω²·r₂ ⇒ g×μ = ω²·r₂
But r₁ = 2×r₂
Therefore substituting the values of r₁ =2×r₂ we have
g×μ = ω²·r₁ = g×μ = ω²·2·r₂
Therefore ω²·r₂ = 0.5×g×μ for the ladybug. That is the ladybug has to overcome half the frictional force experienced by the gentleman bug before it start to slide
The ladybug begins to slide
