Answer:
<u>Magnitude</u>
Explanation:
Each value in nature has a number part, called its magnitude and a dimension called its unit.
For example,
The length of an object is 10 cm. It means that 10 shows the magnitude of length and cm shows its unit.
At the given erro in angle, the error in the measurement of sin 90 degrees would be 0.001.
<h3>
Percentage error</h3>
The percentage error of any measurement is obtained from the ratio of the error to the actual measurement.
The error of sin 90 degrees is calculated as follows;
sin 90 = 1
error in measurement = sin(90 - 0.5)
error in measurement = sin(89.5) = 0.999
<h3>Error in sin 90 degrees</h3>
Error in sin 90 degrees = 1 - 0.999
Error in sin 90 degrees = 0.001
Thus, at the given erro in angle, the error in sin 90 degrees would be 0.001.
Learn more about error in measurement here: brainly.com/question/26668346
Chapter name please then I can answer
Answer:
Explanation:
We know that,
Neptune is 4.5×10^9 km from the sun
And given that,
Earth is 1.5×10^8km from sun
Then,
Let P be the orbital period and
Let a be the semi-major axis
Using Keplers third law
Then, the relation between the orbital period and the semi major axis is
P² ∝ a³
Then,
P² = ka³
P²/a³ = k
So,
P(earth)²/a(earth)³ = P(neptune)² / a(neptune)³
Period of earth P(earth) =1year
Semi major axis of earth is
a(earth) = 1.5×10^8km
The semi major axis of Neptune is
a (Neptune) = 4.5×10^9km
So,
P(E)²/a(E)³ = P(N)² / a(N)³
1² / (1.5×10^8)³ = P(N)² / (4.5×10^9)³
Cross multiply
P(N)² = (4.5×10^9)³ / (1.5×10^8)³
P(N)² = 27000
P(N) =√27000
P(N) = 164.32years
The period of Neptune is 164.32years
