Answer:
<em> The object has frequency of 2 Hz and time period of 0.5 s.</em>
Explanation:
<em>Frequency</em> is defined as number of oscillation per second ie back and forth swings done in single second.
Here it is given that the object oscillates 20 times in 10 seconds.
So f = = 2Hz
The <em>time period</em> is defined as time taken by the object to complete one full oscillation.
T =
T= =0.5 s
<em>Thus the object has frequency of 2 Hz and time period of 0.5 s.</em>
Answer:
(orbital speed of the satellite) V₀ = 3.818 km
Time (t) = 4.5 × 10⁴s
Explanation:
Given that:
The radius of the Earth is 6.37 × 10⁶ m; &
the acceleration of gravity at the satellite’s altitude is 0.532655 m/s
We can calculate the orbital speed of the satellite by using the formula:
Orbital Speed (V₀) = √(r × g)
radius of the orbit (r) = 21000 km + 6.37 × 10⁶ m
= (2.1 × 10⁷ + 6.37 × 10⁶) m
= 27370000
= 2.737 × 10⁷m
Orbital Speed (V₀) = √(r × g)
Orbital Speed (V₀) = √(2.737 × 10⁷ × 0.532655 )
= 3818.215
= 3.818 × 10³
= 3.818 Km
To find the time it takes to complete one orbit around the Earth; we use the formula:
Time (t) = 2 π ×
= 2 × 3.14 ×
= 45019.28
= 4.5 × 10 ⁴ s