<span>Px = 0
Py = 2mV
second, Px = mVcosφ
Py = –mVsinφ
add the components
Rx = mVcosφ
Ry = 2mV – mVsinφ
Magnitude of R = âš(Rx² + Ry²) = âš((mVcosφ)² + (2mV – mVsinφ)²)
and speed is R/3m = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
simplifying
Vf = (1/3m)âš((mVcosφ)² + (2mV – mVsinφ)²)
Vf = (1/3)âš((Vcosφ)² + (2V – Vsinφ)²)
Vf = (V/3)âš((cosφ)² + (2 – sinφ)²)
Vf = (V/3)âš((cos²φ) + (4 – 2sinφ + sin²φ))
Vf = (V/3)âš(cos²φ) + (4 – 2sinφ + sin²φ))
using the identity sin²(Ď)+cos²(Ď) = 1
Vf = (V/3)âš1 + 4 – 2sinφ)
Vf = (V/3)âš(5 – 2sinφ)</span>
If I'm correct, gold does not tarnish so C.
This acceleration is directed towards the center of thecircle<span>. ... So for an object moving in a </span>circle<span>, there must be an inward </span>force<span> acting upon it in order to</span>cause<span> its inward acceleration. This is sometimes referred to as the </span>centripetal force<span> requirement.</span>
Answer:
7661.06 m/s
Explanation:
R = Radius of Earth = 
h = Distance from the Earth = 420000 m
G = Gravitational constant = 
M = Mass of Earth = 

The orbital speed of the satellite is 7661.06 m/s