Answer:
Oops i put the answer in a comment,
-10 to the left m/s^2
Explanation:
Using Kepler's 3rd law which is: T² = 4π²r³ / GM
Solved for r :
r = [GMT² / 4π²]⅓
Where G is the universal gravitational constant,M is the mass of the sun,T is the asteroid's period in seconds, andr is the radius of the orbit.
Change 5.00 years to seconds :
5.00years = 5.00years(365days/year)(24.0hours/day)(6... = 1.58 x 10^8s
The radius of the orbit then is computed:
r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.58 x 10^8s)² / 4π²]⅓ = 4.38 x 10^11m
Answer:
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Answer: mechanical efficieny.
Efficieny is also expressend as percent. The formula for mechanical efficiency as percent is the ratio work output to wor input times 100.
The ideal mechanical efficiency for a machine would be 1 or 100% which means that all the input work is converted into output work. But this is just an idealization as the friction and other losses of energy make it imposible to reach 100% efficiency in reality, so the mechanical efficiency of real machinces is less than 100% or 1.
Answer:
W = 53.6648 J
Explanation:
W = ∫ F * dr
F = < 4*x i , 3*y j > Newtons
dr = < dx, 0 >
Take the dot product:
F * dr = 4* x * dx
Now replacing numeric
W = ∫ 4 * x * dx , ║ x = ( 0 m ⇒ 5.18 m )
W = ¹ / ₂ * (4 N/m) * x ²
W = (2 N/m) * (5.18 m)²
W = 53.6648 J
