Sound waves....................................
Answer:
The units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.
Explanation:
P² = a³ is the simplified version of Kepler's third law which governs the orbital motion of large bodies that orbit around a star. The orbit of each planet is an ellipse with the star at the focal point.
Therefore, if you square the year of each planet and divide it by the distance that it is from the star, you will get the same number for all the other planets.
Thus, the units of the orbital period P is <em>years </em> and the units of the semimajor axis a is <em>astronomical units</em>.
The speed of light is: c
= 3x10^8 m/s <span>
or
c = 186,000,000 miles/sec = 1.86x10^8 mi/s
1 furlong = 0.125 mile
1 fortnight = 2 weeks(7d/wk)(24h/d)(3600s/h)
= 1209600s = 1.2096x10^6 s
Therefore,
c =1.86x10^8 mi/s(1furl/0.125mi)(1.2096x10^6s/fort)
<span>c = 18x10^14 furlong/fortnight = 18x10^8 Mfurlong/fortnight</span></span>
Answer:
a) I = 13.38 kg m / s, b) F = 1,373 10³ N
Explanation:
The impulse is given by the relation
I = ∫ F dt = Δp
I = p_f -p₀
I = m (v_f - v₀)
take the ball's exit direction as positive, whereby the ball velocities
v₀ = -90mph, the final velocity v_f = + 54 m / s
Let's reduce the units to
I = 0.142 [54- (-40.23) ]
the SI system
v₀ = - 90 mph (1609.34 m / 1 mile) (1h / 3600 s = -40.23 m / s
m = 142 g (1kg / 1000) = 0.142 kg
we calculate
I = 0.142 [54- (-40) ]
I = 13.38 kg m / s
b) let's use the definition of momentum
I = ∫ F .dt
I = F ∫ dt
F = I / t
F = 13.38 / 0.008
F = 1,373 10³ N
