Answer:
As each mower presumably needs the same torque to start, and torque is a product of force and moment arm, the longer moment arm of 10.42 cm on Uwi's mower means lower force is required when compared to Urippe's shorter moment arm of 1.35 cm
350 rev/min = 350(2π) / 60 = 36.652 rad/s
36.652 rad/s / 0.294 s = 124.66... <u>125 rad/s²</u>
a = αR = 125(0.1042) = 12.990... <u>13 m/s²</u>
a = αR = 125(0.0135) = 1.68299... <u>1.7 m/s²</u>
I am GUESSING that we are supposed to model these mowers as a uniform disk
τ = Iα
FR = (½mr²)α
F = mr²α/2R
Urippe's pull = (3.56)(0.2041²)(124.66) / (2(0.0135)) = 702.008... <u>702 N</u>
Usi's pull = (3.56)(0.2041²)(124.66) / (2(0.1042)) = 90.9511...<u>91.0 N</u>
L = Iω = (½(3.56)(0.2041²))36.652 = 2.71771...<u>2.72 kg•m²/s down</u>
using the right hand rule
may vary depending on the organization.
Answer:
G M m / R^2 = m v^2 / R gravitational force = centripetal force
G M = v^2 R = constant
As v increases R will must decrease
Take the moon as an example
S = 2 pi R where R is about 240,000 miles for one orbit
S / 1 day = 54,000 miles/day for a 28 day circuit
S / 1 hr = 54000 / 24 = 2200 mph which is much less than a satellite in orbit
Answer:
Pluto
because the the gravitation strength is small, so it will be accelerating slow
Answer:
s = 3 m
Explanation:
Let t be the time the accelerating car starts.
Let's assume the vehicles are point masses so that "passing" takes no time.
the position of the constant velocity and accelerating vehicles are
s = vt = 40(t + 2) cm
s = ½at² = ½(20)(t)² cm
they pass when their distance is the same
½(20)(t)² = 40(t + 2)
10t² = 40t + 80
0 = 10t² - 40t - 80
0 = t² - 4t - 8
t = (4±√(4² - 4(1)(-8))) / 2(1)
t = (4± 6.928) / 2 ignore the negative time as it has not occurred yet.
t = 5.464 s
s = 40(5.464 + 2) = 298.564 cm
300 cm when rounded to the single significant digit of the question numerals.