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
Answer:
D) 25 m/s
Explanation:
In order to solve this problem we must use the following kinematics equation.

where:
Vf = final speed [m/s]
Vi = initial speed = 0
a = acceleration = 5[m/s^2]
t = time = 5[s]
After 5 seconds the acceleration is equal to 5 [m/s^2]
Now replacing the values in the equation:
Vf = 0 + (5*5)
Vf = 25[m/s]
(i) |α| = 235.6rad.s / 0.502s = 469 rad/s²
(ii) tang a = α*r = 469rad/s² * 0.12m / 2*11 = 2.56 m/s²
Answer:

Explanation:
Acceleration is defined as the change in velocity divided by the time it took to produce such change. The formula then reads:

Where Vf is the final velocity of the object, (in our case 80 m/s)
Vi is the initial velocity of the object (in our case 0 m/s because the object was at rest)
and t is the time it took to change from the Vi to the Vf (in our case 0.05 seconds.
Therefore we have:

Notice that the units of acceleration in the SI system are
(meters divided square seconds)