Urippe Went to start his lawnmower and was having trouble. At the same moment, Uwi was starting her identical lawnmower in the y
ard next door without any trouble. There was one difference between the two mowers. Urippe's has the starter cord wrapped around a hub 1.35 cm from the center of the flywheel. Uwi's starter cord is wrapped around a hub 10.42 cm from the center of the flywheel. Explain why Urippe is having a harder time pulling his starter cord
In order to start you must reach a minimum of 350 rpm by pulling the starter cord. Determine the angular acceleration in rad/s required to start the lawnmower if it takes 0.294 seconds to pull the starter cord.
What is the acceleration of the pull cord when Uwi starts her lawn mower?
What is the acceleration of the pull cord when Urippe starts his lawnmower?
The rest is on the paper. Thanks so much
In order to start you must reach a minimum of 350 rpm by pulling the starter cord. Determine the angular acceleration in rad/s required to start the lawnmower if it takes 0.294 seconds to pull the starter cord.
What is the acceleration of the pull cord when Uwi starts her lawn mower?
What is the acceleration of the pull cord when Urippe starts his lawnmower?
The rest is on the paper. Thanks so much
1 answer:
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Answer:
Induced emf in the coil, E = 0.157 volts
Explanation:
It is given that,
Number of turns, N = 100
Diameter of the coil, d = 3 cm = 0.03 m
Radius of the coil, r = 0.015 m
A uniform magnetic field increases from 0.5 T to 2.5 T in 0.9 s.
Due to this change in magnetic field, an emf is induced in the coil which is given by :
E = -0.157 volts
Minus sign shows the direction of induced emf in the coil. Hence, the induced emf in the coil is 0.157 volts.
Answer:
a) the maximum shear stress τ the bar is 16T
/πd³
b) the angle of twist between the ends of the bar is 16tL² / πGd⁴
Explanation:
Given the data in the question, as illustrated in the image below;
d is the diameter of the prismatic bar of length AB
t is the intensity of distributed torque
(a) Determine the maximum shear stress tmax in the bar
Maximum Applied torque T_max = tL
we know that;
shear stress τ = 16T/πd³
where d is the diameter
so
τ = 16T
/πd³
Therefore, the maximum shear stress τ the bar is 16T
/πd³
(b) Determine the angle of twist between the ends of the bar.
let theta () be the angle of twist
polar moment of inertia = πd⁴/32
now from the second image;
lets length dx which is at distance of "x" from "B"
Torque distance x
T(x) = tx
Elemental angle twist = d = T(x)dx / G
so
d = tx.dx / G(πd⁴/32)
d = 32tx.dx / πGd⁴
so total angle of twist will be;
=
=
32tx.dx / πGd⁴
= 32t / πGd⁴
= 32t / πGd⁴ [ L²/2]
= 16tL² / πGd⁴
Therefore, the angle of twist between the ends of the bar is 16tL² / πGd⁴
