5) You pull a 10.0 kg wagon along a flat road. You exert a force of 80.0 N at an angle of 30.0 degrees above the horizontal whil
1 answer:
Consult the attached free body diagram. The only forces doing work on the wagon are the frictional force opposing the wagon's motion and the horizontal component of the applied force.
By Newton's second law, the net vertical force is
• ∑ F [v] = n + (80.0 N) sin(30.0°) - mg = 0
where a is the acceleration of the wagon.
Solve for n (the magnitude of the normal force) :
n = (10.0 kg) g - (80.0 N) sin(30.0°) = 58.0 N
Then
f = 0.500 (58.0 N) = 29.0 N
Meanwhile, the horizontal component of the applied force has magnitude
(80.0 N) cos(30.0°) ≈ 69.3 N
Now calculate the work done by either force.
• friction: -(29.0 N) (10.0 m) = -290. J
• pull: (69.3 N) (10.0 m) = 693 J
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Answer:
The positive displacement from the midpoint of its motion at the speed equal one half of its maximum speed is 3.56 cm.
Explanation:
Maximum speed is :
v (max) = Aω
Speed v at any displacement y is given by
=
(
-
) ........................................................ i
And,
v = v (max)
or, 2 × v = Aω .................................................... ii
Eliminating ω from equations i and ii,
=
(
-
)
or, = (
)
=(
)
or, y = 3.56 cm.
Answer:
Temperature, T = 3.62 kelvin
Explanation:
It is given that,
Total number of gas molecules,
Her body is converting chemical energy into thermal energy at a rate of 125 W, P = 125 W
Time taken, t = 6 min = 360 s
Energy of a gas molecules is given by :
, k is Boltzmann constant
T = 3.62 K
So, the temperature increases by 3.62 kelvin. Hence, this is the required solution.