A group of students must conduct an experiment to determine how the location of an applied force on a classroom door affects the
1 answer:
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Answer:
Q=1005 J
t= 0.67 sec
Explanation:
Lets take condition of room is 1 atm and 25°C.
Heat capacity ,c = 21 J /K.mol
If we assume that air is ideal gas that
P V = n R T
V= 107250 L
At STP number of moles given as
V=22.4 L at S.T.P.
n=4787.94 moles
n= 4.784 Kmoles
So heat required to raise 10°C temperature
Q = n x c x ΔT
Q = 4.78794 x 21 x 10
Q=1004.64 J
Time t
t= Q/P
P= 1.5 KW
t = 1.004.64 /1.5
t= 0.66 sec
The hoop is attached.
Consider that the friction force is given by:
F = μ·N
= μ·m·g·cosθ
We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·<span>g·sinθ
Therefore:
</span>-(1/2)·m·g·sinθ + m·g·sinθ = <span>μ·m·g·cosθ
</span>(1/2)·m·g·sinθ = <span>μ·m·g·cosθ
</span>μ = (1/2)·m·g·sinθ / <span>m·g·cosθ
= </span>(1/2)·tanθ
Now, solve for θ:
θ = tan⁻¹(2·μ)
= tan⁻¹(2·0.9)
= 61°
Therefore, the maximum angle <span>you could ride down without worrying about skidding is 61°.</span>
Consider that the friction force is given by:
F = μ·N
= μ·m·g·cosθ
We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·<span>g·sinθ
Therefore:
</span>-(1/2)·m·g·sinθ + m·g·sinθ = <span>μ·m·g·cosθ
</span>(1/2)·m·g·sinθ = <span>μ·m·g·cosθ
</span>μ = (1/2)·m·g·sinθ / <span>m·g·cosθ
= </span>(1/2)·tanθ
Now, solve for θ:
θ = tan⁻¹(2·μ)
= tan⁻¹(2·0.9)
= 61°
Therefore, the maximum angle <span>you could ride down without worrying about skidding is 61°.</span>