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denis23 [38]
3 years ago
7

A cat chases a mouse across a 1.2 m high table. The mouse steps out of the way, and the cat slides off the table and strikes the

floor 1.3 m from the edge of the table.
Physics
1 answer:
Drupady [299]3 years ago
7 0
The cat has two directions of motions:
The horizontal motion = Dx = 2.2 m
The vertical motion = Dy = -1.3 m (negative sign indicates that the cat is falling)
a = 9.8 m/sec^2
Vy = zero (since you are not moving up)

From the laws of motion:
<span>Dy = Vyt + 0.5ayt^2 
</span>-1.3 = 0(t) + 0.5(-9.8)t^2
<span>t = 0.52s
</span>
Then, again using the laws of motion (but for the horizontal direction this time)
Dx = Vxt 
<span>2.2 = Vx0.52 </span>
<span>Vx = 2.2/0.52 </span>
<span>= 4.23 m/s 
</span>
<span>Therefore the cat's speed when it slid off the table is 4.23 m/s horizontally.</span>



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Question 9 In an RC series circuit, ε = 12.0 V, R = 1.07 MΩ, and C = 2.66 µF. (a) Calculate the time constant. (b) Find the maxi
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Answer:

a.) τ = 2.85 s b.) Q = 3.19 * 10^-5 C c.) t = 1.691 s

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a.) The time constant for RC circuit, τ = RC. Substituting our known values we get:

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τ = (1.07 * 10 ^ 6)Ω * (2.66 * 10 ^ -6) F = 2.8462 s ≈ 2.85 s

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b.) The relationship between capacitance, potential, charge is given:

Q = CV[1-e^{-t/RC} ]

The capacitor is fully charge when t approaches infinity, therefore:

Q =  \lim_{t \to \infty} a_n CV[1-e^{-t/RC} ]

When t approaches infinity, the term e becomes very small (e^-∞ = 0), therefore we can simplify the equation and plug in our values

Q = (2.66*10^{-6}) F * (12.0)V *[1 - 0] = 3.192 * 10^{-5}

Q = 3.19 * 10^-5 C

c.) Using the same equation as before, we can substitute Q in and solve for Q:

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t = 1.691 s

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