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Novosadov [1.4K]
4 years ago
10

After training for a marathon, you would like to complete your first attempt of the 26.22 mile race in under 4.800 hours. At wha

t pace is the slowest you could run and still finish by your goal time (provide your answer in miles/s)?
Physics
1 answer:
melamori03 [73]4 years ago
8 0

Answer:

<em>1.52 x 10^-3 miles per sec</em>

<em></em>

Explanation:

distance to be covered = 26.22 miles

targeted time = 4.8 hours

4.8 hours in seconds = 4.8 x 60 x 60 = 17280 sec

Speed = distance/time

speed = 26.22/17280 = <em>1.52 x 10^-3 miles per sec</em>

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Explain what happens with the light when you see black and white opaque objects
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No, the bungee jumper is not at equilibrium.

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Each shot of the laser gun favored by Rosa the Closer, the intrepid vigilante of the lawless 22nd century, is powered by the dis
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Answer:

a) 4.94e9 J  b) 1.07e10 J

Explanation:

The electric potential energy stored in a capacitor, expressed in terms of the value of the capacitance C, and the voltage between its terminals V, is as follows:

U =\frac{1}{2}*C*V^{2}

a) For the original capacitor, we can find directly U as follows:

U =\frac{1}{2}*1.81F*(73.9e3)^{2} V2 = 4.94e9 J  

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b) Prior to find the electric potential energy of the upgraded capacitor, we need to find out the value of the capacitance C of this capacitor, which is identical to the original, except that has a different dielectric constant.

As the capacitance is proportional to the dielectric constant, we can write the following proportion:

ε₂ / ε₁ = \frac{943}{435}= \frac{C2}{C1} =\frac{Cx}{1.81F}

Cx =\frac{1.81F*943}{435} = 3.92 F

Once calculated the new value of the capacitance, as V remains the same, we can find the electric potential energy for the upgraded capacitor as follows:

U =\frac{1}{2}*3.92F*(73.9e3)^{2} V2 = 1.07e10 J

⇒ U = 1.07*10¹⁰ J

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