Answer:
The correct answer is - 63.61 miles/hour.
Explanation:
Meter and miles both are the measuring units of distance and speed units are meter per second and miles per hour. The numerical relation between these two units are as follows:
1 meter per second = 2.24 miles per hour
The car travels 28.4 m/s so in miles per hour it would be:
So, 28.4 meters per second = 28.4 * 2.24 miles per hour
28.4 meters per second = 63.61 miles per hour.
Thus, the correct answer is - 63.61 miles per hour.
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:

a) For the original capacitor, we can find directly U as follows:
U = 4.94*10⁹ J
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 = 

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 = 1.07*10¹⁰ J
Answer:
$ 0.17
Explanation:
From the question given above, the following data were obtained:
Power (P) = 90 Watts
Time (t) = 1 day
Cost per KWh = $ 0.08
Cost of operation =?
Next, we shall convert 90 W to KW. This can be obtained as follow:
1000 W = 1 KW
Therefore,
90 W = 90 W × 1 KW / 1000 W
90 W = 0.09 KW
Next, we shall express 1 day in hour. This is shown below:
1 day = 24 hours
Next, we shall determine the energy consumption. This can be obtained as follow:
Power (P) = 0.09 KW
Time (t) = 24 h
Energy (E) =?
E = Pt
E = 0.09 × 24
E = 2.16 KWh
Finally, we shall determine the cost of operation. This can be obtained as follow:
Energy (E) = 2.16 KWh
Cost per KWh = $ 0.08
Cost of operation =?
Cost of operation = Energy × Cost per KWh
Cost of operation = 2.16 × 0.08
Cost of operation = $ 0.17
Thus, the cost of operating the light bulb for one day is $ 0.17