False because currents do not flow easily through insulators. If it only said conductors, then it would be true.
efficiency = (useful energy transferred ÷ energy supplied) × 100
It's easy to use this formula, but we have to know both the useful energy and the energy supplied. The drawing doesn't tell us the useful energy, so we have to find a clever way to figure it out. I see two ways to do it:
<u>Way #1:</u>
We all know about the law of conservation of energy. So we know that the total energy coming out must be 250J, because that's how much energy is going in. The wasted energy is 75J, so the rest of the 250J must be the useful energy . . . (250J - 75J) = 175J useful energy.
(useful energy) / (energy supplied) = (175J) / (250J) = <em>70% efficiency</em>
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<u>Way #2: </u>
How much of the energy is wasted ? . . . 75J wasted
What percentage of the Input is that 75J ? . . . 75/250 = 30% wasted
30% of the input energy is wasted. That leaves the other <em>70%</em> to be useful energy.
Answer:
2081.65 m
Explanation:
We'll begin by calculating the time taken for the load to get to the target. This can be obtained as follow:
Height (h) = 3000 m
Acceleration due to gravity (g) = 10 m/s²
Time (t) =?
h = ½gt²
3000 = ½ × 10 × t²
3000 = 5 × t²
Divide both side by 5
t² = 3000 / 5
t² = 600
Take the square root of both side
t = √600
t = 24.49 s
Finally, we shall determine the distance from the target at which the load should be released. This can be obtained as follow:
Horizontal velocity (u) = 85 m/s
Time (t) = 24.49 s
Horizontal distance (s) =?
s = ut
s = 85 × 24.49
s = 2081.65 m
Thus, the load should be released from 2081.65 m.
Answer:
Explanation:
The average pressure at mean sea-level (MSL) in the International Standard Atmosphere (ISA) is 1013.25 hPa, or 1 atmosphere (atm), or 29.92 inches of mercury. Pressure (p), mass (m), and the acceleration due to gravity (g), are related by P = F/A = (m*g)/A, where A is surface area.