Physics, 10.08.2021 14:00 rahmao2773

coin 1 is thrown upward from the top of 100m tower with a speed of 15m/s. coin 2 is dropped from the top of the tower 2.0second later. assume g is 10m/s. how far below the top of the tower des coin 1 pass coin 2

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Answer from: supercutedonuf3753

The height below the tower at which coin 1 pass coin 2 is 89.04 m.

The given parameters:

height of the tower, h = 100 m

initial velocity of coin 1, v = 15 m/s

time spent in air by coin 1 before coin 2 was dropped = 2s

To find:

the height below the tower at which coin 1 passes coin 2

Find the maximum height attained by coin 1 before falling to the ground:

$v^2 = u^2 - 2gh\\\\where;\\\\v \ is \ the \ final \ velocity \ of \ coin \ 1 \ at \ maximum \ height, v \ = 0\\\\0 = (15^2) - 2(10)h\\\\20h = 225\\\\h = \frac{225}{20} \\\\h = 11.25 \ m$

Find the time taken for coin 1 to fall to the ground:

Total height of coin 1 above the ground, H = 11.25 m + 100 m = 111.25 m

$t = \sqrt{\frac{2H}{g} } \\\\t = \sqrt{\frac{2\times 111.25}{10} } \\\\t = 4.72 \ s$

But the time taken for the coin 1 to reach 11.25 m above the tower:

$t_1 = \sqrt{\frac{2h}{g} } \\\\t_1 = \sqrt{\frac{2\times 11.25}{10} } \\\\t_1 = 1.5 \ s$

Total time spent by coin 1 before reaching ground with respect to coin 2:

time = (1.5 s + 4.72 s) - 2 s

time = 4.22 s

Note: the 2 s was subtracted to keep both coins at a fair starting time below the tower.

Find the total time taken for coin 2 to fall to the ground:

Height of coin 2 above the ground = 100 m

Total time taken by coin 2 before falling to the ground is calculated as:

$t_2 = \sqrt{\frac{2(100)}{10} } \\\\t_2 = 4.47s$

The time at which coin 1 will pass coin 2 is 4.22 s.

Find the height below the tower when the time is 4.22 s.

$h = \frac{1}{2} (10)(4.22)^2\\\\h = 89.04 \ m$

Thus, the height below the tower at which coin 1 pass coin 2 is 89.04 m.

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Answer from: Quest

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).

Answer from: Quest

temp rises, and expansion

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