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3 years ago

If your runnin at full speed for an average human, and lept over a 10 foot brook would you make it?

2 answers:

7 0

Unless you are a mutant....I don't think you would make it. I'm 5'8" and most people are 6'7" or less. It's bsically impossible to do. :)

LiRa [457]3 years ago

6 0

This seems a bit too trivial for this site but I'll bite anyways. An average speed for a human would be around 8 mph, so running would probably be around 12 mph. The longest long jump ever is 26 feet, but the average is 7' 3" to 7' 6.5", which is quite far away from the 10 ft brook. Therefore nope!

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3 years ago

A flat roof is very susceptible to wind damage during a thunderstorm and/or tornado. If a flat roof has an area of 500 m2 and wi

Evgesh-ka [11]

Answer: The magnitude of the force exerted on the roof is 490522.5 N.

Explanation:

The given data is as follows.

Below the roof, $v_{1}$ = 0 m/s

At top of the roof, $v_{2}$ = 39 m/s

We assume that $P_{1}$ is the pressure at lower surface of the roof and $P_{2}$ be the pressure at upper surface of the roof.

Now, according to Bernoulli's theorem,

$P_{1} + 0.5 \times \rho \times v^{2}_{1} = P_{2} \times 0.5 \rho \times v^{2}_{2}$

$P_{1} - P_{2} = 0.5 \times \rho \times (v^{2}_{2} - v^{2}_{1})$

= $0.5 \times 1.29 \times [(39)^{2} - (0)^{2}]$

= $0.645 \times 1521$

= 981.045 Pa

Formula for net upward force of air exerted on the roof is as follows.

F = $(P_{1} - P_{2})A$

= $981.045 \times 500$

= 490522.5 N

Therefore, we can conclude that the magnitude of the force exerted on the roof is 490522.5 N.

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3 years ago

How would we be able to detect a large asteroid if it were heading straight for earth?

Stolb23 [73]

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4 years ago

A mass moves back and forth in simple harmonic motion with amplitude A and period T.

Sever21 [200]

a. 0.5 T

- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position

- The period T is the time the system takes to complete one oscillation

During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.

So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion

$1 T : 4 A = t : 2 A$