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yawa3891 [41]
4 years ago
7

A tank is shaped like an upside-down square pyramid, with a base with sides that are 4 meters in length and a height of 12 meter

s. If water is being pumped into the pyramid at a rate of 23m3sec, at what rate is the height of the water increasing when the water is 2 meters deep
Physics
1 answer:
xeze [42]4 years ago
4 0
Hello!

I included a link that might help you in terms of how the problem is broken down.

https://www.slader.com/discussion/question/a-tank-is-shaped-like-an-upside-down-square-pyramid-with-base-of-4m-by-4m-and-a-height-of-12m-how-fa/

However, the answer is:

The height increases at 3/2 m/sec
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Physicist S. A. Goudsmit devised a method for measuring the mass of heavy ions by timing their period of revolution in a known m
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170.36 amu

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Ion makes 5 revolutions in 1.11 ms

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B = 50 mT = 0.05 T

q = 1.6 x 10^-19 C

Let m be the mass in kg

Time period is given by

T = (2 π m) / (B q)

Frequency is the reciprocal of time period.

f = 1 / T = B q / 2πm

So,

m = B q / 2 π f

m = (0.05 x 1.6 x 10^-19) / ( 2 x 3.14 x 4504.5) = 2.828 x 10^-25 kg

as we know that

1 amu = 1.66 x 10^-27 kg

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4 0
4 years ago
A ball is thrown upward from the ground with an initial speed of 20.6 m/s; at the same instant, another ball is dropped from a b
zmey [24]

Answer:

t= 0.68 s

Explanation:

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  • Assuming that the upward direction is the positive one, g must be negative as it always points downward.
  • Taking the ground as our zero reference for the vertical axis (y axis), the equation for the ball thrown upward can be written as follows:

        y = v_{o}* t -\frac{1}{2} * g * t^{2}   (1)

  • As the second ball is dropped, its initial velocity is 0. Taking the height of the building as the initial vertical position (y₀), we can write the equation for the vertical displacement as follows:

        y = y_{o} - \frac{1}2}*g*t^{2}  (2)

  • As the left sides of  (1) and (2) are equal each other (the height  of both balls above the ground must be the same), the time must be the same also.
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        y -y_{o} = -\frac{1}{2}*g* t^{2} (3)

  • Replacing the right side of (3) in (1), we get:

       y = v_{o}*t + (y- y_{o})

       ⇒ t =\frac{y_{o} }{v_{o} } =\frac{14 m}{20.6 m/s} = 0.68 s

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A military jet cruising at an altitude of and speed of burns fuel at the rate of . How would you calculate the amount of fuel th
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Answer:

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<em>A military jet cruising at an altitude of 12.0 km and speed of 1200 km/h burns fuel at the rate of 80.1 L/min. How would you calculate the amount of fuel the jet consumes on a 750 km mission?</em>

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time = distance/speed

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1 hr = 60 minutes

0.625 h = 0.625 x 60 = 37.5 minutes

It takes 37.5 minutes to cover 750 km and the fuel burning rate is 80.1 liters per minute, hence, <u>the amount of fuel consumed by the jet would be;</u>

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