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yawa3891 [41]
3 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]3 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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If the wavelength<span> is given, the energy can be determined by first using the wave equation (c = λ × ν) to </span>find<span> the frequency, then using Planck's equation to </span>calculate<span> energy. Use the equations above to answer the following questions. 1. Ultraviolet radiation has a frequency of 6.8 × 1015 1/s.</span>
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A positive charge +q1 is located to the left of a negative charge -q2. On a line passing through the two charges, there are two
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please the answer below

Explanation:

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V_1=k\frac{q_1}{r-1.0}-k\frac{q_2}{1.0}=0\\\\V_2=k\frac{q_1}{r+5.2}-k\frac{q_2}{5.2}=0\\\\

k=8.89*10^9

For both cases we have

k\frac{q_1}{r-1.0}=k\frac{q_2}{1.0}\\\\q_1(1.0)=q_2(r-1.0)\\\\r=\frac{q_1+q_2}{q_2}\\\\k\frac{q_1}{r+5.2}=k\frac{q_2}{5.2}\\\\q_1(5.2)=q_2(r+5.2)\\\\r=\frac{5.2q_1-5.2q_2}{q_2}

(b) by replacing this values of r in the expression for V we obtain

k\frac{q_1}{\frac{5.2(q_1-q_2)}{q_2}+5.2}=k\frac{q_2}{5.2}\\\\\frac{q_1}{q_2}=\frac{(q_1-q_2)}{q_2}-1.0=\frac{q_1-q_2-q_2}{q_2}=\frac{q_1-2q_2}{q_2}

hope this helps!!

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A disk rotates freely on a vertical axis with an angular velocity of 50 rpm . An identical disk rotates above it in the same dir
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Final angular velocity is 35rpm

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A person throws a pumpkin at a horizontal speed of 4.0 — off a cliff. The pumpkin travels 9.5 m horizontally
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Complete Question

A person throws a pumpkin at a horizontal speed of   4.0 m/s off a cliff. The pumpkin travels 9.5m horizontally before it hits the ground. We can ignore air resistance.What is the pumpkin's vertical displacement during the throw? What is the pumpkin's vertical velocity when it hits the ground?

Answer:

The  pumpkin's vertical displacement  is  H = 27 .67 \ m

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Explanation:

From the question we are told that

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substituting values

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Now the vertical displacement is mathematically represented as

        H  =  v_v t  +  \frac{1}{2} a_v t^2

now the vertical velocity before the throw is  zero

    So

          H =  0 +  \frac{1}{2} (9.8) * (2.375)^2

          H = 27 .67 \ m

Now the final vertical velocity  is mathematically represented as

          v_v__{f}} =  v_v + at

  substituting values

             v_v__{f}} =  0 + (9.8)* (2.375)

            v_v__{f}} = 23.298 \  m/s

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