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Serhud [2]
3 years ago
10

What sometimes occurs when reclaimed water is used in agriculture?

Physics
1 answer:
ioda3 years ago
6 0
I remember this from IPC! If farmers used reclaimed water, there is a chance of chemical contamination in the water can damage the plants, if not kill them entirely! :D
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Satellite A revolves around Earth 5 times a day. What is the radius of its orbit, measured from Earth's center? Assume that Eart
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Answer: 1.44. You can set centripetal force equal to gravitational force caused by the earth because the grav. force is what causes the satellite to orbit in a circle

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4 years ago
Water is pouring into a conical tank at the rate of 8 cubic feet per minute. If the height of the tank is 10 feet and the radius
iren [92.7K]

Answer:

\frac{25}{8\pi}ft/min

Explanation:

We are given that

\frac{dV}{dt}=8 ft^3/min

Height of tank=h=10 ft

Radius of tank, r=4 feet

We have to find the dh/dt when the water is 4 feet deep.

\frac{r}{h}=\frac{4}{10}=\frac{2}{5}

r=\frac{2}{5}h

Volume of cone , V=\frac{1}{3}\pi r^2 h

Substitute the values

Volume of cone , V=\frac{1}{3}\pi(\frac{2}{5}h)^2h=\frac{4}{75}\pi h^3

Differentiate w.r.t t

\frac{dV}{dt}=\frac{12}{75}\pi\times h^2\times \frac{dh}{dt}

Substitute the values

8=\frac{12}{75}\times\pi (4)^2\times \frac{dh}{dt}

\frac{dh}{dt}=\frac{8\times 75}{12\times 16\pi}=\frac{25}{8\pi}ft/min

3 0
4 years ago
John doe gets on the highway in his 1967 Shelby 427 Cobra starting from the dead stop at the bottom of the on ramp of it can be
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It will take john 4.477 seconds

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3 years ago
A source charge of 5.0 µC generates an electric field of 3.93 × 105 at the location of a test charge. How far is the test charge
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I definitely guessed and got the right answer so :))

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4 years ago
Read 2 more answers
A projectile's horizontal range on level ground is r=v20sin2θ/g. at what launch angle or angles will the projectile land at half
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As given the formula of range is

R = \frac{v^2sin2\theta}{g}

now for the maximum range

\theta = 45

R_{max} = \frac{v^2}{g}

now for half of this maximum range we will have

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\frac{v^2}{2g} = \frac{v^2sin2\theta}{g}

now from above we have

sin2\theta = \frac{1}{2}

so two possible values for above is given as

\theta = 15 degree, 75 degree

so above two angle we will have half of maximum range

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4 years ago
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