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Citrus2011 [14]

3 years ago

15

Suppose that water is pouring into a swimming pool in the shape of a right circular cylinder at a constant rate of 5 cubic feet

per minute. If the pool has radius 6 feet and height 9 feet, what is the rate of change of the height of the water in the pool when the depth of the water in the pool is 5 feet

1 answer:

Marina86 [1]3 years ago

7 0

Answer:

$\frac{dH}{dt}=0.044\;\;feet\;per\;min$

Explanation:

Given,

Radius R = 6 feet

Height H = 9 feet

Depth of the water = 5 feet

Volume of right circular cylinder,

$V=\pi R^2 H\\$

Differentiate with respect to the H

$\frac{dV}{dH}=\pi R^2\\\frac{dV}{dH}=36\pi$

Now,

$\frac{dV}{dH}.\frac{dt}{dt} =36\pi\\\frac{dV}{dt}.\frac{dt}{dH}=36\pi\\ 5.\frac{dt}{dH}=36\pi\\\frac{dH}{dt}=\frac{5}{36\pi}\\\frac{dH}{dt}=0.044\;\;feet\;per\;min$

Hence, the rate of chage of height of water is 0.044 feet per min.

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A parallel-plate vacuum capacitor is connected to a battery and charged until the stored electric energy is . The battery is rem

Viktor [21]

Answer:

A

The energy dissipated in the resistor ${U_k} = \frac{U}{k}$

B

The energy dissipated in the resistor ${U_k} = kU$

Explanation:

In order to gain a good understanding of the solution above it is necessary to understand that the concept required to solve the question is energy stored in the parallel plate capacitor.

Initially, take the first case. In that, according to the formula for energy stored in parallel plate capacitor with the dielectric inserted between the two plates, find the energy stored. Then, find the energy stored in the parallel plate capacitor when no dielectric is present. Then, write the equation of energy stored in the capacitor with the dielectric present in the form of the energy stored in the capacitor without the dielectric present. The equation must not be in the form of voltage as battery is removed in this case.

For part B, use the equation of the energy dissipated in the resistor. Write it in the form of the equation for energy stored in the parallel plate capacitor without dielectric in it. The equation must be in the form of voltage as battery is kept connected. Looking at the fundamentals

The energy stored in the parallel plate capacitor with the dielectric is given by,

$U _k = \frac{1}{2} \frac{q ^2}{kC}$

Here, the energy stored in the capacitor will be equal to the energy dissipated in the resistor. In this equation, Uk is the energy dissipated in the resistor, q is charge, k is the dielectric constant, and C is the capacitance.

Now, the equation of the energy stored in the parallel plate capacitor without dielectric is,

$U= \frac{1}{2} \frac{q ^2}{C}$

In this equation, U is the energy stored in the parallel plate capacitor without dielectric, q is charge, and C is the capacitance.

For part B, the battery is still connected. Thus, the equation $q = CV$ is used to modify the above equation.

Thus, the energy stored in the parallel plate capacitor with the dielectric is given by,

$U_ k = \frac{1}{2} \frac{k ^{2} C^ 2 V ^2}{kC} \\\\= \frac{1}{2} kCV ^2$

In this equation, $Uk$ is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

The equation of the energy stored in the parallel plate capacitor without dielectric is,

$U= \frac{1}{2} \frac{C^ 2 V ^2}{C} \\\\= \frac{1}{2} CV ^2$

In this equation, U is the energy dissipated in the resistor, V is voltage, k is the dielectric constant, and C is the capacitance.

(A)

The equation for energy dissipated in the resistor is,

$U _k = \frac{1}{2} \frac{q ^2}{kC}$

Substitute $U = \frac{1}{2}\frac{{{q^2}}}{C}$ in the equation of ${U_k}$

$U _k = \frac{1}{2} (\frac{1}{k} )\frac{q ^2}{C} \\\\= (\frac{1}{k} ) \frac{q^2}{C}\\\\ U_{k} = \frac{U}{k}$

Note :

If the resistance relates to the capacitor, the energy stored in the capacitor is dissipated through the resistance. Thus, by substituting the equation of U, the expression is found out.

(B)

The equation for energy dissipated in the resistor is

$U_{k} = \frac{1}{2}kCV^2$

Here, V is voltage in the circuit.

Substitute $U =\frac{1}{2} CV^2$ in the equation of ${U_k}$

So,

$U_{k} = \frac{1}{2} kCV^2\\$

$= k(\frac{1}{2} CV^2)$

$U_{k} = kU$

4 0

3 years ago

when the sun and moon are in a line with the earth, the: group of answer choices arrival of high tide will be delayed. gravitati

Alecsey [184]

When the sun, moon, and Earth are lined up (during a new or full moon), the solar tide adds to the lunar tide to produce extremely high tides and very low tides, both of which are known as spring tides.

Basically describes a situation in astronomy where three celestial bodies align in a straight line as part of a gravitational system. The phrase is frequently used to describe how the Sun, Moon, and Earth are in a straight line.

The moon is responsible for causing high and low tides. The tidal force is produced by the moon's gravitational pull. Earth and its water protrude outward on both the side that is closest to and farthest from the moon as a result of the tidal force. These watery peaks are high tide

To know more about high tides

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1 year ago

What is the acceleration of a proton moving with a speed of 6.5 m/s at right angles to a magnetic field of 1.5 T?

Brilliant_brown [7]

Answer:

The acceleration of the proton is 9.353 x 10⁸ m/s²

Explanation:

Given;

speed of the proton, u = 6.5 m/s

magnetic field strength, B = 1.5 T

The force of the proton is given by;

F = ma = qvB(sin90°)

ma = qvB

where;

m is mass of the proton, = 1.67 x 10⁻²⁷ kg

charge of the proton, q = 1.602 x 10⁻¹⁹ C

The acceleration of the proton is given by;

$a = \frac{qvB}{m}\\\\a = \frac{(1.602*10^{-19})(6.5)(1.5)}{1.67*10^{-27}}\\\\a = 9.353*10^8 \ m/s^2$

Therefore, the acceleration of the proton is 9.353 x 10⁸ m/s²

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3 years ago

what is the main difference between a substance going through a physical change and one going through a chemical ?

Ainat [17]

Answer: Physical changes only change the appearance of a substance, not its chemical composition. Chemical changes cause a substance to change into an entirely substance with a new chemical formula. Chemical changes are also known as chemical reactions.

Explanation:

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Answer:

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Explanation:

i got it right

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