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FrozenT [24]
2 years ago
15

A bathtub contains 65 gallons of water and the total weight of the tub and water is approximately 931.925 pounds. You pull the p

lug and the water begins to drain. Let v v represent the number of gallons of water that has drained from the tub since the plug was pulled. Note that water weights 8.345 pounds per gallon. Write an expression in terms of v v that represents the weight of the water that has drained from the tub (in pounds).
Physics
1 answer:
levacccp [35]2 years ago
8 0

Answer:

Q = 8,345 * v

Explanation:

So, we are looking for an expression of the amount of water that has been drained from the tub. The expression is in terms of v that represent the number of gallons of water drained since the plug was pulled. Since we are interested in the pounds of water that has been drained from the tub we need to take into account that for every gallon of water drained, 8.345 pounds have left the tub. Therefore, the expression for the weight of water Q that has been drained from the tub in terms of v is simply :

Q = 8,345 * v

Where v is the amount of gallons that has been drained from the tub.

Have a nice day. let me know if I can help with anything else

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Suppose you are designing an amplifier and loudspeaker system to use at a rock concert. You want to make it as loud as possible.
OverLord2011 [107]

Answer is given below

Explanation:

  • Audio power amplifiers are found in all types of sound systems, including sound reinforcement, public address and home audio systems, as well as musical instrument amplifiers such as guitar amplifiers.
  • This is the last electronic step in the general audio playback series before sending the signal to the loudspeaker.  So when we want maximum volume or loud sound, we have to get it with maximum output and high input and low output impedance
4 0
2 years ago
An ideal gas is contained in a vessel at 300 K. The temperature of the gas is then increased to 900 K. (i) By what factor does t
Dahasolnce [82]

The question is missing some parts. Here is the complete question.

An ideal gas is contained in a vessel at 300K. The temperature of the gas is then increased to 900K.

(i) By what factor does the average kinetic energy of the molecules change, (a) a factor of 9, (b) a factor of 3, (c) a factor of \sqrt{3}, (d) a factor of 1, or (e) a factor of \frac{1}{3}?

Using the same choices in part (i), by what factor does each of the following change: (ii) the rms molecular speed of the molecules, (iii) the average momentum change that one molecule undergoes in a colision with one particular wall, (iv) the rate of collisions of molecules with walls, and (v) the pressure of the gas.

Answer: (i) (b) a factor of 3;

              (ii) (c) a factor of \sqrt{3};

              (iii) (c) a factor of \sqrt{3};

             (iv) (c) a factor of \sqrt{3};

              (v) (e) a factor of 3;

Explanation: (i) Kinetic energy for ideal gas is calculated as:

KE=\frac{3}{2}nRT

where

n is mols

R is constant of gas

T is temperature in Kelvin

As you can see, kinetic energy and temperature are directly proportional: when tem perature increases, so does energy.

So, as temperature of an ideal gas increased 3 times, kinetic energy will increase 3 times.

For temperature and energy, the factor of change is 3.

(ii) Rms is root mean square velocity and is defined as

V_{rms}=\sqrt{\frac{3k_{B}T}{m} }

Calculating velocity for each temperature:

For 300K:

V_{rms1}=\sqrt{\frac{3k_{B}300}{m} }

V_{rms1}=30\sqrt{\frac{k_{B}}{m} }

For 900K:

V_{rms2}=\sqrt{\frac{3k_{B}900}{m} }

V_{rms2}=30\sqrt{3}\sqrt{\frac{k_{B}}{m} }

Comparing both veolcities:

\frac{V_{rms2}}{V_{rms1}}= (30\sqrt{3}\sqrt{\frac{k_{B}}{m} }) .\frac{1}{30} \sqrt{\frac{m}{k_{B}} }

\frac{V_{rms2}}{V_{rms1}}=\sqrt{3}

For rms, factor of change is \sqrt{3}

(iii) Average momentum change of molecule depends upon velocity:

q = m.v

Since velocity has a factor of \sqrt{3} and velocity and momentum are proportional, average momentum change increase by a factor of

(iv) Collisions increase with increase in velocity, which increases with increase of temperature. So, rate of collisions also increase by a factor of \sqrt{3}.

(v) According to the Pressure-Temperature Law, also known as Gay-Lussac's Law, when the volume of an ideal gas is kept constant, pressure and temperature are directly proportional. So, when temperature increases by a factor of 3, Pressure also increases by a factor of 3.

4 0
1 year ago
A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra
skelet666 [1.2K]

Answer:

So the acceleration of the child will be 8.05m/sec^2

Explanation:

We have given angular speed of the child \omega =1.25rad/sec

Radius r = 4.65 m

Angular acceleration \alpha =0.745rad/sec^2

We know that linear velocity is given by v=\omega r=1.25\times 4.65=5.815m/sec

We know that radial acceleration is given by a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2

Tangential acceleration is given by

a_t=\alpha r=0.745\times 4.65=3.464m/sec^

So total acceleration will be a=\sqrt{7.2718^2+3.464^2}=8.05m/sec^2

7 0
2 years ago
Write a hypothesis about how the height of the cylinder affects the temperature of the water. Use the "if . . . then . . . becau
IgorLugansk [536]

If the mass of the cylinder increases, the temperature of the water increases, because a greater mass means the cylinder has more potential energy that can be converted to thermal energy, increasing the temperature of the water.


4 0
2 years ago
One end of a string is fixed. An object attached to the other end moves on a horizontal plane with uniform circular motion of ra
sveticcg [70]

Answer:

If both the radius and frequency are doubled, then the tension is increased 8 times.

Explanation:

The radial acceleration (a_{r}), measured in meters per square second, experimented by the moving end of the string is determined by the following kinematic formula:

a_{r} = 4\pi^{2}\cdot f^{2}\cdot R (1)

Where:

f - Frequency, measured in hertz.

R - Radius of rotation, measured in meters.

From Second Newton's Law, the centripetal acceleration is due to the existence of tension (T), measured in newtons, through the string, then we derive the following model:

\Sigma F = T = m\cdot a_{r} (2)

Where m is the mass of the object, measured in kilograms.

By applying (1) in (2), we have the following formula:

T = 4\pi^{2}\cdot m\cdot f^{2}\cdot R (3)

From where we conclude that tension is directly proportional to the radius and the square of frequency. Then, if radius and frequency are doubled, then the ratio between tensions is:

\frac{T_{2}}{T_{1}} = \left(\frac{f_{2}}{f_{1}} \right)^{2}\cdot \left(\frac{R_{2}}{R_{1}} \right) (4)

\frac{T_{2}}{T_{1}} = 4\cdot 2

\frac{T_{2}}{T_{1}} = 8

If both the radius and frequency are doubled, then the tension is increased 8 times.

5 0
1 year ago
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