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Reil [10]
2 years ago
10

For a machine with 35-cm -diameter wheels, what rotational frequency (in rpm) do the wheels need to pitch a 85 mph fastball?

Physics
1 answer:
Inessa05 [86]2 years ago
6 0

Answer:

The rotational frequency must be 2073.56 rpm

Explanation:

Notice that we need to obtain a rotational frequency in "rpm" (revolutions per minute), so we better start by converting all the given information into the appropriate units:

The magnitude of the velocity for the pitch is given in miles per hour, while the diameter of the machine's wheels is given in cm. Let's reduce all units of length into meters(using the metric system), and the units of time into minutes.

Conversion of the 85 mph  speed into meters per minute:

Recall that 1 mile equals 1609.34 meters, and that 1 hour equals 60 minutes, so we write:

85\,\frac{miles}{hour} = 85\,\frac{1609.34\,m}{60\,min} =2279.898\,\frac{m}{min}

which can be rounded to approximately 2280 m/min.

We also convert the 35 cm diameter into meters:

diameter = 0.35 m

Now we use the equation that relates angular velocity (w) and the radius (R) of the circular movement, with tangential velocity (v_t), in order to obtain the angular velocity of the wheel:

v_t=w*R\\w=\frac{v_t}{R}

but recall that this angular velocity is given in radians per unit of time. So first find the radius of the wheel (half its diameter). R = 0.175 m

So we have:

w=\frac{2280}{0.175}\frac{radians}{min} \\w=13028.57\,\frac{radians}{min}

And now, recalling that 2\pi radians equal one revolution, we convert the angular velocity ot revolutions per minute by dividing the "w" we found by 2\pi :

rotational frequency = \frac{13028.57}{2\pi} \frac{rev}{min} = 2073.56 \frac{rev}{min}

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Harman [31]
Given:
Ca = 3Cb                      (1)
where
Ca =  heat capacity of object A
Cb =  heat capacity f object B

Also,
Ta = 2Tb                     (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.

Let
Tf =  final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.

Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb)           (3)

Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb

Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb

Answer:
T_{f} =( \frac{1+6k}{1+3k} )T_{b}= \frac{1}{2}( \frac{1+6k}{1+3k})T_{a}
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7 0
2 years ago
Given three capacitors, c1 = 2.0 μf, c2 = 1.5 μf, and c3 = 3.0 μf, what arrangement of parallel and series connections with a 12
Lesechka [4]

Answer:

Connect C₁ to C₃ in parallel; then connect C₂ to C₁ and C₂ in series. The voltage drop across C₁ the 2.0-μF capacitor will be approximately 2.76 volts.

-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-.

Explanation:

Consider four possible cases.

<h3>Case A: 12.0 V.</h3>

-\begin{array}{c}-{\bf 2.0\;\mu\text{F}-}\\-1.5\;\mu\text{F}- \\-3.0\;\mu\text{F}-\end{array}-

In case all three capacitors are connected in parallel, the 2.0\;\mu\text{F} capacitor will be connected directed to the battery. The voltage drop will be at its maximum: 12 volts.

<h3>Case B: 5.54 V.</h3>

-3.0\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-1.5\;\mu\text{F}-\end{array}]-

In case the 2.0\;\mu\text{F} capacitor is connected in parallel with the 1.5\;\mu\text{F} capacitor, and the two capacitors in parallel is connected to the 3.0\;\mu\text{F} capacitor in series.

The effective capacitance of two capacitors in parallel is the sum of their capacitance: 2.0 + 1.5 = 3.5 μF.

The reciprocal of the effective capacitance of two capacitors in series is the sum of the reciprocals of the capacitances. In other words, for the three capacitors combined,

\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_3}+ \dfrac{1}{C_1+C_2}} = \frac{1}{\dfrac{1}{3.0}+\dfrac{1}{2.0+1.5}} = 1.62\;\mu\text{F}.

What will be the voltage across the 2.0 μF capacitor?

The charge stored in two capacitors in series is the same as the charge in each capacitor.

Q = C(\text{Effective}) \cdot V = 1.62\;\mu\text{F}\times 12\;\text{V} = 19.4\;\mu\text{C}.

Voltage is the same across two capacitors in parallel.As a result,

\displaystyle V_1 = V_2 = \frac{Q}{C_1+C_2} = \frac{19.4\;\mu\text{C}}{3.5\;\mu\text{F}} = 5.54\;\text{V}.

<h3>Case C: 2.76 V.</h3>

-1.5\;\mu\text{F}-[\begin{array}{c}-{\bf 2.0\;\mu\text{F}}-\\-3.0\;\mu\text{F}-\end{array}]-.

Similarly,

  • the effective capacitance of the two capacitors in parallel is 5.0 μF;
  • the effective capacitance of the three capacitors, combined: \displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_2}+ \dfrac{1}{C_1+C_3}} = \frac{1}{\dfrac{1}{1.5}+\dfrac{1}{2.0+3.0}} = 1.15\;\mu\text{F}.

Charge stored:

Q = C(\text{Effective}) \cdot V = 1.15\;\mu\text{F}\times 12\;\text{V} = 13.8\;\mu\text{C}.

Voltage:

\displaystyle V_1 = V_3 = \frac{Q}{C_1+C_3} = \frac{13.8\;\mu\text{C}}{5.0\;\mu\text{F}} = 2.76\;\text{V}.

<h3 /><h3>Case D: 4.00 V</h3>

-2.0\;\mu\text{F}-1.5\;\mu\text{F}-3.0\;\mu\text{F}-.

Connect all three capacitors in series.

\displaystyle C(\text{Effective}) = \frac{1}{\dfrac{1}{C_1} + \dfrac{1}{C_2}+\dfrac{1}{C_3}} =\frac{1}{\dfrac{1}{2.0} + \dfrac{1}{1.5}+\dfrac{1}{3.0}} =0.667\;\mu\text{F}.

For each of the three capacitors:

Q = C(\text{Effective})\cdot V = 0.667\;\mu\text{F} \times 12\;\text{V} = 8.00\;\mu\text{C}.

For the 2.0\;\mu\text{F} capacitor:

\displaystyle V_1=\frac{Q}{C_1} = \frac{8.00\;\mu\text{C}}{2.0\;\mu\text{F}} = 4.0\;\text{V}.

6 0
1 year ago
one horsepower is a unit of power equal to 746w. how much energy can a 150-horsepower engine transform in 10.0s?
Dafna1 [17]

1 watt = 1 joule/second

1 horsepower = 746 watts = 746 joule/second

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=  (150 x 746 x 1 x 10)  joule  =  1,119,000 joules .   
if correct plz mark brainly
8 0
1 year ago
Suppose you're on a hot air balloon ride, carrying a buzzer that emits a sound of frequency f. If you accidentally drop the buzz
Olin [163]

Answer:

The correct answer is option 'd': The frequency decreases and the intensity of the sound decreases.

Explanation:

1) <u>Effect on Frequency </u>

According to Doppler's effect of sound we have

for a source of sound moving away from the observer the relation between the observed and the original frequency is given by

f_{app}=\frac{c-v_{rec}}{c+v_{s}}\times f_{original}

where

c = speed of sound in air

v_{rec} is the velocity of observer of sound

v_{s} is the velocity of source of sound

f_{o} is the original frequency of sound

As we see the ratio is less than 1 thus the frequency of sound that the observer receives is less than that of source.

2) <u>Effect on Intensity:</u>

At a distance 'r' from source emitting a wave of Power 'P' is given by

I=\frac{P}{4\pi r^{2}}

As we see on increasing 'r' intensity of sound decreases.

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6 0
1 year ago
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