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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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Answer:

Explanation:

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This torque of weight

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2 years ago
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<h2><u>Answer:</u></h2>

The simulation kept track of the variables and automatically recorded data on object displacement, velocity, and momentum. If the trials were run on a real track with real gliders, using stopwatches and meter sticks for measurement, the data compared by the following statements:

1. (There would be variables that would be hard to control, leading to less reliable data.)

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before colliding, the momentum of block A is +15.0 kg m/s. after, block A has a momentum -12.0 kg*m/s. what is the momentum of b
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Answer:

The momentum of block B = 27 Kg m/s

Explanation:

Given,

The initial momentum of block A, MU = 15 Kg m/s

The final momentum of block A, MV = -12 Kg m/s

Consider the block B is initially at rest.

Therefore, the initial momentum of block B, mu = 0

According to the laws of conservation of linear momentum, the momentum of the body before impact is equal to the momentum of the body after impact.

                               <em> MU + mu = MV + mv</em>

                                15  +  (0) = (-12) + mv

                                         mv = 15 + 12

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Hence, the momentum of the block B after impact is, mv = 27 Kg m/s

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2 years ago
If a scale on Earth reads 650 N, what is your mass?
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If the scale reads 650N, then the mass of whoever it is standing on the scale is

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Answer: -2.5

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