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Reil [10]
4 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]4 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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You and a highway patrolman are driving at constant speeds in opposite directions on a straight highway. The patrolman is drivin
kompoz [17]

Answer:  75 mph

Explanation:

The Relative Speed for a mobile is equal to the diference between the object and the observer:

Relative Speed (Rs) = Object's Velocity  - Observer's Velocity

Thinking on those terms, we would need to have a universal observer to do any understandable measurement on daily basics. This is why we all use earth as a Static Observer for every measurement we do everyday.

Using Earth as an observer, the Velocity for the Patrolman is:

Patrolman Velocity (Vp) = 60 mph

Because the radar gun does measure the Relative Speed for the object, which is 135 mph, we need to work with the equation to find the Velocity using Earth as a reference.

Object's Relative Velocity = Object's Velocity - Patrolman's Velocity

Object's Velocity = Object's Relative Velocity + Patrolman's Velocity

We need to keep in mind, the Patrolman is going on the opposite direction. Because of this the sign for his velocity should be negative.

Object's Velocity = 135 mph + ( -60 mph)

Object's Velocity = 135 mph - 60 mph

Object's Velocity = 75 mph

3 0
3 years ago
Define the fundamental difference between kinematics and dynamics. .
Tcecarenko [31]

In kynematics you describe the motion of particles using vectors and their change in time. You define a position vector r for a particle, and then define velocity v and acceleration a as

v=\frac{dr}{dt} \\

a=\frac{dv}{dt}

In dynamics Newton's laws predict the acceleration for a given force. Knowing the acceleration, and the kynematical relations defines above, you can solve for the position as a function of time: r(t)

5 0
3 years ago
The speed of sound in air is around 330 m/s. If a bat emits a single high-pitched ‘click’ of sound in a cave that is 25m wide, c
Bingel [31]

Answer:

0.15 s

Explanation:

From the question given above, the following data were obtained:

Speed of sound (v) = 330 m/s

Distance (x) = 25 m

Time (t) =?

The time taken for the echo of the sound to the bat can be obtained as follow:

v = 2x / t

330 = 2 × 25 / t

330 = 50 / t

Cross multiply

330 × t = 50

Divide both side by 330

t = 50 / 330

t = 0.15 s

Thus, it will take 0.15 s for the echo of the sound to the bat

4 0
3 years ago
Interference occurs with not only light waves but also all frequencies of electromagnetic waves and all other types of waves, su
telo118 [61]

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

4 0
4 years ago
1. You have a cat who has a mass of 10 kg and is
disa [49]

Answer:

1) F = 100N

2) a = 2 m/s²

3) m = 25 kg

Explanation:

1) F = ma ( F = ?, m = 10 kg, a = 10 m/s² )

  F = 10×10

  F = 100 N

2) F = ma ( F = 20N, m = 10 kg, a = ? )

   20 = 10×a

   10a = 20

   a = 20/10

   a = 2 m/s²

3)F = ma ( F = 100N, m = ?, a = 4 m/s² )

  100 = m×4

  4m = 100

  m = 100/4

  m = 25 kg

Hope that helps! Good luck!

 

7 0
3 years ago
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