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3 years ago

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Pulleys and Belts. In a pulley system, pulley A is moving at 1500 rpm and has a diameter of 15 in. Three pulleys, B, C, and D, a

ll of different sizes, are attached to a single output axle. Speed and torque output are changed within the system by moving the drive belt between pulleys B, C, and D.. 1.A speed of 1750 rpm is required when the drive belt is connected to pulley B. What is the diameter of pulley B? . 2.A speed of 2000 rpm is required when the drive belt is connected to pulley C. What is the diameter of pulley C? . 3.A speed of 3250 rpm is required when the drive belt is connected to?

2 answers:

NeTakaya3 years ago

8 0

For this question I would use the similarities between pulley A, to each pulley of the question.

<span>1. A speed of 1750 rpm is required when the drive belt is connected to pulley B. What is the diameter of pulley B?</span>

15inches / 1500rpm = (x<span>) inches / 1750rpm</span>

(x) = 15 inches * 1750 rpm / 1500rpm

(x) = 17.5 inches

2.A speed of 2000 rpm is required when the drive belt is connected to pulley C. What is the diameter of pulley C?

15inches / 1500rpm = (x) inches / 2000rpm

(x) = 15 inches * 2000 rpm / 1500rpm

(x) = 20 inches

3. A speed of 3250 rpm is required when the drive belt is connected to?

15inches / 1500rpm = (x) inches / 3250rpm

(x) = 15 inches * 3250 rpm / 1500rpm

(x) = 32.5 inches

son4ous [18]3 years ago

3 0

Answer:

1. A speed of 1750 rpm is required when the drive belt is connected to pulley B. What is the diameter of pulley B?

15inches / 1500rpm = (x) inches / 1750rpm

(x) = 15 inches * 1750 rpm / 1500rpm

(x) = 17.5 inches

2.A speed of 2000 rpm is required when the drive belt is connected to pulley C. What is the diameter of pulley C?

15inches / 1500rpm = (x) inches / 2000rpm

(x) = 15 inches * 2000 rpm / 1500rpm

(x) = 20 inches

3. A speed of 3250 rpm is required when the drive belt is connected to?

15inches / 1500rpm = (x) inches / 3250rpm

(x) = 15 inches * 3250 rpm / 1500rpm

(x) = 32.5 inches

Explanation:

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If we have a new moon today, when we will have the next full moon?.

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2 years ago

A sinusoidal wave traveling on a string has a period of 0.20 s, a wavelength of 32 cm, and an amplitude of 3 cm. The speed of th

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Answer:

$v = 1.6 \frac{m}{s} *\frac{100cm}{1m}= 160 \frac{cm}{s}$

Explanation:

If we have a periodic wave we need to satisfy the following basic relationship:

$v = \lambda f$

From the last formula we see that the velocity is proportional fo the frequency.

For this case we have the following info given by the problem:

$T= 0.2 s, \lambda =32 cm* \frac{1m}{100cm} =0.32 m, A= 3cm*\frac{1m}{100 cm}=0.03 m$

We know that the frequency is the reciprocal of the period so we have this formula:

$f = \frac{1}{T}$

And if we replace we got:

$f =\frac{1}{0.2 s}= 5Hz$

Now since we have the value for the wavelength we can find the velocity like this:

$v = 0.32 m * 5Hz = 1.6 \frac{m}{s}$

And if we convert this into cm/s we got:

$v = 1.6 \frac{m}{s} *\frac{100cm}{1m}= 160 \frac{cm}{s}$

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2 years ago

The curved section of a horizontal highway is a circular unbanked arc of radius 740 m. If the coefficient of static friction bet

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3 years ago

Three parallel wires each carry current I in the directions shown in (Figure 1). The separation between adjacent wires is d.

jasenka [17]

(a) The magnitude of the net magnetic force per unit length on the top wire is μI²/πd.

(b) The magnitude of the net magnetic force per unit length on the middle wire is zero.

(c) The magnitude of the net magnetic force per unit length on the bottom wire is 3μI²/4πd.

<h3>Force per unit length</h3>

The magnitude of the net magnetic force per unit length on the top wire is calculated as follows;

F₁/L = (μI₁/2π) x (I₂/d + I₃/d)

F₁/L = (μI/2π) x (I/d + I/d)

F₁/L = (μI/2π) x (2I/d)

F₁/L = μI²/πd

The magnitude of the net magnetic force per unit length on the middle wire is calculated as follows;

F₂/L = (μI₂/2π) x (I₃/d - I₁/d)

F₂/L = (μI/2π) x (I/d - I/d) = 0

The magnitude of the net magnetic force per unit length on the middle bottom is calculated as follows;

F₃/L = (μI₂/2π) x (I₁/d + I₂/d)

F₃/L = (μI/2π) x (I/2d + I/d)

F₃/L = (μI/2π) x (3I/2d)

F₃/L = 3μI²/4πd

Thus, the magnitude of the net magnetic force per unit length on the top wire is μI²/πd.

The magnitude of the net magnetic force per unit length on the middle wire is zero.

The magnitude of the net magnetic force per unit length on the bottom wire is 3μI²/4πd.

Learn more about magnetic force here: brainly.com/question/13277365

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2 years ago