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3 years ago

When calculating gear ratios, which formula is different from the others?

Engineering

1 answer:

BartSMP [9]3 years ago

6 0

Answer:

The number of threads in a worm is the number of teeth in a worm. The speed transmission ratio of a worm and worm gear set is obtained by dividing the number of teeth of the worm gear by the number of threads of the worm.

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A transmission line (TL) of length L and conductance per unit length G' is connected to an ideal constant voltage generator V. T

Andru [333]

Answer:

The Current will decrease by a factor of 2

Explanation:

Given the conditions, it should be noted that the current in the circuit is determined by the LOAD. In other words, the amount of current generator will be producing depends upon the load connected to it.

Now, as the question says, the load is reduced to half its original value, we can write:

$P1 = \sqrt{3} (V) (I1) Cos\alpha ----- (1)$

$P2 = \sqrt{3} (V) (I2) Cos\alpha\\$

Since, P2 = P1/2,

$P1/2 = \sqrt{3} (V) (I2) Cos\alpha ----- (2)\\$

Dividing equations (1) and (2), we get,

P1 / (P1/2) = I1/ I2

$I2 = I1 / 2\\$

Hence, it is proved that the current in the transmission line will decrease by a factor of 2 when load is reduced to half.

7 0

2 years ago

(a) What is the distinction between hypoeutectoid and hypereutectoid steels? (b) In a hypoeutectoid steel, both eutectoid and pr

Gnoma [55]

Answer:

See explanation below

Explanation:

Hypo-eutectoid steel has less than 0,8% of C in its composition.

It is composed by pearlite and α-ferrite, whereas Hyper-eutectoid steel has between 0.8% and 2% of C, composed by pearlite and cementite.

Ferrite has a higher tensile strength than cementite but cementite is harder.

Considering that hypoeutectoid steel contains ferrite at grain boundaries and pearlite inside grains whereas hypereutectoid steel contains a higher amount of cementite, the following properties are obtainable:

Hypo-eutectoid steel has higher yield strength than Hyper-eutectoid steel

Hypo-eutectoid steel is more ductile than Hyper-eutectoid steel

Hyper-eutectoid steel is harder than Hyper-eutectoid steel

Hypo-eutectoid steel has more tensile strength than Hyper-eutectoid steel.

When making a knife or axe blade, I would choose Hyper-eutectoid steel alloy because

1. It is harder

2. It has low cost

3. It is lighter

When making a die to press powders or stamp a softer metals, I will choose hypo-eutectoid steel alloy because

1. It is ductile

2. It has high tensile strength

3. It is durable

7 0

2 years ago

Read 2 more answers

A 4.4 HP electric motor spins a shaft at 2329 rpm. Find: The torque load carried by the shaft is closest to: Select one: a)-27.0

harina [27]

Answer:

Load carried by shaft=9.92 ft-lb

Explanation:

Given: Power P=4.4 HP

P=3281.08 W

<u><em>Power: </em></u>Rate of change of work with respect to time is called power.

We know that P= $Torque\times speed$

$\omega=\frac{2\pi N}{60}$ rad/sec

So that P= $\dfrac{2\pi NT}{60}$

So 3281.08= $\dfrac{2\pi \times 2329\times T}{60}$

T=13.45 N-m (1 N-m=0.737 ft-lb)

So T=9.92 ft-lb.

Load carried by shaft=9.92 ft-lb

3 0

2 years ago

A vehicle is considered to be legally parked if it is parked _____ or more from a pedestrian crosswalk or a marked or unmarked i

Yuki888 [10]

Answer:

A vehicle is considered to be legally parked if it is parked 20 feet (6 m) or more from a pedestrian crosswalk or a marked or unmarked intersection.

Explanation:

Hello!

I obtained the provided data from the New York State Driver's Manual. I wish it was useful to help you.

Success in your homework!

7 0

3 years ago

The output voltage of a power supply is normally distributed with mean 12 V and standard deviation 0.11 V. If the upper and lowe

podryga [215]

Answer:

82.62%

Explanation:

The z score is a score used in statistics to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

$z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean\ and\ \sigma=standard\ deviation.\\\\Given \ that\ \mu=12V, \sigma=0.11V.\\\\For\ x11.85V:\\\\z=\frac{11.85-12}{0.11} =-1.36\\\\$

From the normal distribution table, P(11.85 < x < 12.15) = P(-1.36 < z < 1.36) = P(z < 1.36) - P(z < -1.36) = 0.9131-0.0869 = 0.8262 = 82.62%

4 0

2 years ago