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3 years ago

How can we calculate the speed of the output gear in a simple gear train? Explain with the help of an example.

Engineering

1 answer:

Snowcat [4.5K]3 years ago

3 0

Answer:

$N_3=\dfrac{T_1}{T_3}N_1$

Explanation:

In the diagram there three gears in which gear 1 is input gear ,gear 2 is idle gear and gear 3 is out put gear.

Lets take

$Speed\ of\ gear 1=N_1$

$Number\ of\ teeth\ of\ gear 1=T_1$

$Speed\ of\ gear 3=N_3$

$Number\ of\ teeth\ of\ gear 3=T_3$

All external matting gears will rotates in opposite direction with respect to each other.

So the speed of gear third can be given as follows

$\dfrac{T_1}{T_3}=\dfrac{N_3}{N_1}$

$N_3=\dfrac{T_1}{T_3}N_1$

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A 225 MPa conducted in which the mean stress was 50 MPa and the stress amplitude was (a) Compute the maximum and (b) Compute the

tamaranim1 [39]

Answer:

Explanation:

Given data in question

mean stress = 50 MPa

amplitude stress = 225 MPa

to find out

maximum stress, stress ratio, magnitude of the stress range.

solution

we will find first maximum stress and minimum stress

and stress will be sum of (maximum +minimum stress) / 2

so for stress 50 MPa and 225 MPa

$\sigma _{m}$ = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2

50 = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2 ...........1

and

225 = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2 ...........2

from eqution 1 and 2 we get maximum and minimum stress

$\sigma _{maximum}$ = 275 MPa ............3

and $\sigma _{minimum}$ = -175 MPa ............4

In 2nd part we stress ratio is will compute by ratio of equation 3 and 4

we get ratio = $\sigma _{minimum}$ / $\sigma _{maximum}$

ratio = -175 / 227

ratio = -0.64

now in 3rd part magnitude will calculate by subtracting maximum stress - minimum stress i.e.

magnitude = $\sigma _{maximum}$ - $\sigma _{minimum}$

magnitude = 275 - (-175) = 450 MPa

3 0

2 years ago

Assume the availability of an existing class, ICalculator, that models an integer arithmetic calculator and contains: an instanc

shtirl [24]

We connect with computers through coding, often known as computer programming.

We connect with computers through coding, often understood as computer programming.
Coding exists similar to writing a set of instructions because it instructs a machine what to do.
You can instruct computers what to do or how to behave much more quickly by learning to write code.

class ICalculator {

int currentValue;

int add(int value) {

this.currentValue = currentValue + value;

return currentValue;

}

int sub(int value) {

this.currentValue = currentValue - value;

return currentValue;

}

int mul(int value) {

this.currentValue = currentValue * value;

return currentValue;

}

int div(int value) {

this.currentValue = currentValue / value;

return currentValue;

}

public class ICalculator2 extends ICalculator {

int negate() {

if (currentValue != 0)

this.currentValue = -currentValue;

return currentValue;

}

public static void main(String[] args) {

ICalculator2 ic = new ICalculator2();

ic.currentValue=5;

System.out.println(ic.add(2));

System.out.println(ic.sub(5));

System.out.println(ic.mul(3));

System.out.println(ic.div(3));

System.out.println(ic.negate());

}

To learn more about code, refer to

brainly.com/question/22654163

#SPJ4

3 0

1 year ago

If it is desired to lay off a distance of 10,000' with a total error of no more than Â± 0.30 ft. If a 100' tape is used and the

Ira Lisetskai [31]

Answer:

± 0.003 ft

Explanation:

Since our distance is 10,000 ft and we need to use a full tape measure of 100 ft. We find that 10,000 = 100 × 100.

Let L' = our distance and L = our tape measure

So, L' = 100L

Now by error determination ΔL' = 100ΔL

Now ΔL' = ± 0.30 ft

ΔL = ΔL'/100

= ± 0.30 ft/100

= ± 0.003 ft

So, the maxim error per tape is ± 0.003 ft

3 0

3 years ago

A program is seeded with 30 faults. During testing, 21 faults are detected, 15 of which are seeded faults and 6 of which are ind

Vesna [10]

Answer:

Estimated number of indigenous faults remaining undetected is 6

Explanation:

The maximum likelihood estimate of indigenous faults is given by,

$N_F=n_F\times \frac{N_S}{n_S}$ here,

$n_F$ = the number of unseeded faults = 6

$N_S$ = number of seeded faults = 30

$n_s$ = number of seeded faults found = 15

So NF will be calculated as,

$N_F=6\times \frac{30}{15}=12$

And the estimate of faults remaining is $N_F-n_F$ = 12 - 6 = 6

8 0

3 years ago

Why might construction crews want to install pipes before the foundation is poured?

denpristay [2]

I’m a concrete mason myself and I can tell you it is a pain in the butt to Roto hammer a hole into the concrete to put the pipe in it’s a lot easier to just pour the concrete around it

6 0

2 years ago