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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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The European Space Agency launched a probe called Rosetta in March 2004. In August 2014, Rosetta reached its destination: a co

Art [367]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The of the lander would be 75.66 pounds

Explanation:

From the question we are given that the

Weight of the Philae on Earth is = 220 pound-mass

gravity of Mars is = 3.71 m/ $s^{2}$

Weight of the Philae on Mars is = ?

Now Mass is quantity of matter in an object so it is always constant for a particular object

Generally Weight = Mass × Gravity :

$\frac{Weight Of Earth}{Gravity Of Earth} = \frac{Weight Of Mars}{Gravity Of Mars}$

Hence

Weight of the Philae on Mars is = $\frac{Weight Of Earth *Gravity Of Mars}{Gravity Of Earth}$

$=\frac{200 *3.71}{9.81}$

$= 75.66 pounds$

6 0

3 years ago

What happens if you leave your car on while pumping gas

Natali5045456 [20]

Answer:

the air you pumped would be off

5 0

3 years ago

One kilogram of air, initially at 5 bar, 350 K, and 3 kg of carbon dioxide (CO2), initially at 2 bar, 450 K, are confined to opp

pentagon [3]

Answer:

Check the explanation

Explanation:

Energy alance of 2 closed systems: Heat from CO2 equals the heat that is added to air in

$m_{a} c_{v,a}(T_{eq} -T_{a,i)} =m_{co2} c_{v,co2} (T_{co2,i} -T_{eq)}$

1x0.723x $(T_{eq} -350)$ =3x0.780x $(450-T_{eq} )$ ⇒ $T_{eq}$ = 426.4 °K

The initail volumes of the gases can be determined by the ideal gas equation of state,

$V_{a,i} = \frac{mRT_{a,i} }{P_{a,i} }$ = $\frac{1x (8.314 28.97 kJ kg • °K)x 350°K}{5 bar x 100KPa bar}$ = 0.201 $m^{3}$

The equilibrium pressure of the gases can also be obtained by the ideal gas equation

$P_{eq=\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,eq}+V_{CO2,eq)} } =\frac{(m_{a}R_{a}T_{eq})+(m_{a}R_{a}T_{eq} ) }{(V_{a,i}+V_{CO2,i)} }$

$P_{eq}$ = 1x(8.314 28.97)x426.4+3x(8.314 44)x426.4

(0.201+1.275)

= 246.67 KPa = 2.47 bar

6 0

3 years ago

An aluminum part will be subjected to cyclic loading where the maximum stress will be 300 MPa and the minimum stress will be-100

Dominik [7]

Answer:

a) The mean stress experimented by the aluminium part is 100 megapascals, b) The stress amplitude of the aluminium part is 400 megapascals, c) The stress ratio of the aluminium part is 4.

Explanation:

a) The mean stress is determined by this expression:

$\sigma_{m} = \frac{\sigma_{min}+\sigma_{max}}{2}$

Where:

$\sigma_{m}$ - Mean stress, measured in megapascals.

$\sigma_{min}$ - Minimum stress, measured in megapascals.

$\sigma_{max}$ - Maximum stress, measured in megapascals.

If we know that $\sigma_{min} = -100\,MPa$ and $\sigma_{max} = 300\,MPa$ , the mean stress is:

$\sigma_{m} = \frac{-100\,MPa+300\,MPa}{2}$

$\sigma_{m} = 100\,MPa$

The mean stress experimented by the aluminium part is 100 megapascals.

b) The stress amplitude is given by the following difference:

$\sigma_{a} = |\sigma_{max}-\sigma_{min}|$

Where $\sigma_{a}$ is the stress amplitude, measured in megapascals.

If we know that $\sigma_{min} = -100\,MPa$ and $\sigma_{max} = 300\,MPa$ , the stress amplitude is:

$\sigma_{a} = |300\,MPa-(-100\,MPa)|$

$\sigma_{a} = 400\,MPa$

The stress amplitude of the aluminium part is 400 megapascals.

c) The stress ratio ( $R$ ) is the ratio of the stress amplitude to mean stress. That is:

$R = \frac{\sigma_{a}}{\sigma_{m}}$

If we know that $\sigma_{m} = 100\,MPa$ and $\sigma_{a} = 400\,MPa$ , the stress ratio is:

$R = \frac{400\,MPa}{100\,MPa}$

$R = 4$

The stress ratio of the aluminium part is 4.

3 0

3 years ago

20. What is a "whipping motion and why is it<br> used?

Reptile [31]

Whipping is done in a forward direction. You deposit a dab or puddle of metal down, and then whip the rod forward and up a little to let that dab cool a bit, then you move back and deposit another dab slightly ahead of the previous dab.
Cellulose rod like 6010 and 6011 are known as "fast freeze" electrodes, also known to be deep penetrating rods. The "whip and stitch" motion is used so that you get the full benefit of the cellulose characteristics. The forward motion basically gouges out the base metal, and it gets filled in with the back-step.
Hope this helped:)

6 0

4 years ago