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2 years ago

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

Engineering

1 answer:

BartSMP [9]2 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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Software that is released to have users test out the "bugs" is known as Ransomeware O Break-in software 2 O Flim flam software O

Sophie [7]

Answer:

Beta software

Explanation:

5 0

2 years ago

Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7 m/s. Assume the width of the plate (into the paper)

mezya [45]

Complete Question

Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7m/s. Assume the width of the plate (into the paper) is 0.5 m. If the plate is at a co temperature of 100C,find:

The total heat transfer rate from the plate to the air

Answer:

$q=1.7845$

Explanation:

From the question we are told that:

Air Temperature $T_1=40c$

Length $l=2m$

Velocity $v=7m/s$

Width $w=0.5$

Constant temperature $T_t= 100C$

Generally the equation for Total heat Transfer is mathematically given by

$q=hA(T_s-T_\infty)$

Where

h=Convective heat transfer coefficient

$h=29.9075w/m^2k$

Therefore

$q=h(L*B)(T_s-T_\infty)$

$q=29.9075*(2*0.5)(100+273-(40+273))$

$q=1794.45w$

$q=1.7845$

5 0

2 years ago

Additional scals apply to the

REY [17]

Location of the class depends on satiation

4 0

1 year ago

An Ideal gas is being heated in a circular duct as while flowing over an electric heater of 130 kW. The diameter of duct is 500

Assoli18 [71]

Answer: The exit temperature of the gas in deg C is $32^{o}C$ .

Explanation:

The given data is as follows.

$C_{p}$ = 1000 J/kg K, R = 500 J/kg K = 0.5 kJ/kg K (as 1 kJ = 1000 J)

$P_{1}$ = 100 kPa, $V_{1} = 15 m^{3}/s$

$T_{1} = 27^{o}C = (27 + 273) K = 300 K$

We know that for an ideal gas the mass flow rate will be calculated as follows.

$P_{1}V_{1} = mRT_{1}$

or, m = $\frac{P_{1}V_{1}}{RT_{1}}$

= $\frac{100 \times 15}{0.5 \times 300}$

= 10 kg/s

Now, according to the steady flow energy equation:

$mh_{1} + Q = mh_{2} + W$

$h_{1} + \frac{Q}{m} = h_{2} + \frac{W}{m}$

$C_{p}T_{1} - \frac{80}{10} = C_{p}T_{2} - \frac{130}{10}$

$(T_{2} - T_{1})C_{p} = \frac{130 - 80}{10}$

$(T_{2} - T_{1})$ = 5 K

$T_{2}$ = 5 K + 300 K

$T_{2}$ = 305 K

= (305 K - 273 K)

= $32^{o}C$

Therefore, we can conclude that the exit temperature of the gas in deg C is $32^{o}C$ .

8 0

3 years ago

Which option should the engineers focus on as they develop the train in the following scenario?

pav-90 [236]

Answer:

Engineers can design a train with a regenerative braking system

Explanation:

Assuming the point of the question is that the engineers want to focus on using energy efficiently when starting and stopping, they would likely want to consider a regenerative braking system. Such a system can store energy during braking so that it can be used during starting, reducing the amount of energy that must be supplied by an outside power source.

5 0

2 years ago