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8090 [49]
3 years ago
13

Describe a gear train that would transform a counterclockwise input rotation to a counterclockwise output rotation where the dri

ven gear rotates three times for every rotation of the driver gear.
Engineering
1 answer:
masya89 [10]3 years ago
3 0

Answer:

For a gear train that would train that transform a counterclockwise input into a counterclockwise output such that the gear that is driven rotates three times when the driver rotates once, we have;

1) The number of gears in the gear train = 3 gears with an arrangement such that there is a gear in between the input and the output gear that rotates clockwise for the output gear to rotate counter clockwise

2) The speed ratio of the driven gear to the driver gear = 3

Therefore, we have;

Speed \ Ratio =\dfrac{Speed \ of \ Driven \ Gear}{Speed \ of \ Driver \ Gear} = \dfrac{The \ Number \ of \ Teeth \ of \ Driver \ Gear}{The \ Number \ of \ Teeth \ of \ Driven \ Gear}

Therefore, for a speed ratio of 3, the number of teeth of the driver gear, driving the output gear, must be 3 times, the number of teeth of the driven gear

Explanation:

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In the figure below, block A weighs 20 lb , while block B weighs 10 lb . Friction between the surfaces of the two blocks may be
scoray [572]

Answer:

As P is continually increased, the block will now slip, with the friction force acting on the block being: f = muK*N, where muK is the coefficient of kinetic friction, with f remaining constant thereafter as P is increased.

8 0
2 years ago
Calculate KI for a rectangular bar containing an edge crack loaded in three-point bending where P=35.0 kN, W=50.8 mm, B=25 mm, a
Katyanochek1 [597]

Answer:

K_{I}=5.21 MPa\sqrt{m}

Explanation:

given data

Load P = 35 kN

Width of bar W = 50.8 mm

Breadth of bar B = 25 mm

Ratio of crack length to width α = a/W = 0.2

solution

we get here KI for a rectangular bar that is express as

K_{I} = \frac{6P}{BW}Y\sqrt{\pi a}   ................................1

here Y is the geometrical function

so

Y = \frac{1.12+\alpha (3.43\alpha -1.89)}{1-0.55\alpha}

Y = \frac{1.12+0.2(3.43\times 0.2-1.89)}{1-0.55\times 0.2}  

Y = \frac{0.8792}{0.89}  

Y = 0.9878

so put here value in equation 1

K_{I} = \frac{6\times 35\times 10^{3}}{0.025\times 0.0508}\times 0.9878\times \sqrt{3.1415\times (0.2\times 0.0508)}    

K_{I} = 165354.33\times 10^{3}\times 0.9878\times 0.0319

K_{I} = 5210.45 × 10³  Pa\sqrt{m}  

K_{I} = 5.21 MPa \sqrt{m}

5 0
3 years ago
Pls help!!! will give brainly!!!
gogolik [260]
Short-term goals are planned for the near future while long-term goals will take longer to complete. An example of a short-term goal is trying to get into a certain college. A long-term goal is wanting to have a big family.
3 0
3 years ago
A cartridge electrical heater is shaped as a cylinder of length L=200 mm and outer diameter D=20 mm. Under normal operating cond
Setler79 [48]

Answer:

When water is surrounding T_s = 34.17 degree C

When air surrounding T_S = 1434.7 degree C

from above calculation we can conclude that air is less effective than water  as heat transfer agent

Explanation:

Given data:

length  = 300 mm

Outer diameter  = 30 mm

Dissipated energy = 2 kw = 2000 w

Heat transfer coefficient IN WATER = 5000 W/m^2 K

Heat transfer coefficient in air  = 50 W/m^2 K

we know that q_{convection} =  P

From newton law of coding we have

q_{convection} =  hA(T_s -  T_{\infity})

T_s is surface temp.

T - temperature at surrounding

P = hA(T_s -  T_{\infity})[tex]\frac{P}{\pi hDL} =  (T_s -  T_{\infity})

solving for[/tex] T_s [/tex] w have

T_s = T_{\infty} + \frac{P}{\pi hDL}

T_s = 20 + \frac{2000}{\pi 5000\times 0.03\times 0.3}

T_s = 34.17 degree C

When air is surrounding we have

T_s = T_{\infty} + \frac{P}{\pi hDL}

T_s = 20 + \frac{2000}{\pi 2000\times 0.03\times 0.3}

T_s = 1434.7 degree C

from above calculation we can conclude that air is less effective than water  as heat transfer agent

5 0
3 years ago
"Given a nodal delay of 84.1ms when there is no traffic on the network (i.e. usage = 0%), what is the effective delay when netwo
shtirl [24]

Answer:

Explanation:

effective delay = delay when no traffic x \frac{100}{100- network\r usage}

effective delay = 84.1 \times \frac{100}{100-39.3}=138.55024711697ms

4 0
3 years ago
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