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zzz [600]
2 years ago
5

What is relation of crankshaft and camshaft

Engineering
2 answers:
Monica [59]2 years ago
6 0
Mierda me dices solo necesito puntos
amid [387]2 years ago
6 0

Answer:

A camshaft uses egg-shaped “cams” to open and close engine valves (one cam per valve), while a crankshaft converts “cranks” (the up/down motion of the pistons) to rotational motion.

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Multiply. Write the answer in simplest form. 1 3/10×1/8
kicyunya [14]

9514 1404 393

Answer:

  13/80

Explanation:

The product is ...

  (1 3/10)×(1/8) = (13/10)×(1/8) = (13×1)/(10×8) = 13/80

4 0
2 years ago
Refrigerant-134a enters a 28-cm-diameter pipe steadily at 200 kPa and 20°C with a velocity of 5 m/s. The refrigerant gains heat
Alexandra [31]

Answer:

V = 0.30787 m³/s

m = 2.6963 kg/s

v2 =  0.3705 m³/s

v2 = 6.017 m/s

Explanation:

given data

diameter = 28 cm

steadily =200 kPa

temperature = 20°C

velocity = 5 m/s

solution

we know mass flow rate is

m = ρ A v

floe rate V = Av

m = ρ V

flow rate = V = \frac{m}{\rho}

V = Av = \frac{\pi}{4} * d^2 * v1

V = \frac{\pi}{4} * 0.28^2 * 5

V = 0.30787 m³/s

and

mass flow rate of the refrigerant is

m = ρ A v

m = ρ V

m = \frac{V}{v} = \frac{0.30787}{0.11418}

m = 2.6963 kg/s

and

velocity and volume flow rate at exit

velocity = mass × v

v2 = 2.6963 × 0.13741 = 0.3705 m³/s

and

v2 = A2×v2

v2 = \frac{v2}{A2}

v2 = \frac{0.3705}{\frac{\pi}{4} * 0.28^2}

v2 = 6.017 m/s

7 0
3 years ago
When a rubber is stretched during a tensile test, its elongation is initially proportional to the applied force, but as it reach
Softa [21]
I don’t feel like reading
4 0
2 years ago
Create a flowchart that describes the following algorithm and then write Python code to implement the algorithm. Write the Pytho
Naya [18.7K]

Answer:

<em>Python code is as follows: </em>

********************************************************************************

#function to get number up to any number of decimal places

def toFixed(value, digits):

return "%.*f" % (digits, value)

print("Enter the price: ", end='', flush=True) #prompt for the input of price

price = float(input()) #taken input

totalCost = price + 0.05 * price #calculating cost

print("Total Cost after the sales tax of 5% is applied: " + toFixed(totalCost,2)) #print and output totalCost

************************************************************************************

3 0
3 years ago
A circular bar will be subjected to an axial force (P) of 2000 lbf. The bar will be made of material that has a strength (S) of
schepotkina [342]

Answer:

n = 2.36

Explanation:

The stress experimented by the circular bar is:

\sigma = \left[\frac{2000\, lbf}{\frac{\pi}{4}\cdot (0.5\,in)^{2}}\right]\cdot \left(\frac{1\,kpsi}{1000\,psi} \right)

\sigma = 10.186\,kpsi

The safety factor is:

n = \frac{24\,kpsi}{10.186\,kpsi}

n = 2.36

5 0
3 years ago
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