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tamaranim1 [39]
3 years ago
5

Explain the 3 examples of technology transfer

Engineering
1 answer:
xxTIMURxx [149]3 years ago
8 0

Answer:

ht f of tevno animation de espain ingleis

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7 0
3 years ago
The best penetration is achieved with DCEN current.<br> True or false
GenaCL600 [577]

Answer:

True

Explanation:

The best penetration is achieved with dcen current. Mild steel is expensive and requires the most amount of cleaning. Copper-coated steel welding rods are not used for gas tungsten arc welding because they will continue the weld or electrode.

6 0
3 years ago
Burn rate can be affected by: A. Variations in chamber pressure B. Variations in initial grain temperature C. Gas flow velocity
Digiron [165]

Answer: D) All of the above

Explanation:

Burn rate can be affected by all of the above reasons as, variation in chamber pressure because the pressure are dependence on the burn rate and temperature variation in initial gain can affect the rate of the chemical reactions and initial gain in the temperature increased the burning rate. As, gas flow velocity also influenced to increasing the burn rate as it flowing parallel to the surface burning. Burn rate is also known as erosive burning because of the variation in flow velocity and chamber pressure.

4 0
3 years ago
Durante el segundo trimestre de 2001, Tiger Woods fue el golfista que más dinero ganó en el PGATour. Sus ganancias sumaron un to
ehidna [41]

Answer: a. 0.4667

b. 0.4667 and C 0.0667

Explanation:

Given Data:

N = population size (10)

n = random selection (2)

r = number of observations = 7

Therefore

f(y) = ( r/y ) ( N - r / n - y ) / ( N /n )

When y = 1

f(1) = ( 7/1 ) ( 10 - 7 / 2 -1 ) / ( 10/2 )

= 7 / 15

= 0.4667

When y = 2

f(2) = ( 7/2 ) ( 10 - 7 / 2 -2 ) / ( 10/2 )

= 7 / 15

= 0.4667

When y = 0

f(0) = ( 7/0 ) ( 10 - 7 / 2 -0) / ( 10/2 )

= 1 / 15

= 0.0667

8 0
3 years ago
A rigid tank having 25 m3 volume initially contains air having a density of 1.25 kg/m3, then more air is supplied to the tank fr
Hoochie [10]

Answer:

\Delta m = 102.25\,kg

Explanation:

The mass inside the rigid tank before the high pressure stream enters is:

m_{o} = \rho_{air}\cdot V_{tank}

m_{o} = (1.25\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})

m_{o} = 31.25\,kg

The final mass inside the rigid tank is:

m_{f} = \rho \cdot V_{tank}

m_{f} = (5.34\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})

m_{f}= 133.5\,kg

The supplied air mass is:

\Delta m = m_{f}-m_{o}

\Delta m = 133.5\,kg-31.25\,kg

\Delta m = 102.25\,kg

4 0
3 years ago
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