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3 years ago

A continuously variable transmission:

Engineering

1 answer:

Reil [10]3 years ago

8 0

Answer:

A continuously variable transmission (CVT) is an automatic transmission that can change seamlessly through a continuous range of gear ratios. This contrasts with other transmissions that provide a limited number of gear ratios in fixed steps. The flexibility of a CVT with suitable control may allow the engine to operate at a constant RPM while the vehicle moves at varying speeds.that means the ans is A.

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Resistance produces____<br> in a conductor.<br> •Gas<br> •Energy<br> •Heat

Zepler [3.9K]

Answer:

The answer would be Heat

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3 years ago

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What gage pressure does a skin diver experience when they dive to 35 ft in the ocean with a water temperature of 55 °F? Report y

TEA [102]

Answer:

p=15.097lbf/in^2

Explanation:

the manometric pressure is that in which the atmospheric pressure is not taken into account, so for this case only the pressure exerted by the water on the bus is calculated using the following equation.

P=ρgh

where

ρ=density of water at 55°F=999.4kg/m^3

g=9.81m/s^2

h=35ft=10.668m

Solving

P=(994.4)(9.81)(10.668)=104067.02Pa = 15.097lbf/in^2

the gage pressure does a skin diver experience when they dive to 35 ft is 15.097lbf/in^2

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3 years ago

You could receive an electric shock if you try to use water to put out a class fire

laila [671]

Answer:

Yes that's why they made fire extinguishers

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3 years ago

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A 2-pole, 50Hz, 20 kV turbo generator is rated 120 MVA and .95 power factor lagging. The machine rotor has a moment of inertia o

notsponge [240]

Answer:

782.290 MJ

Explanation:

Given Data:

number of poles ( p ) = 2-pole

Frequency ( f ) = 50 Hz

voltage ( v ) = 20 kV

power = 120 MVA

power factor (cos ∅ ) = 0.95

moment of inertia ( J ) = 11000 kg*m^2

First we calculate the value of

Ns = ( 120 * f ) / p

= ( 120 * 50 ) / 2 = 3000 rpm

Ws = ( 2 $\pi$ Ns ) / f

= (2 $\pi$ *3000 ) / 50 = 377.14 rad/sec

determine the kinetic energy of the machine

K.E = $\frac{1}{2} * Jw^{2} _{s}$

= 1/2 * 11000 * ( 377.14 ) ^2

= 5500 * 142234.58 = 782.290 MJ

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3 years ago

Consider a rocket engine in which the combustion chamber pressure and temperature are 30 atm and 3756 K, respectively. The area

Lerok [7]

Answer:

379.48 s ( specific impulse )
3678.75 m/s ( exit velocity )
263.21 kg/s ( mass flow )
979855.3 N ( thrust )
0.168 $m^{2}$ ( throat area )

Explanation:

A) Specific impulse

standard altitude = 25 km

take free steam pressure ( $p_{0}$ ) = 2527.3 N/ $m^{2}$

Exit pressure = ambient pressure. therefore exit pressure ( $p_{e}$ ) = 2527.3 N/ $m^{2}$

Applying specific impulse equation

$I_{sp}$ = $\frac{1}{g_{0} }$ [ $\frac{2Y T_{0} }{Y - 1} (\frac{R}{M} )[ 1 - (\frac{P_{e} }{p_{0} })^{\frac{y-1}{1} } ]]^{\frac{1}{2} }$ equation 1

given

$g_{0}$ = 9.81 m/ $s^{2}$ , M = 20 kg/kmol, R = 8314.47 j/kmol.k

y = 1.18, $T_{0}$ = 3756 k, $p_{e }$ = 2527.3 N/ $m^{2}$ , $p_{0}$ = 30 atm

substitute the given data into equation 1

$I_{sp}$ ( specific impulse ) = 379.48 s

B) exit velocity

Relating the equation for isentropic process, pressure and temperature at the combustion chamber , the exit temperature (Te) can be calculated using this formula :

$\frac{p_{e} }{p_{0} }$ = $(\frac{T_{e} }{T_{0} })^{\frac{y}{y-1} }$