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Mars2501 [29]
2 years ago
6

True or False Most coolants you put into your car are toxic.

Engineering
2 answers:
swat322 years ago
8 0

Answer: True.

Explanation: Coolant is flammable and toxic.

vaieri [72.5K]2 years ago
5 0
True
Anything you put in your car is toxic
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A Geostationary satellite has an 8kW RF transmission pointed at the earth. How much force does that induce on the spacecraft? (N
soldier1979 [14.2K]

Answer:

The force induced on the aircraft is 2.60 N

Solution:

As per the question:

Power transmitted, P_{t} = 8 kW = 8000 W

Now, the force, F is given by:

P_{t} = Force(F)\times velocity(v) = Fv               (1)

where

v = velocity

Now,

For a geo-stationary satellite, the centripetal force, F_{c} is provided by the gravitational force, F_{G}:

F_{c} = F_{G}

\frac{mv^{2}}{R} = \frac{GM_{e}m{R^{2}}

Thus from the above, velocity comes out to be:

v = \sqrt{\frac{GM_{e}}{R}}

v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.979\times 10^{24}}{42166\times 10^{3}}} = 3075.36 m/ s

where

R = R_{e} + H

R = \sqrt{GM_{e}(\frac{T}{2\pi})^{2}}

where

G = Gravitational constant

T = Time period of rotation of Earth

R is calculated as 42166 km

Now, from eqn (1):

8000 = F\times 3075.36

F = 2.60 N

6 0
3 years ago
I want a problems and there solutions of The inception of cavitation?​
Ugo [173]

Answer:

The overview of the given scenario is explained in explanation segment below.

Explanation:

  • The inception of cavitation, that further sets the restriction for high-pressure and high-free operation, has always been the matter of substantial experimental study over the last few generations.
  • Cavitation inception would be expected to vary on the segment where the local "PL" pressure mostly on segment keeps falling to that are below the "Pv" vapor pressure of the fluid and therefore could be anticipated from either the apportionment of the pressure.

    ⇒  A cavitation number is denoted by "σ" .

4 0
3 years ago
NO LINK plsssss just answer
solmaris [256]

Answer:

Explanation:

4 0
2 years ago
Write equations used to calculate the diode reverse saturation current, the voltage at which diode goes into resistive behavior,
laiz [17]

Answer:

Diode equation for reverse saturation current

I_o = A\times e^{\frac{-Eg}{KT}} + B\times e^{\frac{-Eg}{2KT}}  

Voltage at which diode goes into Resistive region:V=-5 volts

Voltage at which high level injection occurs:Va=0.55 volt

Voltage at which avalanche multiplication occurs:V=5volts

Explanation:

we take here forward and reverse 0.7 volt and -5 volt  

As Diode current equation is express as

I_D = I_o \times (e^{\frac{V_D}{\eta V_T}} -1 )   ....................1

here I_D is total current through the diode and I_o is reverse saturated current and V_D is  voltage drop across diode and \eta is idealized factor and V_T is thermal voltage

so here we know that when Bios is forward than

V_D  = V_T     .................2

ans Bios is Reverse than  

V_D  = V_R      ..................3

so here

1.  diode reverse saturation current is express as

I_o = A\times e^{\frac{-Eg}{KT}} + B\times e^{\frac{-Eg}{2KT}}  

and

2. Voltage at which diode go into Reverse behavior will be

V_D  = V_R    = -5 volt

and

3. voltage at which high level injection occur that is

Va = 0.55 volt

and

4. voltage at which avalanche multiplication occurs is

Va = 5 volt

6 0
3 years ago
zener shunt regulator employs a 9.1-V zener diode for which VZ = 9.1 V at IZ = 9 mA, with rz = 40 and IZK= 0.5 mA. The available
gulaghasi [49]

Answer:

V_z=9.1v

V_{zo}=8.74V

I=10mA

R=589 ohms

Explanation:

From the question we are told that:

Zener diode Voltage V_z=9.1-V

Zener diode Current I_z=9 .A

Note

rz = 40\\\\IZK= 0.5 mA

Supply Voltage V_s=15

Reduction Percentage P_r= 50 \%

Generally the equation for Kirchhoff's Voltage Law is mathematically given by

V_z=V_{zo}+I_zr_z

9.1=V_{z0}+9*10^{-3}(40)

V_{zo}=8.74V

Therefore

At I_z-10mA

V_z=V_{z0}+I_zr_z

V_z=8.74+(10*10^{-3}) (40)

V_z=9.1v

Generally the equation for Kirchhoff's Current Law is mathematically given by

-I+I_z+I_l=0

I=10mA+\frac{V_z}{R_l}

I=10mA+\frac{9.1}{0}

I=10mA

Therefore

R=\frac{15V-V_z}{I}

R=\frac{15-9.1}{10*10^{-3}}

R=589 ohms

5 0
3 years ago
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