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erma4kov [3.2K]
3 years ago
5

What is the net force on this object?​

Physics
1 answer:
Sergio039 [100]3 years ago
6 0
200N

Explanation:
600N-400N = 200N
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3. When driving a car, the chemical energy in gasoline is converted to heat (thermal energy) in the engine. Most of this thermal
tekilochka [14]

Explanation:

As we know that the maximum of thermal energy will be converted into mechanical energy.( The thermal energy will be given to us by engine.)

But the total thermal energy will not be converted into the mechanical energy,  there will be some lost of energy. The lost of mechanical energy will be in multiple forms. 2 of them are

1.Heat : The mechanical energy heats up the engine and then it will lost some energy.

2.Noise : Some energy also lost in the sound or the noise comes out from the engine. It will consume some energy.

Also, the engine never be 100% efficient some amount of energy lost in environment also.

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3 years ago
Clouds in space can begin as what​
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They begin as dark clouds and cosmic dust

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3 years ago
A composite load consists of three loads connected in parallel. One draws 100 W at a PF of 0.92 lagging, another takes 250 W at
fenix001 [56]

Answer:

a) I_{RMS} = 4.79 A

b) PF = 0.908

Explanation:

Get the reactive powers for each of the loads:

Reactive power = Real Power * tanθ

For load 1

Active power, P₁ = 100 W

Power factor, cos \theta_{1} = 0.92

\theta_{1} = cos^{-1} 0.92\\\theta_{1} = 23.074

Q_{1}= P_{1} tan \theta_{1} \\Q_{1}= 100tan 23.074\\Q_{1}= 42.60 W

For load 2

Active power, P₂ = 250 W

Power factor, cos \theta_{2} = 0.8

\theta_{2} = cos^{-1} 0.8\\\theta_{2} = 36.87

Q_{2}= P_{1} tan \theta_{2} \\Q_{2}= 250tan 36.87\\Q_{2}= 187.5 W

For load 3

Active power, P₃ = 250 W

Power factor, cos \theta_{3} = 1

\theta_{3} = cos^{-1} 1\\\theta_{3} =0

Q_{2}= P_{1} tan \theta_{3} \\Q_{3}= 150tan 0\\Q_{3}= 0 W

Calculate the total reactive power, Q_{net} = 42.6 + 187.5 + 0

Q_{net} = 230.1 W

Calculate the total active power, P_{net} = 100 + 250 + 150 = 500 W

S_{net} = P_{net} + Q_{net} \\S_{net} = 500 + j230.1

P_{net} = IVcos \theta_{net}

\theta_{net} = tan^{-1} \frac{230.1}{500} \\\theta_{net} = 24.712

V = 115 V_{rms}

500 = I_{RMS}  * 115 cos 24.712\\I_{RMS} = 500/104.47\\ I_{RMS} = 4.79 A

b) Power factor of the composite load is cos\theta_{net}

\theta_{net}  = 24.712\\PF = cos 24.712\\PF = 0.908

4 0
3 years ago
Two parallel metal plates are at a distance of 8.00 m apart.The electric field between the plates is uniform directed towards th
yuradex [85]

Answer:

C

Explanation:

Formula E=F/C also E=V/d

In this case use the second formula; E=V/d

Data given; E=4N/C d=8m

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4 years ago
The ratio of the wavelength of AM radio waves traveling in a vacuum to the wavelength of FM radio waves traveling in a vacuum is
Debora [2.8K]

This question is a big fat non sequitur !

The wavelength of radio waves traveling through vacuum only depends on the frequency that the radio station is licensed to broadcast on, (which had better be the frequency of the transmitter that they buy and use, or they're in big trouble).

The wavelength does NOT depend on the type of modulation that's used to put information onto the signal.  

An amateur radio (ham) operator may very well start out using FM to talk over his radio to somebody else, and then for some reason they may decide to switch to AM.  They can do that without ANY change in the wavelength of their transmissions.

Now, in the USA and many other countries, it so happens that all AM stations are licensed by their governments to transmit their programs on a channel somewhere between 500 KHz and 1.6 MHz, and all FM stations are licensed by their governments to transmit their programs on a channel somewhere between 88 MHz and 108 MHz.  (And THAT's what the radio receivers in these countries are built to receive.)

Then we might say that all of the AM stations are grouped around 1 MHz, and all of the FM stations are grouped around 100 MHz.  The FM frequencies are very roughly 100 times the AM frequencies, so the AM wavelengths are very roughly  100 times the FM wavelengths.  That's <em>choice (3)</em> .

But please don't get the idea that it has anything to do with using AM or FM technology.  It's just a matter of where in the spectrum the government decided to put the AM stations and where they put the FM stations.

For that matter . . . An analog TV station uses an AM signal for the picture and an FM signal for the sound, and it all goes in the same channel, with just about the same wavelengths !

3 0
3 years ago
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