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3 years ago

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What is the Net Force of the applied forces? *

1 answer:

Oduvanchick [21]3 years ago

3 0

All of anove
Bc it make sense

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Which of the following is correct concerning the uncontrolled burn phase?

beks73 [17]

The option that is the correct one concerning the uncontrolled burn phase is:

The uncontrolled burn phase is characterized by uncontrolled combustion in a cylinder until fuel accumulated during ignition delay is burned.

<h3>What is uncontrolled combustion?</h3>

Uncontrolled Combustion is known to be the the time and place in which a kind of an ignition will stop and it is said to be never fixed by anything in regards to the compression ignition engine as seen in SI engines.

Note that the four Stages of combustion are:

1. Pre-flame combustion

2. Uncontrolled combustion

3. Controlled combustion and

4. After burning

Hence, The uncontrolled burn phase is characterized by uncontrolled combustion in a cylinder until fuel accumulated during ignition delay is burned as all the fuel need to burn out.

Learn more about burn phase from

brainly.com/question/14414049

#SPJ1

7 0

2 years ago

What will happen when the sun blows up?

RSB [31]

We will all chaotically burn to death!

Hope this helps!!

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3 years ago

Read 2 more answers

If the CD rotates clockwise at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angu

gavmur [86]

Answer:

Explanation:

Given that,

Initial Angular velocity w=500rpm

Converting from rpm to rad/s

1rev =2πrad

1minutes =60secs

500rpm=500rev/mins

w = 500×2π/60

wi=52.36rad/s

The final angular velocity wf=0rad/s

Time to stop is t=2.6sec

We want to find angular acceleration α

Using the equation of angular motion

wf = wi + αt.

0 = 52.36 + 2.6α

-52.36=2.6α

α = -52.36/2.6

α = -20.14rad/s²

The angular acceleration is negative because it is decelerating.

Then, α=20.14rad/s²

8 0

3 years ago

Read 2 more answers

You and your friend are going bungee jumping! You wait directly below them with a camera. When they leap from the bridge they be

Alex

Answer:

The amplitude is $A = 90.2 \ m$

Explanation:

From the question we are told that

The frequency of when sound is approaching observer is $f = 392 Hz$

The frequency as the move away from observer is $f_ a = 330 \ Hz$

The time between the pitch are $t = 10 \ s$

Here you are the observer and your friends are the source of the sound

The period is mathematically evaluated as

$T = 2 t$

as it is the time to complete one oscillation which from on highest pitch to the next highest pitch

Now T can also be mathematically represented as

$T = \frac{2 \pi}{w}$

Where $w$ is the angular velocity

=> $\frac{2 \pi}{w} = 2 * 10$

=> $w = 0.314 \ rad/sec$

Now using Doppler Effect,

The source of the sound is approaching the observer

The

$f = f_o (\frac{v}{v- wA} )$

$392 = f_o (\frac{v}{v- wA} )$

Where A is the amplitude

So when the source is moving away from the observer

$f_a = f_o (\frac{v}{v+ wA} )$

$330 = f_o (\frac{v}{v+ wA} )$

Here $f_o$ is the fundamental frequency

Dividing the both equation we have

$\frac{392}{330} = \frac{f_o(\frac{v}{v-wA} )}{f_o(\frac{v}{v+wA}}$

$1.1878 = \frac{v+wA}{v-wA}$

$1.1878 v - 1.1878 wA = v+wA$

$1.1878 v = 2.1878 wA$

=> $A = \frac{(0.1878 * (330))}{(2.1878)* (0.314)}$

$A = 90.2 \ m$

7 0

3 years ago

Why is the gravitational potential energy of an object 1 meter above Earth's surface more than its potential energy 1 meter abov

Umnica [9.8K]

I Believe The Answer Is C

3 0

3 years ago