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1 year ago

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

Physics

1 answer:

beks73 [17]1 year ago

7 0

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

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Why can't fossils form in igneous rocks Magma flows too quickly for it to settle in The extreme heat destroys all of the remains

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Answer:

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Explanation:

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3 years ago

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Jada is rowing a boat across a river that has a current of 5 m/s in the ˆ j direction. Leanne, standing on the shore, observes J

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Answer: d. 8.25 m/s

Explanation:

We are given that Current= 5 m/s in j direction

Velocity= 8 m/s i + 3 m/s j

Now, we have to find Jada's speed with respect to the water.

First we find Jada's velocity with respect to water

v= (8 i + 3 j) - (5 j)

v= 8i - 2 j

To find the speed, we take the magnitude of this velocity vector we have

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So option d is correct.

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3 years ago

A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk

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Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration $\frac{m}{s^{2} }$

at: truck acceleration $\frac{m}{s^{2} }$ )

$ac = at= \frac{vf-vi}{t-ti}$ equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 $\frac{m}{s}$ , t = final time =5 s

We replaced the known information in the equation(1):

$ac = at = \frac{13-0}{15-0}$

$ac=ac=\frac{13}{15} \frac{m}{s}$

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

$Wc=mc*g$

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 $\frac{m}{s^{2} }$

$Wc= 2230*9.8$

$Wc=21854 N$

Normal force calculation:

Newton's first law

$sum Fy= 0$

$N-W=0$

$N=W$

$N=21854 N$

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

$Ff=0.373*21854$

$Ff=8151.54 N$

Calculation of the tension force in the rope (T):

Newton's Second law

$sum Fx= mc*ac$

$T-Ff=mc*ac$

$T=2230(\frac{13}{15}) + 8151.54$

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

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3 years ago

Juan makes an adjustment to an electromagnet that causes the electromagnet to lose some of its strength. What did Juan most like

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Answer:

<h3>The answer is 12 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

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m is the mass

From the question we have

$a = \frac{600}{50} = \frac{60}{5} \\$

We have the final answer as

Hope this helps you

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3 years ago