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n200080 [17]
3 years ago
7

When fuel and air are compressed in the compression stroke, ...... a. each molecule of fuel is heated to its flash point b. each

molecule of air is heated to its flash point c. the exhaust valve opens to release excessive pressure
Physics
1 answer:
alexira [117]3 years ago
7 0

None of the choices is an appropriate response.

There's no such thing as the temperature of a molecule.  Temperature and
pressure are both outside-world manifestations of the energy the molecules
have.  But on the molecular level, what it is is the kinetic energy with which
they're all scurrying around.

When the fuel/air mixture is compressed during the compression stroke,
the temperature is raised to the flash point of the mixture.  The work done
during the compression pumps energy into the molecules, their kinetic
energy increases, and they begin scurrying around fast enough so that
when they collide, they're able to stick together, form a new molecule,
and release some of their kinetic energy in the form of heat.


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A firecracker in a coconut blows the coconut into three pieces. Two pieces of equal mass fly off south and west, perpendicular t
Lisa [10]

Explanation:

Let us assume that piece 1 is m_{1} is facing west side. And, piece 2 is m_{2} facing south side.

Let m_{1} \text{and} m_{2} = m and m_{3} = 2m. Hence,

        2mv_{3y} = m(21)

          v_{3y} = \frac{21}{2}

                     = 10.5 m/s

2mv_{3x} = m(21)

          v_{3x} = \frac{21}{2}

                     = 10.5 m/s

  v_{3} = \sqrt{v^{2}_{3x} + v^{2}_{3y}}

              = \sqrt{(10.5)^{2} + (10.5)^{2}}

              = 14.84 m/s

or,          = 15 m/s

Hence, the speed of the third piece is 15 m/s.

Now, we will find the angle as follows.

          \theta = tan^{-1} (\frac{v_{3y}}{v_{3x}})

                      = tan^{-1}(\frac{10.5}{10.5})

                      = tan^{-1} (45^{o})

                      = 45^{o}

Therefore, the direction of the third piece (degrees north of east) is north of east.

5 0
3 years ago
How is work involved in stopping a car?
vichka [17]
You have to move your foot to stop the car so I guess that would be considered work by moving your foot
5 0
3 years ago
Read 2 more answers
What is the frequency of a photon with an energy of 4. 56 x 10^-19 j
Sauron [17]

The frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

<h3>What is a frequency?</h3>

The number of waves that travel through a particular point in a given length of time is described by frequency. So, if a wave takes half a second to pass, the frequency is 2 per second.

Given that the energy of the photon is 4.56 x 10⁻¹⁹ J. Therefore, the frequency of the photon can be written as,

\rm \gamma = \dfrac{E}{h} = \dfrac{4.56x10^{-19} J}{6.626 \times 10^{-34}\ Jsec^{-1}}\\\\\\\gamma  = 6.88 \times 10^{14}\ s^{-1}

Hence, the frequency of a photon with an energy of 4.56 x 10⁻¹⁹ J is 6.88×10¹⁴ s⁻¹.

Learn more about Frequency:

brainly.com/question/5102661

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5 0
2 years ago
Read 2 more answers
Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus
tino4ka555 [31]

Answer:

1)4.7334J

2)225.4m/s

Explanation:

v= the Velocity of both the bullet and the block after collision=?

H= Height of the bullet along circular arc= 10cm=0.1m

g= acceleration due to gravity= 9.81m/s^2

R= Radius of the circular arc= 18cm= 0.18m

m= Mass of the bullet= 30g= 0.03kg

M= Mass of the block = 4.8 kg

Using the law of conservation of energy

Potential energy of the system= Kinectic energy of the system

1/2 mv^2= mgh..............eqn(1)

But we have two mass m and M

We can write eqn(1) as

0.5(m+M)v^2= (m+M)gh ...........eqn(2)

If we make "v" subject of the formula we have

v = √2gh

Then substitute the values we have

= √2 x 9.81 x 0.1 = 1.40m/s

1) We can now calculate the total energy of the system after collision as

KE = 1/2(m+M)v^2

= 1/2 x (0.03+4.8) x (1.40)^2

KE = 4.7334J

Hence, the total energy of the composite system at any time after the collision is 4.7334J

2)to determine the initial velocity of the bullet.

From law of momentum conservation, which can be expressed as

m1u1+m2u2=(m1+m2)v

Where the initial Velocity of the bullet u1= ?

Final velocity of the bullet = 0

the Velocity of both the bullet and the block after collision=v= 1.40m/s

(0.03×u1) +(u×0)= (4.8+0.03)1.4

0.03u1=6.762

U1=225.4m/s

Hence, the initial velocity of the bullet is 225.4m/s

3 0
2 years ago
An electric fan is running on HIGH. After fan has been running for 1.3 minutes, the LOW button is pushed. The fan slows down to
mafiozo [28]

Answer:

    wo = 18.75 rev / s

Explanation:

This is an exercise in endowment kinematics, it indicates that the final angular velocity is w_f = 109 rad / s, the time to reach this velocity is t = 1.87 s and the deceleration a = 4.7 rad / s²

         w_f = w₀ - a t

         w₀ = w_f + a t

         w₀ = 109 + 4.7 1.87

         w₀ = 117.8 rad / s

let's reduce to revolutions / s

         w₀ = 117.8 rad / s (1 rev / 2pi rad)

         w₀ = 18.75 rev / s

8 0
3 years ago
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