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kirza4 [7]
2 years ago
10

During the compression stroke of an internal combustion engine, _____

Physics
1 answer:
Leona [35]2 years ago
6 0

easy, The fuel is ignited

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10. Discrimination refers to behaviors that
salantis [7]

Answer:

d)

Explanation:

the only one that makes sense

4 0
3 years ago
Consider a short time span just before and after the spark plug in a gasoline engine ignites the fuel-air mixture and releases 1
Tju [1.3M]

Answer:

Temperature after ignition=7883.205 K

Explanation:

The number of moles is,

n=PV/RT

=(1.18x10^6)(47.9x10^-6)/8.314(325)

= 0.0209 moles

a) In this process volume is constant

Q=U

=nCv.dT

dT= Q/nCv

=1970/(1.5x8.314)(0.0209)

= 7558.205 K

The final temperature is,

= 7558.205+325

= 7883.205 K

5 0
3 years ago
Read 2 more answers
If the coefficient of static friction is 0.357, and the same ladder makes a 58.0° angle with respect to the horizontal, how far
zavuch27 [327]

Answer: d= 0.57* l

Explanation:

We need to check that before ladder slips the length of ladder the painter can climb.

So we need to satisfy the equilibrium conditions.

So for ∑Fx=0, ∑Fy=0 and ∑M=0

We have,

At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal

At the top of ladder, N₂ acting horizontal

And Between somewhere we have the weight of painter acting downward equal to= mg

So, we have N₁=mg

and also mg*d*cosФ= N₂*l*sin∅

So,

d=\frac{N2}{mg}*l * tan∅

Also, we have f₁=N₂

As f₁= чN₁

So f₁= 0.357 * 69.1 * 9.8

f₁= 241.75

Putting in d equation, we have

d= \frac{241.75}{69.1*9.8} *l * tan 58

d= 0.57* l

So painter can be along the 57% of length before the ladder begins to slip

3 0
3 years ago
Read 2 more answers
A diverging lens (f1 = -12.0cm) is located 50.0 cm to the left of a converging lens (f2 = 34.0 cm). A 2.0 cm-tall object stands
pishuonlain [190]

Answer:

The final image relative to the converging lens is 34 cm.

Explanation:

Given that,

Focal length of diverging lens = -12.0 cm

Focal length of converging lens = 34.0 cm

Height of object = 2.0 cm

Distance of object = 12 cm

Because object at focal point

We need to calculate the image distance of diverging lens

Using formula of lens

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}

\dfrac{1}{v}=\dfrac{1}{-12}-\dfrac{1}{-12}

v=\infty

The rays are parallel to the principle axis after passing from the diverging lens.

We need to calculate the image distance of converging lens

Now, object distance is ∞

Using formula of lens

\dfrac{1}{v}=\dfrac{1}{34}-\dfrac{1}{\infty}

v=34

The image distance is 34 cm right to the converging lens.

Hence, The final image relative to the converging lens is 34 cm.

5 0
3 years ago
The aorta pumps blood out of the heart to the rest of the body. Please select the best answer from the choices provided. T or f
Daniel [21]

The left ventricle relaxes and fills up with blood before squeezing and pumping the oxygen-rich blood through the aortic valve into the aorta — the main artery that carries blood to your body.<u>TRUE</u>

3 0
3 years ago
Read 2 more answers
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