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3 years ago

7

Technician a says multiple discharge ignition system fires the spark plug during each of the engine's four cycle strokes. Techni

an B says that a dual spark plug ignition system ensures better ignition and combustion of the air- fuek mixture that a sigle spark plug per cylinder system who is right?

1 answer:

patriot [66]3 years ago

7 0

Answer:

Technician B only is correct

Explanation:

Here we have that the multiple discharge ignition system is a system that fires the spark plug multiple times during each power stroke to provide ample spark to complete the ignition. Therefore, technician A is not correct.

The dual spark plug ignition system is a method if obtaining the ideal combustion and improved fuel consumption than a single spark plug ignition system.

Therefore, only technician B is correct.

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3 years ago

Nancys airplane trip took 4 hour. For one-fourth of that time, the airplane flew at a speed of 880 km/h, it flew at a speed of 6

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4 0

2 years ago

An ancient club is found that contains 100 g of pure carbon and has an activity of 6.5 decays per second. Determine its age assu

never [62]

Answer:

The age of living tree is 11104 years.

Explanation:

Given that,

Mass of pure carbon = 100 g

Activity of this carbon is = 6.5 decays per second = 6.5 x60 decays/min =390 decays/m

We need to calculate the decay rate

$R=\dfrac{-dN}{dt}=\lambda N=\dfrac{0.693}{t_{\frac{1}{2}}}N$ ....(I)

Where, N = number of radio active atoms

$t_{\frac{1}{2}}$ =half life

We need to calculate the number of radio active atoms

For $N_{12_{c}}$

$N_{12_{c}}=\dfrac{N_{A}}{M}$

Where, $N_{A}$ =Avogadro number

$N_{12_{c}}=\dfrac{6.02\times10^{23}}{12}$

$N_{12_{c}}=5.02\times10^{22}\ nuclie/g$

For $N_{c_{14}}$

$N_{c_{14}}=1.30\times10^{-12}N_{12_{c}}$

$N_{c_{14}}=1.30\times10^{-12}\times5.02\times10^{22}$

$N_{c_{14}}=6.526\times10^{10}\ nuclei/g$

Put the value in the equation (I)

$R=\dfrac{0.693\times6.526\times10^{10}\times60}{5700\times3.16\times10^{7}}$

$R=15.0650\ decay/min g$

100 g carbon will decay with rate

$R=100\times15.0650=1507\ decay/min$

We need to calculate the total half lives

$(\dfrac{1}{2})^{n}=\dfrac{390}{1507}$

$2^n=\dfrac{1507}{390}$

$2^n=3.86$

$n ln 2=ln 3.86$

$n=\dfrac{ln 3.86}{ln 2}$

$n =1.948$

We need to calculate the age of living tree

Using formula of age

$t=n\times t_{\frac{1}{2}}$

$t=1.948\times5700$

$t=11103.6 =11104\ years$

Hence, The age of living tree is 11104 years.

5 0

3 years ago