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Firdavs [7]
3 years ago
6

Name three engine industries which use the 12 volt battery-ignition system with solid state components

Engineering
1 answer:
sukhopar [10]3 years ago
6 0

Answer:THE DISTRIBUTOR AUTOMOTIVE IGNITION SYSTEM

THE DISTRIBUTOR-LESS AUTOMOTIVE IGNITION SYSTEM

COIL-ON-PLUG IGNITION SYSTEM

Explanation:

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ludmilkaskok [199]

Answer: That you are dressed appropriately, to speak in a formal manner, and to be confident in your answers.

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3 years ago
A 5000-lb truck is being used to lift a 1000-lb boulder B that is on a 200-lb pallet A. Knowing that the truck starts from rest
artcher [175]

Answer:

Explanation:

Total weight being moved = 5000+1000+200

= 6200 lb .

Force applied = 700 lb

= 700 x 32 = 22400 poundal .

acceleration (a) = 22400 / 6200

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To know velocity after 6 ft we apply the formula

v² = u² + 2as

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3 years ago
A 5-mm-thick stainless steel strip (k = 21 W/m•K, rho = 8000 kg/m3, and cp = 570 J/kg•K) is being heat treated as it moves throu
Drupady [299]

Answer:

The temperature of the strip as it exits the furnace is 819.15 °C

Explanation:

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L_c = \frac{V}{A} = \frac{LA}{2A} = \frac{5*10^{-3}}{2} = 0.0025 \ m

The Biot number is given as;

B_i = \frac{h L_c}{k}\\\\B_i = \frac{80*0.0025}{21} \\\\B_i = 0.00952

B_i < 0.1,  thus apply lumped system approximation to determine the constant time for the process;

\tau = \frac{\rho C_p V}{hA_s} = \frac{\rho C_p L_c}{h}\\\\\tau = \frac{8000* 570* 0.0025}{80}\\\\\tau = 142.5 s

The time for the heating process is given as;

t = \frac{d}{V} \\\\t = \frac{3 \ m}{0.01 \ m/s} = 300 s

Apply the lumped system approximation relation to determine the temperature of the strip as it exits the furnace;

T(t) = T_{ \infty} + (T_i -T_{\infty})e^{-t/ \tau}\\\\T(t) = 930 + (20 -930)e^{-300/ 142.5}\\\\T(t) = 930 + (-110.85)\\\\T_{(t)} = 819.15 \ ^0 C

Therefore, the temperature of the strip as it exits the furnace is 819.15 °C

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3 years ago
What is true about a point and shoot camera?
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Answer:

What is true about a point and shoot camera?

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Physical properties of minerals
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Most minerals can be characterized and classified by their unique physical properties: hardness, luster, color, streak, specific gravity, cleavage, fracture, and tenacity.

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