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puteri [66]
2 years ago
14

Technician A says that many bearings also have a thin overlay of babbitt. Technician B says that bearings should have spread and

crush to keep them from spinning when the crankshaft rotates. Which technician is correct?
Physics
1 answer:
rosijanka [135]2 years ago
8 0

Answer:

In this case, both technicians are correct

Explanation:

There are several bearings with a thin overlay of babbitt. Also, technician B is also correct because the bearings is expected to be stagnant when the crankshaft rotates, and this can be achieved by ensuring that these bearings have spread and crush capable of ensuring they do not spin during operation.

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choice B

Explanation:i took the test

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A motorcycle accelerates from at a rate of 4m/s 2 while traveling 60m what it’s the motorcycles velocity at the end of this moti
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Answer: C

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2 years ago
What quantity of heat is needed to convert 1 kg of ice at -13 degrees C to steam at 100 degrees C?
Effectus [21]

Answer:

Heat energy needed = 3036.17 kJ

Explanation:

We have

     heat of fusion of water = 334 J/g

     heat of vaporization of water = 2257 J/g

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     specific heat of water = 4.18 J/g·°C

     specific heat of steam = 2.09 J/g·°C

Here wee need to convert 1 kg ice from -13°C to vapor at 100°C

First the ice changes to -13°C from 0°C , then it changes to water, then its temperature increases from 0°C to 100°C, then it changes to steam.

Mass of water = 1000 g

Heat energy required to change ice temperature from -13°C to 0°C

          H₁ = mcΔT = 1000 x 2.09 x 13 = 27.17 kJ

Heat energy required to change ice from 0°C to water at 0°C

          H₂ = mL = 1000 x 334 = 334 kJ

Heat energy required to change water temperature from 0°C to 100°C  

          H₃ = mcΔT = 1000 x 4.18 x 100 = 418 kJ    

Heat energy required to change water from 100°C to steam at 100°C  

          H₄ = mL = 1000 x 2257 = 2257 kJ    

Total heat energy required

          H = H₁ +  H₂ + H₃ + H₄ = 27.17 + 334 + 418 +2257 = 3036.17 kJ

Heat energy needed = 3036.17 kJ

5 0
3 years ago
Two vectors of magnitudes 30 units and 70 units are added to each other. What are possible results of this addition? (section 3.
Yanka [14]

Answer:

the correct answer is option C which is 50 units.

Explanation:

given,

two vector of magnitude = 30 units and of 70 units

to calculate resultants vector = \sqrt{a^2+b^2+2 a b cos\theta}

cos θ value varies from -1 to 1

so, resultant vector

=\sqrt{a^2+b^2-2 a b cos\theta}\ to\ \sqrt{a^2+b^2+ 2 a b cos\theta}

a = 30 units    and  b = 70 units

= \sqrt{30^2+70^2-2\times 30\times 70}\ to\ \sqrt{30^2+70^2+2\times 30\times 70}

=   40 units to 100 units

hence, the correct answer is option C which is 50 units.

                       

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