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3 years ago

Which of the following is not an example of an engine?

1 answer:

4 0

The answer is A. waterfall
To be considered as an engine , it should be a Man-made objects that could be used to produce power that creates motions.
From all the options above could be used to produce such power, but the waterfall is not made by mandkind

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What provides the energy for the conversion from the open complex to chain elongation?

Marat540 [252]

TP hydrolysis distinct from any incorporation into the chain.

7 0

3 years ago

An object is projected from the ground with an upward speed of ų m/s has a speed of 23m/s when it is at a height of 5m above the

vovikov84 [41]

Answer:

25.08m/s

Explanation:

mgh1 + 0.5mv1² = mgh2 + 0.5mv2²

h1 = 0m

v1 = u

h2 = 5m

v2 = 23m/s

putting the values into the formula above;

m(10)(0) + 0.5m(u²) = m(10)(5) + 0.5m(23²)

0 + 0.5mu² = 50m + 264.5m

0.5mu² = 314.5m

dividing through by m

0.5u² = 314.5

u² = 629

u = 25.08m/s

Therefore, the initial speed "u" = 25m/s

6 0

3 years ago

Two shuffleboard disks of equal mass, one orange and the other yellow, are involved in an elastic, glancing collision. The yello

klemol [59]

Answer:

3.62m/s and 2.83m/s

Explanation:

Apply conservation of momentum

For vertical component,

Pfy = Piy

m* Vof (sin38) - m*Vgf (sin52) = 0

Divide through by m

Vof(sin38) - Vgf(sin52) = 0

Vof(sin38) = Vgf(sin52)

Vof (sin38/sin52) = Vgf

0.7813Vof = Vgf

For horizontal component

Pxf= Pxi

m* Vof (cos38) - m*Vgf (cos52) = m*4.6

Divide through by m

Vof(cos38) + Vgf(cos52) = 4.6

Recall that

0.7813Vof = Vgf

Vof(cos38) + 0.7813 Vof(cos52) = 4.6

0.7880Vof + 0.4810Vof = 4.

1.269Vof = 4.6

Vof = 4.6/1.269

Vof = 3.62m/s

Recall that

0.7813Vof = Vgf

Vgf = 0.7813 * 3.62

Vgf = 2.83m/s

3 0

3 years ago

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stepan [7]

Answer:

Explanation:

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5 0

3 years ago

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