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3 years ago

Define mechanical advantage and efficiency of a machine. What factor affects

Chemistry

1 answer:

gulaghasi [49]3 years ago

3 0

Answer:

Explanation:

Thus, the ratio of the resistance force to the effort, called the actual mechanical advantage (AMA), is less than the IMA. The efficiency of any machine measures the degree to which friction and other factors reduce the actual work output of the machine from its theoretical maximum.

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Volume Conversion Problem

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4,997.9

Explanation:

just convert fluid ounces to millilters.

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Hydrogen peroxide (H2O2) decomposes into water (H2O) and oxygen gas (O2). Hydrogen peroxide is a liquid that is stored in dark-c

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3 years ago

You drop a ball from a height of 2.0 m, and it bounces back to a height of 1.6 m. (a) what fraction of its initial energy is los

sweet [91]

Answer : Part a) Fraction of energy lost : 20 %

Part b) Speed before and after bounce = 6.3 m/s and 5.6 m/s

Part c) Energy is lost as thermal energy .

Part A) Fraction of energy lost during bouncing :

The energy possessed by any object when present at any height is potential energy . The formula of potential energy is given as :

PE = mgh

where PE = potential energy

,m = mass pf object , g = gravitational acceleration and h = height

Given : Initial height , h₁ = 2 m final height , h₂ = 1.6 m

Initial potential energy : m * g* h ₁

Final potential energy = m* g* h₂

Energy lost = Initial PE - Final PE

= ( mgh₁ - mgh2 )

Fraction of energy lost : $\frac{energy lost}{initial energy}=\frac{mgh1 - mgh2}{mgh1}$

Plugging value in above formula and taking " mg " common =>

Fraction of energy lost = $\frac{mg(h1-h2)}{mg (h1)} * 100$

$= \frac{(h1-h2)}{h1} * 100$

$= \frac{(2 - 1.6) }{2} * 100$

Fraction of energy lost = 20%

---------------------------------------------------------------------------------------------------

Part B ) Speed of ball just before and after the bounce.

Speed of ball before the bounce :

The potential energy gets converted to kinetic energy when it fall from height of 2m , so

Potential energy = kinetic energy

mgh₁ = $\frac{1}{2}$ m v²

or v ² = 2gh₁

Given : g = 9.8 m/s² h= 2 m

v² = 2 * 9.8 m/s² * 2 m = 39.2 m²/s²

v = 6.3 m/s

Speed of ball after bounce :

Potential energy = kinetic energy

mgh₂ = $\frac{1}{2}$ m v²

or v² = 2gh₂

= 2 * 9.8 m/s² * 1.6 m = 31.36 m²/s²

v = 5.6 m/s

---------------------------------------------------------------------------------------------

Part C) The energy lost due to friction. When the ball touches the ground , there occur friction force between the surface of ground and ball , due to which energy is lost as thermal energy .

5 0

3 years ago

Read 2 more answers

There are 12 liters of nitrogen gas in a sample. How many moles of N2 in this sample?

Firlakuza [10]

Answer:

0.54 moles of N2

Explanation:

First remember to find the volume in moles, 1 mole equals 22.4 liters.

So now use the dimensional analysis to show your work.

12 liters of N2 * 1 mol /22.4 liters of N2

Now calculate this. 12 * 1/22.4 or 12/ 22.4

12/22.4 = 0.535714286.

12 has two significant digits.

With that, The answer rounds to 0.54.

So that the final answer is 0.54 moles.

Hope it helped!

6 0

3 years ago