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PIT_PIT [208]
3 years ago
11

A 1500 kg car, initially traveling at 22.0 m/s, hits its brakes and skids to a stop. Determine the work done by friction.

Physics
1 answer:
Sedbober [7]3 years ago
8 0

Answer:

<em>The work done by the car is 363 kJ</em>

Explanation:

Work : Work is said to be done when a Force moves an object through a certain distance. Work and Energy are interchangeable because they have the same unit. The unit of work is Joules (J).

Mathematically work done can be expressed as,

E = W = 1/2mv²

W =  1/2mv²................................ Equation 1

Where E = Energy, W = work done, m = mass of the car, v = velocity of the car

<em>Given: m=1500 kg, v=22 m/s</em>

<em>Substituting these values into equation 1</em>

<em>W = 1/2(1500)(22)²</em>

<em>W = 750 × 484</em>

<em>W = 363000 J</em>

<em>W = 363 kJ</em>

<em>Thus the work done by the car is 363 kJ</em>

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Energy is a physical system's ability to perform labor. The capital letter E is a typical sign for energy. The joule, denoted by the letter J, is the common unit. The energy produced by one newton's (1 N) worth of force acting over one meter's (1 m) worth of displacement is measured in joules (1 J). Because it is a fundamental human requirement, energy plays a significant role in our daily lives.

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2 years ago
Alcohol of mass 33.2g and density 0.79kg/m³ or 790kg/m³ is mixed with water of 9g. What is the density of the resulting mixture?
KATRIN_1 [288]

Answer:

  0.83 g/cm³

Explanation:

The volume of the alcohol is ...

  (33.2 g)/(0.79 g/cm³) ≈ 42.0253 cm³

The density of water is about 1 g/cm³, so the volume of 9 g of it is ...

  (9 g)/(1 g/cm³) = 9 cm³

The total volume is ...

  42.0253 cm³ +9 cm³ = 51.0253 cm³

The total mass is ...

  33.2 g + 9 g = 42.4 g

So, the resulting density is ...

  (42.4 g)/(51.0253 cm³) ≈ 0.83 g/cm³

The resulting mixture has a density of about 0.83 g/cm³.

_____

<em>Additional comment</em>

Alcohol dissolves in water, so the total volume will be slightly less than 51.0253 cm³. The attached curve shows the result of mixing ethanol and water.

The weight of a mole of ethanol is about 46 g, of water, about 18.02 g. Then the mole fraction of alcohol is ...

  (33.2/46)/(33.2/46 +9/18.02) ≈ 0.59

The volume of the mix is then estimated to be (-1.05 cc/mol)(1.221 mol), or about 1.28 cm³ less than the volume indicated above. That brings the density up to about 0.85 g/cm³.

We're not completely sure of the relevance of this calculation, since many of the applicable parameters are not specified. The point is that <em>the density of the mix will probably be slightly more than the value calculated above</em>. YMMV

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