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DerKrebs [107]
3 years ago
11

A runner stood at the 100-meter mark on a track. When the timer started, the runner sprinted north to the 200-meter mark. It too

k the runner 25 seconds to get to the 200-meter mark. What was the average velocity of the runner while he was sprinting?
4 m/s north

4 m/s south

8 m/s north

8 m/s south
Chemistry
1 answer:
WITCHER [35]3 years ago
7 0

Answer:

4m/sec north

Explanation:

Given the runner sprinted from 100m mark to 200m mark, which means the displacement of the runner is 200-100=100m.

Now given the time taken to travel this 100 distance is 25 seconds

We know that Average velocity = \frac{displacement}{time}

Average velocity = \frac{100}{25}  = 4 m/sec.

We can observe from the above formulae that the direction of the average velocity is in the direction of the displacement which is given north.

Therefore the Average velocity of the runner is 4m/sec north

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Answer: Freezing point of a solution will be -1.16^0C

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

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i= vant hoff factor = 1 (for non electrolyte)

K_f = freezing point constant = 5.12^0C/m

m= molality

\Delta T_f=i\times K_f\times \frac{\text{mass of solute}}{\text{molar mass of solute}\times \text{weight of solvent in kg}}

Weight of solvent (benzene)= 1480 g =1.48 kg

Molar mass of solute (octane) = 114.0 g/mol

Mass of solute (octane) = 220 g

(5.50-T_f)^0C=1\times 5.12\times \frac{220g}{114.0 g/mol\times 1.48kg}

(5.50-T_f)^0C=6.68

T_f=-1.16^0C

Thus the freezing point of a solution will be -1.16^0C

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