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3 years ago

Problem:

1 answer:

8 0

I do have a couple ideas and tips that may help you win. I don’t know how the guidelines are set up so if the ideas won’t be helpful I apologize.

First off put some ice cubes in the container then sprinkle salt on them, The reaction will create an effect and be super cold.

Another idea would be to get some dry ice if you able to, This will freeze it solid within seconds.

The last idea combines the the first. Take a bowl and fill it with with water and ice (Make sure the bowl is insulated) add a small handful of salt into the bowl, Put your drink into the cooler and before shutting stir then well then close and wait for the amount of time left, Your should have a cold water bottle.

I hoped this helped you out and I hope you also win the contest.

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The speed of sound at temperature of________in the air is 360m/s.

Aleks04 [339]

Vo= 331+0.6T

360=331+0.6T

360-331=0.6T

29=0.6T

0.6T/29

T=6/290 so change it to simplest form and us formulas good luck

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3 years ago

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If a tank filled with water contains a block and the height of the water above point A within the block is 0.6 meter what’s the

LenaWriter [7]

The answer is C I believe

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3 years ago

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

$\Delta H=\frac{-81719J}{0.015mole}$

$\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol$

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

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3 years ago

To a good approximate, the only external force that does work on a cyclist moving on level ground is the force of air resistance

mart [117]

Answer:

a. 120 W

b. 28.8 N

Explanation:

To a good approximate, the only external force that does work on a cyclist moving on level ground is the force of air resistance. Suppose a cyclist is traveling at 15 km/h on level ground. Assume he is using 480 W of metabolic power.

a. Estimate the amount of power he uses for forward motion.

b. How much force must he exert to overcome the force of air resistance?

(a) He is 25% efficient, therefore the cyclist will be expending 25% of his power to drive the bicycle forward

Power = efficiency X metabolic power

= 0.25 X 480

= 120 W

(b)

power if force times the velocity

P = Fv

convert 15 km/h to m/s

v = 15 kmph = 4.166 m/s

F = P/v

= 120/4.166

= 28.8 N

definition of terms

power is the rate at which work is done

force is that which changes a body's state of rest or uniform motion in a straight line

velocity is the change in displacement per unit time.

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A sea breeze occurs because warm air rises on land and moves toward the ocean to cool. The cool air then moves from the ocean to

Norma-Jean [14]

Im going to go with a. True

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