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shusha [124]
4 years ago
8

A student designs an experiment where he takes a small wood car with 1.0 inch diameter wheels and rolls it down a ramp. The stud

ent measures the speed of the car when it reaches the bottom of the ramp. The student then takes a small plastic car with 1.5 inch diameter wheels and repeats the process. Finally, the student takes a small metal car with 2.0 inch diameter wheels and repeats the process a final time. The student notes that the metal car had the fastest speed at the bottom of the ramp.
The student can conclude:
Chemistry
2 answers:
aleksley [76]4 years ago
8 0
The metal car airodinamics with body style and weight increased speed
Anastaziya [24]4 years ago
3 0

Answer:

As the radius of a wheel , in rolling motion on a inclined plane, changes , its final speed , does not change . In other words , acceleration of wheel on inclined plane does not depend upon radius of the wheel . It also does not depend upon the mass of the wheel.

Explanation:

In case of perfect rolling on an inclined plane, the acceleration of body is given by the following expression.

a = g/1+k²/r² where k is known as radius of gyration . For a wheel its value is equal to( 1/√2)r.r is radius . From this expression , one can conclude that acceleration and therefore velocity does not depend upon radius of the wheel . Therefore the reason of the metal car having greatest speed  is the resistance of air and friction of surface being the least in its case.

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1c) anion and N-3
1d) cation and Ca+2
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3 years ago
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How many grams of XeF6 are required to react with 0.579 L of hydrogen gas at 6.46 atm and 45°C in the reaction shown below?
ankoles [38]
The chemical reaction equation for this is 

XeF6 + 3H2 ---> Xe + 6HF

Assuming gas behaves ideally, we use the ideal gas formula to solve for number of moles H2 with T = 318.15K (45C), P = 6.46 atm, V = 0.579L. Then we use the gas constant R = 0.08206 L atm K-1 mol-1.

we get n = 0.1433 moles H2

to get the mass of XeF6, 

we divide 0.1433 moles H2 by 3 since 1 mole XeF6 needs 3 moles H2 to react then multiply by the molecular weight of XeF6 which is 245.28 g/mole XeF6.

0.1433 moles H2 x \frac{1 mole XeF6}{3 moles H2} x \frac{245.28 g XeF6}{1 mole XeF6} = 11.71 g XeF6

Therefore, 11.71 g of XeF6 is needed to completely react with 0.579 L of Hydrogen gas at 45 degrees Celcius and 6.46 atm.
3 0
4 years ago
If 1 egg and 1/3 cup of oil are needed for each bag of brownie mix, how many bags of brownie mix do you need if you want to use
Charra [1.4K]
The answer is 3 bags total
5 0
4 years ago
A sample of ice is heated continuously until it becomes a liquid, and then a gas. Its temperature is recorded throughout and a g
padilas [110]

Answer:

The answer to your question is: C. The specific latent heat of fusion

Explanation:

A. The specific latent heat of vaporization  Specific latent heat of vaporization indicates the transition from liquid to vapor, but we are not looking for this definition. This answer is wrong.

B. The specific heat indicates the amount of heat needed to increase the temperature of water 1°C, so this answer is wrong.

C. The specific latent heat of fusion . This heat indicate the transition from solid ie to liquid, so this is the right answer.

D. The internal energy measures the energy of the molecules of a substance, so this answer is wrong.

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3 years ago
The following balanced equation describes the reduction of iron(III) oxide to molten iron within a blast furnace: Fe2O3(s) + 3CO
Irina-Kira [14]

Answer:

Amount of excess Carbon (ii) oxide left over = 23.75 g

Explanation:

Equation of the reaction: Fe₂O₃ + 3CO ----> 2Fe + 3CO₂

Molar mass of Fe₂O₃ = 160 g/mol;

Molar mass of Carbon (ii) oxide = 28 g/mol

From the equation of reaction, 1 mole of Fe₂O₃ reacts with 3 moles of carbon (ii) oxide; i.e. 160 g of iron (iii) oxide reacts with 84 g (3 * 28 g)  of carbon (ii) oxide

450 g of Fe₂O₃ will react with 450 * 84/180) g of carbon (ii) oxide = 236..25 g of carbon (ii) oxide

Therefore the excess reactant is carbon (ii) oxide.

Amount of excess Carbon (ii) oxide left over = 260 - 236.25

Amount of excess Carbon (ii) oxide left over = 23.75 g

5 0
3 years ago
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