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3 years ago

11

While David was riding his bike around the circular cul-de-sac by his house he wondered if the constant circle motion was having

any affect on his tires. what would be the best way for David to investigate this?

1 answer:

Agata [3.3K]3 years ago

4 0

Measure the tire preasure

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Two tons of water is pumped into a tank 15 ft. High in 5 minutes. Determine the power output of the pump. (Note: One ton is equa

Semmy [17]

Answer:

200 lb•ft/s

Explanation:

From the question given above, the following data were obtained:

Force (F) = 2 tons

Time (t) = 5 mins

Height (h) = 15 ft

Power (P) =?

Next, we shall convert 2 tons to pound. This can be obtained as follow:

1 ton = 2000 lb

Therefore,

2 tons = 2 × 2000

2 tons = 4000 lb

Next, we shall convert 5 mins to seconds. This can be obtained as follow:

1 min = 60 s

Therefore,

5 mins = 5 × 60

5 mins = 300 s

Finally, we shall determine the power of the pump. This can be obtained as follow:

Force (F) = 4000 lb

Time (t) = 300 s

Height (h) = 15 ft

Power (P) =?

P = F × h / t

P = 4000 × 15 / 300

P = 60000 / 300

P = 200 lb•ft/s

Thus, the power of the pump is 200 lb•ft/s

5 0

2 years ago

Be sure to answer all parts. Consider the reaction A + B → Products From the following data obtained at a certain temperature, d

worty [1.4K]

Answer : The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+B\rightarrow Products$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b$

where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first observation:

$3.20\times 10^{-1}=k(1.50)^a(1.50)^b$ ....(1)

Expression for rate law for second observation:

$3.20\times 10^{-1}=k(1.50)^a(2.50)^b$ ....(2)

Expression for rate law for third observation:

$6.40\times 10^{-1}=k(3.00)^a(1.50)^b$ ....(3)

Dividing 1 from 2, we get:

$\frac{3.20\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(1.50)^a(2.50)^b}{k(1.50)^a(1.50)^b}\\\\1=1.66^b\\b=0$

Dividing 1 from 3, we get:

$\frac{6.40\times 10^{-1}}{3.20\times 10^{-1}}=\frac{k(3.00)^a(1.50)^b}{k(1.50)^a(1.50)^b}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[A]^1[B]^0$

$\text{Rate}=k[A]$

Thus,

The order of reaction with respect to A is, first order reaction.

The order of reaction with respect to B is, zero order reaction.

The overall order of reaction is, first order reaction.

7 0

3 years ago

Question: If red dye is added to cold water and blue dye is added to warm water, which dye will

vlada-n [284]

Answer:

Blue dye spreads out quicker

Explanation:

This is because the dye gets spread out by water molecules. Warm water has faster moving water molecules than cold water.

6 0

3 years ago

Read 2 more answers

There is a gas at 780 mm of Hg, in a volume of 5 liters and a temperature of 37 C, the volume is changed to 5.5 liters and the

HACTEHA [7]

Answer:

32.8 C

Explanation:

- Use combined gas law formula and rearrange.

- Hope that helped! Please let me know if you need further explanation.

6 0

3 years ago

How to prepare super saturated solution of copper sulphate ? Explain the experiment

expeople1 [14]

Explanation:

Monitor the temperature of the water with the thermometer. Stop heating the water once it nears the boiling point of 100 degrees Celsius. Add copper(II) sulfate and stir until the heated solution is saturated. When the solution is saturated, copper(II) sulfate will not dissolve anymore

6 0

2 years ago