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If a bullet travels at 407.0 m/s, what is its speed in miles per hour? Number click to edit mi/h For the same bullet travelling

kaheart [24]

Answer : The speed in miles per hour is 22 mile/hr.

The speed in yard per min is 26617.8 yard/min

Explanation :

The conversion used for meters to miles is:

$1m=100cm\times \frac{1in}{2.54cm}\times \frac{1ft}{12in}\times \frac{1mile}{5280ft}$

The conversion used for second to hour is:

$1s=\frac{1}{60}min\times \frac{1hr}{60min}$

The conversion used for meter per second to mile per hour is:

$1\frac{m}{s}=\frac{100cm\times \frac{1in}{2.54cm}\times \frac{1ft}{12in}\times \frac{1mile}{5280ft}}{\frac{1}{60}min\times \frac{1hr}{60min}}$

As we are given the speed of 407.0 meter per second. Now we have to determine the speed in miles per hour.

$1m/s=2.2mile/hr$

So, $407.0m/s=\frac{2.2mile/hr}{1m/s}\times 407.0m/s=895.4mile/hr$

Therefore, the speed in miles per hour is 22 mile/hr.

The conversion used for meter to yard

1m = 1.09 yard

The conversion used for second to hour is:

$1s=\frac{1}{60}min$

The conversion used for meter per second to mile per hour is:

$1m/s=\frac{1.09yard}{frac{1}{60min}}$

$1m/s=65.4yard/min$

As we are given the speed of 407.0 meter per second. Now we have to determine the speed in yards per min

$1m/s=65.4yard/min$

So, $407.0m/s=\frac{65.4yard/min}{1m/s}\times 407.0m/s=26617.8yard/min$

Therefore, the speed in yard per min is 26617.8 yard/min

7 0

3 years ago

Give the % composition for each element found in tin (IV) nitride.

drek231 [11]

32.3699% Tin (Sn)

15.2774% Nitrogen (N)

52.3527% Oxygen (O)

4 0

3 years ago

The experiment above is repeated, but instead of extracting once with 6 mL the extraction is done three times with 2 mL of dichl

nevsk [136]

Complete Question

A student is extracting caffeine from water with dichloromethane. The K value is 4.6. If the student starts with a total of 40 mg of caffeine in 2 mL of water and extracts once with 6 mL of dichloromethane

The experiment above is repeated, but instead of extracting once with 6 mL the extraction is done three times with 2 mL of dichloromethane each time. How much caffeine will be in each dichloromethane extract?

Answer:

The mass of caffeine extracted is $P = 39.8 \ mg$

Explanation:

From the question above we are told that

The K value is $K = 4.6$

The mass of the caffeine is $m = 40 mg$

The volume of water is $V = 2 mL$

The volume of caffeine is $v_c = 2 mL$

The number of times the extraction was done is n = 3

Generally the mass of caffeine that will be extracted is

$P = m * [\frac{V}{K * v_c + V} ]^3$

substituting values

$P = 40 * [\frac{2}{4.6 * 2 + 2} ]^3$

$P = 39.8 \ mg$

6 0

3 years ago

Electron affinity is the measure of the attraction of an electron toward an isolated gaseous atom. When an electron is added to

zhuklara [117]

False

Explanation:

Electron affinity is negative when energy is absorbed and it is positive when energy is released.

Electron affinity is defined as the energy released in adding an electron to a neutral atom in the gas phase.

It is a measure of the readiness of an atom to gain an electron.

In a reaction where energy is released, electron affinity is usually positive. These reactions are called exothermic reactions.

Endothermic reactions in which energy is absorbed have negative electron affinity values.

Learn more:

Endothermic reactions brainly.com/question/12964401

#learnwithBrainly

6 0

3 years ago

Two methanol-water mixtures are contained in separate tanks. The first mixture contains 40.0 wt% methanol and the second contain

RoseWind [281]

Answer:

$M_{per}= 52.86$

$W_{per}=47.14$

Explanation:

<u>First mixture</u>:

40 wt% methanol - 60 wt% water 200 kg

$m_{met1}=200 kg * 0.4= 80 kg$

$m_{wat1}=200 kg * 0.6= 120 kg$

<u>Second mixture</u>:

70 wt% methanol - 30 wt% water 150 kg

$m_{met2}=150 kg * 0.7= 105 kg$

$m_{wat2}=150 kg * 0.3= 45 kg$

Final mixture:

$m_{metF=80 kg + 105 kg= 185 kg$

$m_{watF}=120 kg + 45 = 165 kg$

$M_{per}=\frac{185 kg}{185 kg + 165 kg}*100= 52.86$

$W_{per}=\frac{165 kg}{185 kg + 165 kg}*100=47.14$

If, the compositions are constant, the only variables are the mass of each mixture used in the final one, so there can be only one independent balance.

8 0

3 years ago