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kobusy [5.1K]
3 years ago
7

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

at 407,0 m/s, what is its speed in yards per minute?
Chemistry
1 answer:
kaheart [24]3 years ago
7 0

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

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First, we have to get:

1- The heat required to increase T of ice from -50 to 0 °C:

according to q formula:

q1 = m*C*ΔT

when m is the mass of ice = mol * molar mass

                                             =  1 mol * 18 mol/g

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by substitution:

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2- the heat required to melt this mass of ice is :

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by substitution:

q2 = 1 mol * 6.01 KJ/mol

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3- the heat required to increase the water temperature from 0°C to 60 °C is:

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∴the total change of enthalpy = q1+q2+q3

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                                                  = 12.405 KJ


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