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Rainbow [258]
3 years ago
8

Convert 93.6 miles per hour. Convert this to kilometers per hour.

Physics
1 answer:
Readme [11.4K]3 years ago
3 0

Answer:

150.6 km

Explanation:

One mile is about 1.61 km so multiply 93.6 by 1.6 which gives you above 150.6

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2 years ago
A large cylindrical tank contains 0.750 cubic meters of nitrogen gas at 27 degrees celsius and 1.5 e5 pa absolute pressure. the
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<span>3.36x10^5 Pascals The ideal gas law is PV=nRT where P = Pressure V = Volume n = number of moles of gas particles R = Ideal gas constant T = Absolute temperature Since n and R will remain constant, let's divide both sides of the equation by T, getting PV=nRT PV/T=nR Since the initial value of PV/T will be equal to the final value of PV/T let's set them equal to each other with the equation P1V1/T1 = P2V2/T2 where P1, V1, T1 = Initial pressure, volume, temperature P2, V2, T2 = Final pressure, volume, temperature Now convert the temperatures to absolute temperature by adding 273.15 to both of them. T1 = 27 + 273.15 = 300.15 T2 = 157 + 273.15 = 430.15 Substitute the known values into the equation 1.5E5*0.75/300.15 = P2*0.48/430.15 And solve for P2 1.5E5*0.75/300.15 = P2*0.48/430.15 430.15 * 1.5E5*0.75/300.15 = P2*0.48 64522500*0.75/300.15 = P2*0.48 48391875/300.15 = P2*0.48 161225.6372 = P2*0.48 161225.6372/0.48 = P2 335886.7441 = P2 Rounding to 3 significant figures gives 3.36x10^5 Pascals. (technically, I should round to 2 significant figures for the result of 3.4x10^5 Pascals, but given the precision of the volumes, I suspect that the extra 0 in the initial pressure was accidentally omitted. It should have been 1.50e5 instead of 1.5e5).</span>
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Calculate the amount of heat (in kj) required to raise the temperature of a 79.0 g sample of ethanol from 298.0 k to 385.0 k. th
earnstyle [38]
We are given with the specific heat capacity of ethanol, the mass of the sample and the temperature change to determine the total amount of heat to raise the temperature. The formula to be followed is H = mCpΔT. Upon subsituting, H = 79 g * 2.42 J/gC *(385-298)C = 16.63 kJ 
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