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ad-work [718]
3 years ago
14

On a road trip, a driver achieved an average speed of (48.0+A) km/h for the first 86.0 km and an average speed of (43.0-B) km/h

for the remaining 54.0 km. What was her average speed (in km/h) for the entire trip? Round your final answer to three significant figures. A = 15 B = 1
Physics
1 answer:
Bond [772]3 years ago
4 0
A = 15, B = 1
48 + 15 = 63 for the first 86km
43 - 1 = 42 for the remaining 54 km
time = distance/speed
t1 = 86/63 = 1.3651hours
t2 = 54/42 = 1.2857hours
speed = distance/time
s = 140/2.6508
average speed = 52.8km/h
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The vacuum pressure of a condenser is given to be 80 kpa. if the atmospheric pressure is 98 kpa, what is the gage pressure and a
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The absolute pressure is given by the equation,

P_{abs}=P_{atm}-P_{vac}

Here, P_{abs} is absolute pressure,P_{atm} is atmospheric pressure and P_{vac} is vacuum pressure.

Therefore,

P_{abs}=98 kPa-80 kPa=18kPa

The gage pressure is given by the equation,

P_{gage}=P_{abs}-P_{atm}.

Thus,

P_{gage}=18kPa-98 kPa=-80 kPa.

In kn/m^2,

The absolute pressure,

P_{abs}=18kPa(\frac{1kN/m^2}{kPa}) =18\ kN/m^2

The gage pressure,

P_{gage}=-80kPa(\frac{1kN/m^2}{kPa}) =-80\ kN/m^2.

In lbf/in2

The absolute pressure,

P_{abs}=18\ kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=2.6\ lbf/in^2

The gage pressure,

P_{gage}=-80kPa(\frac{1.45\times 10^{-1}\ lbf/in^2 }{1kPa} )=-11.6\ lbf/in^2

In psi,

The absolute pressure,

P_{abs}=18\ kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa})=2.610\ psi.

The gage pressure,

P_{gage}=-80kPa(\frac{1.45037738\times 10^{-1}\ psi }{1kPa} )=-11.6030\ psi

In  mm Hg

The absolute pressure,

P_{abs}=18kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})= 135\ mm\ of\ Hg

The gage pressure,

P_{gage}=-80kPa(\frac{7.5\ mm\ of\ Hg }{1\ kPa})=-600\ mm\ of\ Hg



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Answer

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