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faltersainse [42]
3 years ago
9

Which pair of sentences is describing the same velocity? A car is parked. A car is moving in circles. A bus drives 40 miles per

hour north. A truck drives 50 miles per hour north. A boy walks east at 6 kilometers per hour. A girl walks west at the same speed as the boy. Valerie rides her bike south at 8 kilometers per hour. Owen jogs south at 8 kilometers per hour.
Physics
2 answers:
olya-2409 [2.1K]3 years ago
8 0
The bus and the truck have the same velocity.

Also, Valerie and Owen have the same velocity.
Phoenix [80]3 years ago
7 0

Velocity is a vector, so it contains two pieces of information:

- a magnitude, which is given by the speed

- a direction

Two objects, in order to have same velocity, should have both same speed and same direction. In this problem, only the following pair of sentences describe people with same velocity:

Valerie rides her bike south at 8 kilometers per hour.

Owen jogs south at 8 kilometers per hour.

In fact, both Valerie and Owen are moving at same speed (8 km per hour) in the same direction (south), so they have same velocity.

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Explanation:

Given\\V=3ft^3\\T_{1} =300^oF\\T_{L} =400^oF\\P_{L} =200psia\\V_{f} =0.5V

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a) Taking tank as a system, The energy balance can be define as

E_{in}- E_{out}= E_{sys}

m_{i} h_{L} -Q_{out}=m_{2}  u_{2}-m_{1}  u_{1}

The mass balance could be written as

m_{in}- m_{out}= m_{sys} \\m_{i}= m_{2}- m_{1}

The final pressure in the tank could be defined as following

P_{2} =P_{sat300^oF}

from standard steam table we know at

T_{2}= T_{1}=300^oF\\ P_{2}=67.028psia

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From steam table at

T_{1} =300^oF\\v_{1}= v_{g}=6.4663ft^3/lbm\\v_{1}=  v_{g}=1099.8Btu/lbm\\ v_{f} =0.01745ft^3/lbm\\v_{f} =269.51Btu/lbm\\

at\\P_{L} =200psia\\T_{L}=400^oF\\h_{L}=1210.9Btu/lbm

initial mass in the tank could be define as

m_{1}=\frac{V}{v_{1} }  \\m_{1} =\frac{3}{6.4663} =0.464lbm\\

Final mass in the tank could be define as

m_{2}=\frac{V_{f} }{v_{f} }+\frac{V_{g} }{v_{g} }  \\ m_{2} =\frac{1.5}{0.01745} +\frac{1.5}{6.4663} =86.2lbm

The amount of steam that has entered the tank

m_{i}=m_{2}-  m_{1}\\ m_{i}=86.2-0.464=85.74lbm

c)

The internal energy in final state could be defined as following

U_{2}=m_{f}  u_{f}+ m_{g} u_{g}\\ U_{2} =85.96*269.51+0.232*1099.8=23422Btu\\

The heat transfer could be defined as following

Q_{out}=m_{i}  h_{L}+m_{1}  u_{1}-m_{2}  u_{2}\\ Q_{out}=85.74*1210.9+0.464*1099.8-23422=80910Btu

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