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Evgesh-ka [11]
2 years ago
3

Driver a is cruising along enjoying the fall colors. driver b starts her car at the instant he passes her. their velocities are

shown as functions of time in the graph below. at what instants in time on the graph are drivers a and b side by side?
Physics
2 answers:
LenaWriter [7]2 years ago
8 0

When you look far ahead as you drive, you are looking down at the area just in front of your vehicle. In driving, you do not make steering adjustments based on what is in front of you, and that is looking down at the road. If you do this, your driving will be smoother and safer.

masya89 [10]2 years ago
5 0
Their velocities will be the same at the instant their lines on the graph intersect. 

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Technician a says that using a pressure transducer and lab scope is a similar process to using a vacuum gauge. technician b says
Phantasy [73]

Answer: Both Technician A and B

Explanation:

There is a similar process in using a pressure transducer and lab scope to using a vacuum gauge.

And also, the pressure transducer can be used to tie any issues to individual cylinders if paired with a second trace consisting of the ignition pattern. Therefore, both Technician A and B are correct.

7 0
2 years ago
If the rocket has an initial mass of 6300 kg and ejects gas at a relative velocity of magnitude 2000 m/s , how much gas must it
Rzqust [24]

Answer:

The amount of gas that is to be released in the first second in other to attain an acceleration of  27.0 m/s2  is

      \frac{\Delta m}{\Delta t}   = 83.92 \ Kg/s

Explanation:

From the question we are told that

   The mass of the rocket is m = 6300 kg

   The velocity at gas is being ejected is  u =  2000 m/s

    The initial acceleration desired is a =  27.0 \  m/s

   The time taken for  the gas to be ejected is  t = 1 s

Generally this desired acceleration is mathematically represented as

        a = \frac{u *  \frac{\Delta m}{\Delta t} }{M -\frac{\Delta m}{\Delta t}* t}

Here \frac{\Delta m}{\Delta  t }  is the rate at which gas is being ejected with respect to time

Substituting values

      27 = \frac{2000 *  \frac{\Delta m}{\Delta t} }{6300 -\frac{\Delta m}{\Delta t}* 1}

=>   170100 -27* \frac{\Delta m}{\Delta t} = 2000 *  \frac{\Delta m}{\Delta t}

=>   170100  = 2027 *  \frac{\Delta m}{\Delta t}

=>   \frac{\Delta m}{\Delta t}   = \frac{170100}{2027}

=>   \frac{\Delta m}{\Delta t}   = 83.92 \ Kg/s

     

3 0
2 years ago
Phyllis had always been the one to take care of Douglas, now he feels as if it is his duty and privilege to take care of her, ev
a_sh-v [17]
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7 0
2 years ago
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Two identical objects A and B fall from rest from different heights to the ground. If object B takes twice as long as A to reach
aivan3 [116]
I believe this ratio is 4:1 due to the inverse square law
4 0
2 years ago
Read 2 more answers
refrigerant 134a enters a compressor operating at steady state as saturated vapor at 0.12 MPa and exits at 1.2 MPa and 70 C at a
Afina-wow [57]

Answer:

the power input to the compressor is 7.19Kw

Explanation:

Hello!

To solve this problem follow the steps below.

1. We will call 1 the refrigerant state at the compressor inlet and 2 at the outlet.

2. We use thermodynamic tables to determine enthalpies in states 1 and 2.

(note: Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)  

through prior knowledge of two other properties such as pressure and temperature.  )

h1[quality=1, P=0.12Mpa)=237KJ/Kg

h2(P=1.2Mpa, t=70C)=300.6KJ/kg

3. uses the first law of thermodynamics in the compressor that states that the energy that enters a system is the same that must come out

Q=heat=0.32kJ/s

W=power input to the compressor

m=mass flow=0.108kg/S

m(h1)+W=Q+m(h2)

solving for W

W=Q+m(h2-h1)

W=0.32+0.108(300.6-237)=7.19Kw

the power input to the compressor is 7.19Kw

7 0
2 years ago
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