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KIM [24]
3 years ago
10

A car is stopped at an entrance ramp to a freeway; its driver is preparing to merge. At a certain moment while stopped, this dri

ver observes a platoon of vehicles a distance x0 upstream and initiates the merge maneuver. The platoon approaches the entrance ramp at a constant speed v. The stopped car can accelerate from speed 0 to v at uniform acceleration rate a .
Find the latest time at which the stopped car can safely start the merge maneuver; i.e., derive an expression for the "latest safe start time" in terms of the variables x0, v and a. Assume this latest time enables the platoon to maintain its constant speed and "just touch" the merging car's trajectory. Ignore the physical dimensions of the car.
Engineering
1 answer:
Sophie [7]3 years ago
4 0

Answer:

T = \sqrt{\frac{2X_{0} }{a} }

Explanation:

Given,

Initial Velocity, u=0

As platoon is moving with constant velocity v,

Final Velocity, v=v

The vehicle starts from 0 to v at constant acceleration of a,

Relevent expressions:

v=u+at...........................(1)

v2=u2+2as.....................(2)  

V^{2} = 2aS, as S = X_{0},

V^{2} = 2aX_{0},

From(1)

v=at

Hence

(at)^2=2aX_{0}

T^{2}   =  \frac{2aX_0}{a^{2} }

T = \sqrt{\frac{2X_{0} }{a} }

This is the final expression for time.

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A manometer containing a fluid with a density of 60 lbm/ft3 is attached to a tank filled with air. If the gage pressure of the a
8090 [49]

Answer:

The fluid level difference in the manometer arm = 22.56 ft.

Explanation:

Assumption: The fluid in the manometer is incompressible, that is, its density is constant.

The fluid level difference between the two arms of the manometer gives the gage pressure of the air in the tank.

And P(gage) = ρgh

ρ = density of the manometer fluid = 60 lbm/ft³

g = acceleration due to gravity = 32.2 ft/s²

ρg = 60 × 32.2 = 1932 lbm/ft²s²

ρg = 1932 lbm/ft²s² × 1lbf.s²/32.2lbm.ft = 60 lbf/ft³

h = fluid level difference between the two arms of the manometer = ?

P(gage) = 9.4 psig = 9.4 × 144 = 1353.6 lbf/ft²

1353.6 = ρg × h = 60 lbf/ft³ × h

h = 1353.6/60 = 22.56 ft

A diagrammatic representation of this setup is presented in the attached image.

Hope this helps!

5 0
3 years ago
A gas flows through a one-inlet, one-exit control volume operating at steady state. Considering an adiabatic control volume with
Hunter-Best [27]

Answer:

b. equal to the specific entropy of the gas at the inlet.

Explanation:

Isentropic process is the process in which the entropy of the system remains unchanged. The word isentropic is formed from the combination of the prefix "iso" which means "equal" and the word entropy.

If a process is completely reversible, without the need to provide energy in the form of heat, then the process is isentropic.

3 0
3 years ago
An npn BJT has emitter, base, and collector doping levels of 1019 cm????3, 5 1018 cm????3, and 1017 cm????3, respectively. It is
Darina [25.2K]

Answer:

Explanation:

The answer to the given problem is been solved in the fine attached below.

8 0
3 years ago
Consider three branch prediction schemes: predict not taken, predict taken, and dynamic prediction. Assume that they all have ze
SOVA2 [1]

Answer:

Check the explanation

Explanation:

1. When a branch is taken with 5% frequency, the branch prediction scheme is "predict taken" because the process has to begin fetching and then execute at the target address.

2. When a branch is taken with 95% frequency, the branch prediction scheme is "predict not taken" because the branch outcome is definitely known.

3. When a branch is taken with 70% frequency, the prediction scheme is "dynamic prediction" the branch prediction can dynamically change during the program execution.

3 0
3 years ago
A fuel gas containing 45.00 mole% methane and the balance ethane is burned completely with pure oxygen at 25.00 degree C, and th
Dennis_Churaev [7]

Answer:

A)

- Q ( kw ) for vapor =  -1258.05 kw

- Q ( kw ) for liquid = -1146.3 kw

B )

- Q ( kj ) for vapor  = -1258.05 kJ

- Q ( KJ ) for liquid = - 1146.3 KJ

Explanation:

Given data :

45.00 % mole of methane

55.00 % of ethane

attached below is a detailed solution

A) calculate - Q(kw)

- Q ( kw ) for vapor =  -1258.05 kw

- Q ( kw ) for liquid = -1146.3 kw

B ) calculate  - Q ( KJ )

- Q ( kj ) for vapor  = -1258.05 kJ

- Q ( KJ ) for liquid = - 1146.3 KJ

since combustion takes place in a constant-volume batch reactor

7 0
3 years ago
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