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Yuki888 [10]
3 years ago
8

You must signal _____ before any turn or lane change. A. 5 seconds B. 10 seconds C. 50 ft D. 100 ft

Engineering
2 answers:
Eddi Din [679]3 years ago
8 0

Answer:

Option D. We must signal 100 ft before any turn or lane change.

Explanation:

We must signal 100 ft before changing in turns or any lanes. Mainly we can use the proper indicator of our car to show which way we are trying to get the turn, so that the vehicles behind can understand.

If anyhow the indicators are not working properly, we should use our hands as indicator and the way of turning.  

Nezavi [6.7K]3 years ago
6 0
The answer is D! hope you pass
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List six possible valve defects that should be included in the inspection of a used valve?
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Answer:

Valvular stenosis , Valvular prolapse , Regurgitation,

Explanation:

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3 years ago
A rigid tank having 25 m3 volume initially contains air having a density of 1.25 kg/m3, then more air is supplied to the tank fr
Hoochie [10]

Answer:

\Delta m = 102.25\,kg

Explanation:

The mass inside the rigid tank before the high pressure stream enters is:

m_{o} = \rho_{air}\cdot V_{tank}

m_{o} = (1.25\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})

m_{o} = 31.25\,kg

The final mass inside the rigid tank is:

m_{f} = \rho \cdot V_{tank}

m_{f} = (5.34\,\frac{kg}{m^{3}} )\cdot (25\,m^{3})

m_{f}= 133.5\,kg

The supplied air mass is:

\Delta m = m_{f}-m_{o}

\Delta m = 133.5\,kg-31.25\,kg

\Delta m = 102.25\,kg

4 0
3 years ago
14. A large car fire presents the possibility of
dexar [7]

Answer:

Both of the above

Explanation:

5 0
2 years ago
Read 2 more answers
A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28
Delicious77 [7]

Answer:

5984.67N

Explanation:

A 14 inch diameter pipe is decreased in diameter by 2 inches through a contraction. The pressure entering the contraction is 28 psi and a pressure drop of 2 psi occurs through the contraction if the upstream velocity is 4.0 ft/sec. What is the magnitude of the resultant force (lbs) needed to hold the pipe in place?

from continuity equation

v1A1=v2A2

equation of continuity

v1=4ft /s=1.21m/s

d1=14 inch=.35m

d2=14-2=0.304m

A1=pi*d^2/4

0.096m^2

a2=0.0706m^2

from continuity once again

1.21*0.096=v2(0.07)

v2=1.65

force on the pipe

(p1A1- p2A2) + m(v2 – v1)

from bernoulli

p1 + ρv1^2/2 = p2 + ρv2^2/2

difference in pressure or pressure drop

p1-p2=2psi

13.789N/m^2=rho(1.65^2-1.21^2)/2

rho=21.91kg/m^3

since the pipe is cylindrical

pressure is egh

13.789=21.91*9.81*h

length of the pipe is

0.064m

AH=volume of the pipe(area *h)

the mass =rho*A*H

0.064*0.07*21.91

m=0.098kg

(193053*0.096- 179263.6* 0.07) + 0.098(1.65 – 1.21)

force =5984.67N

4 0
3 years ago
Two small balls A and B with masses 2m and m respectively are released from rest at a height h above the ground. Neglecting air
statuscvo [17]

Answer:

The kinetic energy of A is twice the kinetic energy of B

Explanation:

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3 years ago
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