3 years ago

How do u walk dogs and they have to park in the car

Engineering

1 answer:

alexandr1967 [171]3 years ago

6 0

Answer: walk dog on leash for one and two you csnt park in a car you park the car and walk the dog on a leash

Explanation: did you phrase the question wrong?

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What flow type occurs when the heat exchanger has multiple passes and baffles?

Vanyuwa [196]

I think that the answer would be B or C

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2 years ago

The observations of 124.53, 124.55, 142.51, and 124.52 are obtained when taping the length of a line. What should the observer c

Nesterboy [21]

Answer:

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Explanation:

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2 years ago

Technician A says that the original equipment manufacturer (OEM) offer scan tools to service their specific line of vehicles. Te

Anastasy [175]

Answer:

a. Technician A

Explanation:

Aftermarket parts are replacement parts that are not made by the original equipment manufacturer. They can not guarantee as much functionaluty when compared to OEMs. Original Equipment Manufacturer (OEM) usually their own proprietary scan tools which offer better and functionality to service their special line of vehicles. So Technician A alone is correct.

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3 years ago

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An electric dipole is made of two charges of equal magnitudes and opposite signs. The positive charge, q=1.0 μC, is located at t

vfiekz [6]

Answer:

work done by electric field is 0.06 J

Explanation:

Given data:

Two point charge is $+ 1\mu C and -1 \mu C$

0+1 charge positioned is (0 cm , 1 cm, 0.00 cm)

-1 charge positioned is (0 cm , -1 cm, 0.00 cm)

E = 3.0\times 10^6 N/C

From above information, the distance between given two charges d = 2 cm

then d = 0.02m

work needed is W = q E d

$W = 1.0 \times 10^{-6} \times 3.0 \times 10^6 \times 0.02$

W = 0.06 J

Therefore work done by electric field is 0.06 J

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2 years ago

Luids of viscosities 1  =0.1 N.s/ms and 2  =0.15 N.s/m2 are contained between two plates (each plate 1

Fantom [35]

Answer:

a) 1 / 3 N

b) 5/3 m/s

Explanation:

For constant speed V at the interface of the two fluids the net force is zero.

Hence, F top = F bottom thus the shear stresses at top and bottom must be the same.

$t_{top} = t_{bottom}\\where\\t = u.\frac{dU}{dt} \\Hence\\(\frac{V - 1}{h_{2} })*u_{top} = (\frac{V}{h_{1} })*u_{bottom} \\\\0.3V = 0.5\\\\V = 5/3 m/s$

For force of top plate

$F_{top} = u_{top} * \frac{V-U}{h_{2} }*A=( 0.15)*(\frac{5/3 -1}{0.3 } )*(1)\\F_{top} = \frac{1}{3} N$

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3 years ago