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Rainbow [258]
3 years ago
8

Please hurry i’ll give you 15 points

Engineering
1 answer:
ki77a [65]3 years ago
7 0

Answer:

measures, dissolves, liquid, carbon dioxide, evaporates, water vapor, mold, decompose

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Please help me with this, picture.
Alenkasestr [34]
Maybe try 086 degrees
3 0
3 years ago
To measure the voltage drop across a resistor, a _______________ must be placed in _______________ with the resistor. Ammeter; S
gulaghasi [49]

Answer:

The given blanks can be filled as given below

Voltmeter must be connected in parallel

Explanation:

A voltmeter is connected in parallel to measure the voltage drop across a resistor this is because in parallel connection, current is divided in each parallel branch and voltage remains same in parallel connections.

Therefore, in order to measure the same voltage across the voltmeter as that of the voltage drop across resistor, voltmeter must be connected in parallel.

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3 years ago
Q-) please give me a reference about Tack coat? Pleae i need it please??!!
Arturiano [62]

Answer:

Tack coat is a sprayed application of an asphalt binder upon an existing asphalt or Portland cement concrete pavement prior to an overlay, or between layers of new asphalt concrete.

Explanation:

4 0
3 years ago
True or False: The differential lock in an AWD-equipped vehicle can be used at any time.
Bingel [31]

the answer would be false

7 0
3 years ago
Read 2 more answers
Water flows through a pipe at an average temperature of T[infinity] = 70°C. The inner and outer radii of the pipe are r1 = 6 cm
Paul [167]

Answer:

The differential equation and the boundary conditions are;

A) -kdT(r1)/dr = h[T∞ - T(r1)]

B) -kdT(r2)/dr = q'_s = 734.56 W/m²

Explanation:

We are given;

T∞ = 70°C.

Inner radii pipe; r1 = 6cm = 0.06 m

Outer radii of pipe;r2 = 6.5cm=0.065 m

Electrical heat power; Q'_s = 300 W

Since power is 300 W per metre length, then; L = 1 m

Now, to the heat flux at the surface of the wire is given by the formula;

q'_s = Q'_s/A

Where A is area = 2πrL

We'll use r2 = 0.065 m

A = 2π(0.065) × 1 = 0.13π

Thus;

q'_s = 300/0.13π

q'_s = 734.56 W/m²

The differential equation and the boundary conditions are;

A) -kdT(r1)/dr = h[T∞ - T(r1)]

B) -kdT(r2)/dr = q'_s = 734.56 W/m²

6 0
3 years ago
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