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3 years ago

Based on Nia's notes, what would be the BEST way to investigate the heat transfer based on the movement of the molecules?

2 answers:

5 0

Answer: Nia could measure the temperature of the bottom floor of a house to see if the heat had risen in the house due to convection.

Explanation:

Anon25 [30]3 years ago

4 0

Answer:

Nia could measure the temperature of the bottom floor of a house to see if the heat had risen in the house due to convection.

Explanation:

Because energy transferred by the mass motion of molecules.

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A 20" round duct is 275' in length. The duct is carrying 2,900 CFM at a friction loss of 0.12 inches WG per 100 feet.

ivolga24 [154]

50500000 FACTION : uo/80

3 0

3 years ago

A wire loop of radius 0.37 m lies so that an external magnetic field of magnitude 0.35 T is perpendicular to the loop. The field

Galina-37 [17]

Answer:

168.57 mV

Explanation:

Initial magnetic flux = BA , B magnetic field and A is area of loop

= .35 x 3.14 x .37²

= .15 Weber

Final magnetic flux

= - .2 x 3.14 x .37²

= - .086 Weber

change in flux

.15 + .086

= .236 Weber

rate of change of flux

= .236 / 1.4

= .16857 V

= 168.57 mV

5 0

3 years ago

What is Marie's instantaneous speed at 20 minutes in miles/min?

AVprozaik [17]

Answer:

0.25miles/min

Explanation:

Instantaneous speed of a person or an object is its speed at a particular moment usually at a period of time.

The speedometer of a car reports the instantaneous speed.

It can be mathematically expressed as;

Instantaneous speed = $\frac{distance}{time}$

At 20min the distance covered is 5miles;

Instantaneous speed = $\frac{5 miles }{20mins}$ = 0.25miles/min

8 0

3 years ago

A 43.9-g piece of copper (CCu= 0.385 J/g°C) at 135.0°C is plunged into 254 g of water at 39.0°C. Assuming that no heat is lost t

Semmy [17]

Answer:

$T = 40.501\,^{\textdegree}C$

Explanation:

The interaction of the piece of copper and water means that the first one need to transfer heat in order to reach a thermal equilibrium with water. Then:

$-Q_{out,Cu} = Q_{in,H_{2}O}$

After a quick substitution, the expanded expression is:

$-(43.9\,g)\cdot (0.385\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-135^{\textdegree}C) = (254\,g)\cdot (4.187\,\frac{J}{g\cdot ^{\textdegree}C} )\cdot (T-39\,^{\textdegree}C)$