Ask question

Ask question

All categories

3 years ago

8

What happens when a negatively charged object A is brought near a neutral object B?

2 answers:

Andrews [41]3 years ago

6 0

Object B gets a positive charge

Eva8 [605]3 years ago

5 0

<span>Object B stays neutral but becomes polarized.

Hope this helps.

~Jurgen</span>

You might be interested in

HELP ASAP PLEASE!!!!!!!

pshichka [43]

Answer:a

Explanation: the plates shifts cracking making earthquakes

8 0

2 years ago

What are four pieces of evidence for continental drift?

Tomtit [17]

continents, paleoclimate indicators, truncated geologic features, and fossils:D

5 0

3 years ago

Read 2 more answers

How does the structure of the stigma aid in pollination

natali 33 [55]

<span>Pollination is the process by which pollen is transferred to the female part of a plant. The stigma is the central part of the flower, that is supported by a style and is part of the female reproductive organ within plants. Its structure is optimised to promote pollination by having hairs, sticky surfaces and three-dimensional sculptures that capture and trap pollen.</span>

5 0

3 years ago

What is the equation that defines work

mina [271]

Work=force*displacment

3 0

3 years ago

Water flows through a cast steel pipe (k = 50 W m.K, ε = 0.8) with an outer diameter of 104mm and 2 mm wall thickness. Calculate

masha68 [24]

Answer:

The heat loss per unit length is $\frac{Q}{L} = 2981 W/m$

Explanation:

From the question we are told that

The outer diameter of the pipe is $d = 104mm = \frac{104}{1000} = 0.104 m$

The thickness is $D = 2mm = \frac{2}{1000} = 0.002m$

The temperature of water is $T = 90^oC = 90 + 273 = 363K$

The outside air temperature is $T_a = -10^oC = -10 +273 = 263K$

The water side heat transfer coefficient is $z_1 = 300 W/ m^2 \cdot K$

The heat transfer coefficient is $z_2 = 20 W/m^2 \cdot K$

The heat lost per unit length is mathematically represented as

$\frac{Q}{L} = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1} + \frac{ln [\frac{d}{D} ]}{z_2}}$

Substituting values

$\frac{Q}{L} = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300} + \frac{ln [\frac{0.104}{0.002} ]}{20}}$

$\frac{Q}{L} = \frac{628}{0.2107}$

$\frac{Q}{L} = 2981 W/m$

6 0

3 years ago