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bezimeni [28]
3 years ago
12

Convection is a mode of heat transfer for which types of objects? A. solids and liquids B. gases and solids C. gases only D. liq

uids and gases E. liquids only.
Physics
2 answers:
Airida [17]3 years ago
7 0

The correct answer to the question will be D). Liquids and gases.

EXPLANATION:

Convection is the type of mode of heat transfer in which there will be actual motion of particles from one part to the another part of a fluid.

In case of liquids, the particles which are present at the bottom of the container get heated up first and become lighter. The lighter particles will go up and the top heavier particles will move downward. Again the same process will be repeated. In this way, a convection cycle is produced.

For instance, heating of water.

In case of gases, same thing happens just like liquid. The air particles will be heated up first and becomes lighter. These heated particles will move upward. Due to this, an empty space will be created in that region. In order to occupy the empty space, the air will flow from another region which is at low temperature as compared to the heated region. Again, the same process will be repeated which will results into the formation of convection cycle .

Hence, convection occurs both in liquids and gases.



OLga [1]3 years ago
7 0
The heat transfer involving convection is a process that is dependent in the density gradients of substances. This is better suited for fluids which are composed of the liquids and gases. Hence, the answer for this item is most appropriately letter D. 
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3 0
3 years ago
Water falls without splashing at a rate of 0.200 L/s from a height of 3.60 m into a 0.730 kg bucket on a scale. If the bucket is
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Answer:

15.106 N

Explanation:

From the given information,

The weight of the bucket can be calculated as:

W_b = m_bg =  \\ \\  W_b = (0.730 \  kg) ( 9.80 \ m/s^2) \\ \\ W_b = 7.154 \ N

The mass of the water accumulated in the bucket after 3.20s is:

m_w= (0.20 \ L/s) ( 3.20)s

m _w=0.64 \ kg

To determine the weight of the water accumulated in the bucket, we have:

W_w = m_w g

W_w = ( 0.64  \ kg )(9.80\  m  \  /s^2)

W_w = 6.272 \ N

For the speed of the water before hitting the bucket; we have:

v = \sqrt{2gh}

v = \sqrt{2*9.80 \ m/s^2 * 3.60 \ m}

v = 8.4 m/s

Now, the force required to stop the water later when it already hit the bucket is:

F = v ( \dfrac {dm}{dt} )

F = (8.4 \ m/s)( 0.200 \ L/s)

F = 1.68 N

Finally, the reading scale is:

F_{scale = 7.154 N + 6.272 N + 1.68 N

= 15.106 N

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3 years ago
Read 2 more answers
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