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agasfer [191]
3 years ago
15

Which of the following processes DOES NOT occur during the convective transfer of heat within a container of air? Group of answe

r choices The cooler portion of the air surrounding a warmer part exerts a buoyant force on it. The flow of air molecules results in a flow of heat. As the warmer part of the air moves, it is replaced by cooler air that is subsequently warmed. A continuous flow of warmer and cooler parts of air is established. The volume of a warmed part of the air is reduced and its density increases.
Physics
1 answer:
vovangra [49]3 years ago
7 0

The volume of a warmed part of the air is reduced and its density increases.

Explanation:

In a convective form of heat transfer, the volume of a warmed part of air is not reduced and its density does not increase.

During convection, heat causes the warm part of the air to expand and its volume increases. When volume increases, density is reduced.

  • Convection is a form of heat transfer that involves the actual movement of particles of the medium.
  • It usually occurs in fluids i.e gases and liquids.
  • In convection, the cold part exerts a buoyant force on the warmer air below and causes it to rise.
  • As the warmer part is rising the cooler part replaces it and a convective cell is formed in the process.

Learn more:

Energy transfer in the sun brainly.com/question/1140127

#learnwithBrainly

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An object experiences a net acceleration to the left. Which of the following statements about this object are true? There may be
dedylja [7]

Answer:

  1. When an object experiences acceleration to the left, the net force acting on this object will also be to the left.
  2. If the mass of the object was doubled, it would experience an acceleration of half the magnitude

Explanation:

When an object experiences acceleration to the left, the net force acting on this object will also be to the left.

From Newton's second law of motion, the acceleration of the object is given as;

a = ∑F / m

a = -F / m

The negative value of "a" indicates acceleration to the left

where;

∑F is the net force on the object

m is the mass of the object

At a constant force, F = ma ⇒ m₁a₁ = m₂a₂

If the mass of the object was doubled, m₂ = 2m₁

a₂ = (m₁a₁) / (m₂)

a₂ = (m₁a₁) / (2m₁)

a₂ = ¹/₂(a₁)

Therefore, the following can be deduced from the acceleration of this object;

  1. When an object experiences acceleration to the left, the net force acting on this object will also be to the left.
  2. If the mass of the object was doubled, it would experience an acceleration of half the magnitude

6 0
3 years ago
A block with mass m1 = 4.50 kg and a ball with mass m2 = 7.70 kg are connected by a light string that passes over a frictionless
Allisa [31]
1.6a =  \frac{g(m_2 + m_3 - \mu km_1)}{m_1 + m_2 + m_3}  \\  \\ 1.6a(m_1 + m_2 + m_3) = g(m_2 + m_3 - \mu km_1) \\  \\ (1.6a + \mu kg)m_1 + (1.6a - g)m_2 = (g - 1.6a)m_3 \\  \\ m_3 =  \frac{1.6a +\mu kg}{g - 1.6a} m_1 - m_2 \\  \\ m_3 = 22.57 kg
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3 years ago
Two point charge
timama [110]

Answer:

Showing results for Two point charge q, separated by 1.5cm have change value of +2.0 and -4.0AND/C respectively what is the magnitude of the Electric force midway between them?

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4 0
1 year ago
Khaled said the North star is special because it appears all over the world and its changes position.
saw5 [17]
False the North Star never changes it position
3 0
2 years ago
A bicycle tire has a pressure of 7.00×105 N/m2 at a temperature of 18.0ºC and contains 2.00 L of gas. What will its pressure be
stepan [7]

Answer:

p_2 = 664081 N/m^{2}

Explanation:

from the ideal gas law we have

PV = mRT

P = \rho RT

\rho = \frac{P}{RT}

HERE  R is gas constant for dry air  =  287  J K^{-1} kg^{-1}

\rho = \frac{7.00 10^{5}}{287(18+273)}

\rho = 8.38 kg/m^{3}

We know by ideal gas law

\rho = \frac{m_1}{V_1}

m_1 = \rho V_1 = 8.38 *2*10^{-3}

m_1 = 0.0167 kg

for m_2

m_2 = \rho V_i - V_removed

m_2 = 8.38*(.002 - 10^{-4})

m_2 = 0.0159 kg

WE KNOW

PV = mRT

V, R and T are constant therefore we have

P is directly proportional to mass

\frac{p_2}{p_1}=\frac{m_2}{m_1}

p_2 = p_1 * \frac{m_2}{m_1}

p_2 =7*10^{5} * \frac {.0159}{0.0167}

p_2 = 664081 N/m^{2}

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