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Anna71 [15]
3 years ago
13

The earth's interior heat and pressure drive a process of heat transfer in the mantle called _____.

Physics
2 answers:
Galina-37 [17]3 years ago
5 0
Convection. It is the transfer of heat by the circulation and movement of the molecules that come from liquid or gas.
almond37 [142]3 years ago
5 0
<h3><u>Answer;</u></h3>

Convection

The earth's interior heat and pressure drive a process of heat transfer in the mantle called <em><u>convection</u></em>.

<h3><u>Explanation;</u></h3>
  • <em><u>Convection is the process of heat transfer that occurs in fluid through convection current. This process may occur in a natural setting such as in the earth's mantle.</u></em>
  • <em><u>The plate tectonics in the earth surface are driven by the convection currents in the magma. The intense heat in the core of the earth causes molten rocks in the mantle to move in a pattern called convection cell that results from the cooling of the rising denser warm material which eventually sink.</u></em>
  • This transfer of the heat from the mantle to the surface of the earth by convection is similar to the hot water, which involves convection currents.
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Answer:

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As per the question:

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Angle, \theta = 37.0^{\circ}

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(a) The magnitude of the normal force exerted by the slide on the child:

F_{N} = mgcos\theta

F_{N} = 22\times 9.8cos37^{\circ} = 172.185\ N

Now,

(b) The angle from the horizontal at which the force is directed is:

90^{\circ} - 37^{\circ} = 53^{\circ}

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Chapter 16, Problem 63. A person is standing in a room at 18 ◦C. The exposed surface area and skin temperature of the person are
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Answer:Q=248.011 W

Explanation:

Given

Temperature of Room T_{\infty }=18^{\circ}\approx 291 K

Area of Person A=1.7 m^2

Temperature of skin T=32^{\circ}\approx 305 K

Heat transfer coefficient h=5 W/m^2.k

Emissivity of the skin and clothes \epsilon =0.9

\Delta T=32-18=14^{\circ}

Total rate of heat transfer=heat Transfer due to Radiation +heat transfer through convection

Heat transfer due radiation Q_1=\epsilon \sigma A(T^4-T_{\infty }^4 )

where \sigma =stefan-boltzman\ constant

Q_1=0.9\times 5.687\times 10^{-8}\times 1.7\times (305^4-291^4)

Q_1=129.01 W

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7 0
3 years ago
P6: An object of mass m sits on a spring of constant k in an elevator that is accelerating upwards with acceleration a. a) In te
tankabanditka [31]

Answer:

(a). The spring compressed is \dfrac{ma+mg}{k}.

(b). The acceleration is 1.5 g.

Explanation:

Given that,

Acceleration = a

mass = m

spring constant = k

(a). We need to calculate the spring compressed

Using balance equation

kx-mg=ma

x=\dfrac{ma+mg}{k}....(I)

The spring compressed is \dfrac{ma+mg}{k}.

(b). If the compression is 2.5 times larger than it is when the mass sits in a still elevator,

The compression is given by

x=2.5\times x_{0}

Here, acceleration is zero

So, x=2.5\times\dfrac{mg}{k}

We need to calculate the acceleration

Put the value of x in equation (I)

2.5\times \dfrac{mg}{k}=\dfrac{ma+mg}{k}

2.5\times\dfrac{mg}{k}=\dfrac{m}{k}(a+g)

a=2.5g-g

a=1.5g

Hence, (a). The spring compressed is \dfrac{ma+mg}{k}.

(b). The acceleration is 1.5 g.

8 0
3 years ago
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