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Len [333]
3 years ago
7

The emissivity of galvanized steel sheet, a common roofing material, is 0.13 at temperatures around 300 K, while its absorptivit

y for solar irradiation is S 0.65. Would the neighborhood cat be comfortable walking on a roof constructed of the material on a day when GS 750 W/m2, T 16 C, and h 7 W/m2 K? Assume the bottom surface of the steel is insulated.
Physics
1 answer:
vfiekz [6]3 years ago
4 0

Answer:

Total heat transfer rate is combined convection and radiation transfer rate, multiplied with absorptivity factor. From that equation you can find temperature of the roof using iteration method.  

Explanation:

<u>Known:</u>

steel sheet roof emissivity ∈ = 0.13

steel sheet roof absorptivity \alpha_s = 0.65

air temperature T∞ = 16°C ...= 289K

heat transfer rate G_s = 750 W/m2

natural convection h = 7W/m2K

Total heat transfer rate is combined heat transfer rate of convection and radiation, multiplied with absorptivity factor:  

\alphaG_s=q_{conv} +q_{rad} =hA(T∞-T_m)+∈σA(T_{surf} -T_{s} )

\alphaG_s-h(T_s-T∞)-∈σ(T_{s} -T_{surf} )=0

2562-7T_s-7.371*10^-9=0

Use iteration method to solve for T_s

T_s=350.17 K...=77.17°C

In this particular situation, cat would not be comfortable with walking on a roof since its surface temperature is 77.17°C which may even cause burns to its paws.  

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soldi70 [24.7K]

B) The object's velocity doubled.

Explanation:

The graph is missing: find it in attachment.

The kinetic energy of an object is the energy possessed by the object due to its motion. It is calculated as

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is the velocity of the object

We notice that:

  • The  kinetic energy is directly proportional to the mass
  • The kinetic energy is proportional to the square of the velocity

In the graph, one of the two quantities (either mass or speed) is represented on the x-axis, while the quantity on the y-axis is the kinetic energy.

First of all, we notice that the relationship is not linear: this means that the quantity on the x-axis cannot be the mass, so it must be the velocity.

Moreover, we notice that when the quantity on the x-axis increases from 1 to 2 (so, it doubles), the kinetic energy increases by a factor of 4. This means that the object's velocity has doubled, therefore

B) The object's velocity doubled.

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3 years ago
List the types of electromagnetic radiation in order from lowest energy photons to highest energy photons.
frutty [35]

radio waves,X-rays,

Explanation:

In order from highest to lowest energy, the sections of the EM spectrum are named: gamma rays, X-rays, ultraviolet radiation, visible light, infrared radiation, and radio waves. Microwaves (like the ones used in microwave ovens) are a subsection of the radio wave segment of the EM spectrum.

8 0
2 years ago
A 2 kg rubber ball is thrown at a wall horizontally at 3 m/s, and bounces back the way it came at an equal speed. A 2 kg clay ba
Lyrx [107]

Answer:

THE RUBBER BALL

Explanation:

From the question we are told that

      The mass of the rubber ball is m_r   =  2 \ kg

      The  initial  speed of the rubber ball is  u =  3 \ m/s

      The final speed at which it bounces bank v  - 3 \ m/s

      The mass of the clay ball  is  m_c =  2  \ kg

       The  initial  speed of the clay  ball is u = 3 \ m/s

       The final speed of the clay ball is  v = 0 \  m/s

Generally Impulse is mathematically represented as

       I  =  \Delta p

where \Delta  p is the change in the linear momentum so  

       I  =  m(v-u)

For the rubber  is  

        I_r  =  2(-3 -3)

       I_r  = -12\ kg \cdot  m/s

=>     |I_r|  = 12\ kg \cdot  m/s

For the clay ball

       I_c  =  2(0-3)

        I_c =  -6 \ kg\cdot \ m/s

=>    | I_c| =  6 \ kg\cdot \ m/s

So from the above calculation the ball with the a higher magnitude of impulse is the rubber ball

       

8 0
3 years ago
Lab report 1.04 hypothesis
Zanzabum

Answer:a supposition or proposed explanation made on the basis of limited evidence as a starting point for further investigation.

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5 0
3 years ago
Write an email to a classmate explaining why the velocity of the current in a river has no FN C02-F20-OP11USB effect on the time
anzhelika [568]

The velocity of the current in a river has no effect on the the time it takes to paddle a canoe across the river, given that the boat is pointed perpendicular to the bank of the river, because

The velocity of the river does not change the velocity and therefore the distance traveled in the direction of the boat which is directed perpendicular to it

Reason:

Let \overset \rightarrow {v_y} represent the velocity of the boat across the river in the direction, \overset \rightarrow {d_y}, and let, \overset \rightarrow {v_x}, represent the velocity of the river, we have;

The velocity of the boat perpendicular to the direction of the river = \overset \rightarrow {v_y}

Therefore, the distance covered per unit time in the perpendicular direction

to the flow of the river is \overset \rightarrow {v_y}, such that the time it takes to cross the river in

the perpendicular direction is the same, for every value of the velocity of

the river.

This is so because the velocity in the perpendicular direction to the flow of

the river, which is the velocity of the boat is unchanged by the velocity of

the river, because there is no perpendicular component of velocity in the

velocity of the river.

\overset \rightarrow {v_y} = 3 m/s

\overset \rightarrow {v_x} = 4 m/s

The width of the river, w_y = 6 meters, we have;

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The direction, θ, is given as follows;

\theta = \arctan \left(\dfrac{4}{3} \right) \approx 53.13^{\circ}

The length of the path of the boat, <em>l</em>, is given as follows;

l = \dfrac{6}{cos \left(\arctan \left(\dfrac{4}{3} \right)\right)} = 10

The length of the path the boat takes = 10 m

The time it takes to cross the river, t = \dfrac{l}{v}, therefore;

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Considering only the y-components, we have;

t = \dfrac{w_y}{v_y}

Therefore;

t = \dfrac{6 \ m}{3 \ m/s} = 2 \, s

Which expresses that the time taken is the same and given that the

vectors of the velocities of the river and the boat are perpendicular, the

distance covered in the direction of the boat is unaffected by the velocity

of the river.

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