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34kurt
1 year ago
12

A windowpane is half a centimeter thick and has an area of 1.0 m2. The temperature difference between the inside and outside sur

faces of the pane is 15° C.
What is the rate of heat flow through this window? (Thermal conductivity for glass is 0.84 J/s⋅m⋅°C.)
a. 50 000 J/sb. 2 500 J/sc. 1 300 J/sd. 630 J/s
Physics
1 answer:
polet [3.4K]1 year ago
3 0

To solve this problem it is necessary to apply the concepts related to the heat flux rate expressed in energetic terms. The rate of heat flow is the amount of heat that is transferred per unit of time in some material. Mathematically it can be expressed as:

\frac{Q}{t} = \frac{kA}{L} (T_H - T_C)

Where

k = 0.84 J/s⋅m⋅°C (The thermal conductivity of the material)

A = 1m^2 Area

L = 5*10^{-3}m Length

T_H= Temperature of the "hot"reservoir

T_C= Temperature of the "cold"reservoir

Replacing with our values we have that,

\frac{Q}{t} = \frac{kA}{L} (T_H - T_C)

\frac{Q}{t} = \frac{(0.84)(1)}{0.005} (15)

\frac{Q}{t} = 2520J/s

Therefore the correct answer is B.

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In mammals, the weight of the heart is approximately 0.5% of the total body weight. Write a linear model that gives the heart we
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Answer:

The weight of heart of a human is 0.93 lbs.

Explanation:

Given that,

Approximately weight of heart is 0.5 % of the total body weight.

Weight of human = 185 lbs

Let the the weight of total body is w and weight of heart is w_{h}.

We need to calculate the weight of heart of a human

Using given data

w_{h}=0.5\times w

Where, h = weight of heart

w = weight of human

w_{h}=\dfrac{0.5}{100}\times 185

w_{h}=0.93\ lbs

Hence, The weight of heart of a human is 0.93 lbs.

8 0
1 year ago
A stable orbit is an orbit that repeats indefinitely, without ever changing shape. After running several simulations with differ
Tresset [83]

Answer:

All of the orbits were in the shape of an ellipse, with the orbited body on the inside of the ellipse.

Explanation:

3 0
1 year ago
7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10o ramp. If the coefficient of friction is 0.02, wh
pentagon [3]

Answer:

The frictional force needed to overcome the cart is 4.83N

Explanation:

The frictional force can be obtained using the following formula:

F= \mu R

where \mu is the coefficient of friction = 0.02

R = Normal reaction of the load = mgcos\theta = 25 \times 9.81 \times cos 10 = 241.52N

Now that we have the necessary parameters that we can place into the equation, we can now go ahead and make our substitutions, to get the value of F.

F=0.02 \times 241.52N

F = 4.83 N

Hence, the frictional force needed to overcome the cart is 4.83N

4 0
2 years ago
Charina says that when waves interact with an object, they will interfere with the object, and that when waves interact with oth
s344n2d4d5 [400]
No, she has it backward.  Waves interfere with each other and reflect off objects.  When two waves overlap their amplitudes add.  If they have the same sign this addition is constructive, meaning the amplitudes grow.  If they have opposite signs this constitutes subtraction and the waves can partially, or completely cancel.  This is known as interference.  Reflection occurs when waves travel from one medium to another.  If the wave impedance of the new medium is different (which it generally is) there will be a partial, or even total, reflection.  
7 0
1 year ago
Read 2 more answers
A metal sphere of radius 2.0 cm carries an excess charge of 3.0 μC. What is the electric field 6.0 cm from the center of the sph
Nonamiya [84]

Answer:

The electric field is  E = 7.5 *10^{6} \ N/C

Explanation:

From the question we are told that

    The radius of the metal sphere is  R = 2.0 \ cm  =  0.02 \ m

     The excess charge which the metal sphere carries is  q =  3.0 \mu C  =  3.0*10^{-6} \ C

      The distance of the position being to the center is D = 6.0 \ cm  = 0.06 \ m

       The coulomb constant is   k =9*10^{9} \  N \cdot m^2 /C^2

Generally the electric field is mathematically represented as  

        E = \frac{k *  q}{D^2}

substituting values

        E = \frac{9*10^{9} *  30.*10^{-6}}{(0.06)^2}

      E = 7.5 *10^{6} \ N/C

5 0
1 year ago
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