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disa [49]
3 years ago
7

The inner and outer surfaces of a 5m x 6m brick wall of thickness 30 cm and thermal conductivity 0.69 w/m.0 c are maintained at

temperature of 20 0 c and 50 c, respectively. determine the rate of heat transfer through the wall, in w.
Physics
1 answer:
Volgvan3 years ago
4 0
The working equation to be used here is written below:

Q = kA(T₁ - T₂)/Δx
where
Q is the rate of heat transfer
k is the heat transfer coefficient
A is the cross-sectional area of the wall
T₁ - T₂ is the temperature difference between the sides of the wall
Δx is the thickness of the wall

The solution is as follows:

Q = (0.69 W/m²·°C)(5 m × 6 m)(50°C - 20°C)/(30 cm * 1 m/100 cm)
Q = 2,070 W/m

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What type of mirror can produce both<br> converging and diverging rays?
Ivahew [28]

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A convex mirror is a diverging mirror (f is negative) and forms only one type of image. It is a case 3 image—one that is upright and smaller than the object, just as for diverging lenses.

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A rock is projected upward from the surface of the moon, at time t = 0.0 s, w a velocity of 30 m/s. The acceleration due to grav
Vinvika [58]
<h2>Answer: 277.777 m</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told that the rock was<u> projected upward from the surface</u>, we will only use the equations related to the Y axis.

In this sense, the movement equations in the Y axis are:

y-y_{o}=V_{o}.t+\frac{1}{2}g.t^{2}    (1)

V=V_{o}-g.t    (2)

Where:

y  is the rock's final position

y_{o}=0  is the rock's initial position

V_{o}=30\frac{m}{s} is the rock's initial velocity

V is the final velocity

t is the time the parabolic movement lasts

g=1.62\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of the moon

As we know y_{o}=0 , equation (2) is rewritten as:

y=V_{o}.t+\frac{1}{2}g.t^{2}    (3)

On the other hand, the maximum height  is accomplished when V=0:

V=V_{o}-g.t=0    (4)

V_{o}-g.t=0    

V_{o}=g.t    (5)

Finding t:

t=\frac{V_{o}}{g}    (6)

Substituting (6) in (3):

y=V_{o}(\frac{V_{o}}{g})+\frac{1}{2}g(\frac{V_{o}}{g})^{2}    (7)

y_{max}=\frac{{V_{o}}^{2}}{2g}    (8)  Now we can calculate the maximum height of the rock

y_{max}=\frac{{(30m/s)}^{2}}{(2)(1.62m/s^{2})}   (9)

Finally:

y_{max}=277.777m  

4 0
3 years ago
How much work is done if 10 N is applied to a 5kg object for 10 meters if there is an opposing force of 5 N
BlackZzzverrR [31]

Answer:

50 J

Explanation:

The net force acting on the box is given by the algebraic sum of the two forces, so:

F=10 N -5 N = 5 N

The net work done on the box is equal to (assuming the net force is parallel to the displacement of the object)

W=Fd

where

F = 5 N is the net force on the object

d = 10 m is the displacement of the object

Substituting,

W=(5 N)(10 m)=50 J

5 0
4 years ago
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