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2 years ago

11

Only put coolant into your radiator when the engine is _____.

1 answer:

Arada [10]2 years ago

5 0

Only put ciilant into ur radiator when the engine is cool (D)

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A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickn

Oxana [17]

Answer:

The percentage loss of the window is $E = 97.3$ %

Explanation:

From the question we are told that

The area of pane of glass is $A = 0.15 m^2$

The thickness is $d = 5mm = \frac{5}{1000} = 0.005m$

The thickness of the wall is $D = 0.15m$

The area of the wall is $a = 10m^2$

Generally the heat lost as a result of conduction of the window is

$Q_{window} = \frac{j_{glass} * A * (\Delta T) }{d}$

Where $j_{glass}$ is the thermal conductivity of glass which has a constant value of

$j_{glass} = 0.80 J/(s \cdot m \cdot C^o)$

Substituting values

$Q_{window} = \frac{ 0.80 * 0.15 * (\Delta T) }{0.005}$

$Q_{window} = 24 \Delta T$

Generally the heat lost as a result of conduction of the wall is

$Q_{wall} = \frac{j_{styrofoam} * A * (\Delta T) }{d}$

$j_{styrofoam}$ s the thermal conductivity of Styrofoam which has a constant value of $j_{styrofoam} = 0.010J / (s \cdot m \cdot C^o)$

Substituting values

$Q_{wall} = \frac{ 0.010 * 10 * (\Delta T) }{0.15}$

$Q_{wall} = 0.667 \ \Delta T$

Now the net loss of heat is

$Q_{net} = Q_{window} + Q_{wall}$

Substituting values

$Q_{net} = 24 + 0.667$

$Q_{net} = 24.667$

Now the percentage loss by the window is

$E = \frac{Q_{window} }{Q_{net}} * 100$

Substituting value

$E = \frac{24}{24 .667} * 100$

$E = 97.3$ %

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2019 nissan altima 2.5 SV

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Using your knowledge of how an ATM is used, develop a set of use cases that could serve as a basis for understanding the require

kenny6666 [7]

Answer:

Explanation:

bces

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german

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Rufina [12.5K]

Answer:

https://quizlet.com/148993376/the-nine-distinct-periods-of-time-flash-cards/

Explanation:

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