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Korvikt [17]
1 year ago
15

Can you please help me with #6 (the 6 was wrong in the question it meant #5 as you can see I tried to change it)

Physics
1 answer:
Harman [31]1 year ago
5 0
\begin{gathered} \text{Question 5)} \\ Q=52,000J \\ m=2kg \\ C=130\text{ J/Kg K} \\ \Delta T=\text{?} \\ Q=Cm\Delta T \\ \text{Solving }\Delta T \\ \Delta T=\frac{Q}{Cm} \\ \Delta T=\frac{52,000J}{(130\text{ J/Kg K)(}2kg\text{)}} \\ \Delta T=200K,\text{ but Kelving and }degree\text{ Celsius have the same change} \\ \text{Hence} \\ \Delta T=200\text{ \degree{}C} \\  \\ \text{Question 6} \\ T1=350\text{ K} \\ \Delta T=200\text{ K} \\ T2=\text{?} \\ \Delta T=T2-T1 \\ \text{Sol}v\text{ing T2} \\ T2=\Delta T+T1 \\ T2=200\text{ K}+350\text{ K} \\ T2=550\text{ K} \\ \text{The final temperature is }550\text{ K} \end{gathered}

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Answer:

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3 years ago
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8 0
3 years ago
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3 0
3 years ago
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If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front o
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The correct answer is letter A. 6 millimeters. <span>If an object 18 millimeters high is placed 12 millimeters from a diverging lens and the image is formed 4 millimeters in front of the lens, the height of the image is 6 millimeters.
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