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ankoles [38]
3 years ago
15

Для вимірювання температури води, що має масу 66 г, у неї занурили термометр, який показав температуру 32,4 °С. Якою була дійсна

температура води, якщо теплоємність термометра 1,9 Дж\кг•°с і перед зануренням у воду він показував температуру навколишнього середовища рівну 17,8°С? Питома теплоємність води 4200 Дж\кг•°С
Physics
1 answer:
Aleksandr-060686 [28]3 years ago
4 0

Answer:

я не знаю відповіді

Explanation:

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Answer:

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Explanation:

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The islands of Java and Sumatra in Indonesia are an example of A. a rift zone B.a hot spot C. an ocean ridge D. a volcanic islan
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A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c
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Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = 71\times 10^{- 9} C

Q' = + 42 nC = 42\times 10^{- 9} C

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E} = 708.03 N/C

Now,

Electric field at the mid-point due to charge Q' is given by:

\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}

\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}

\vec{E'} = 418.84 N/C

Now,

The net Electric field is given by:

\vec{E_{net}} = \vec{E} - \vec{E'}

\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C

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