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3 years ago

You add 800 ml of water at 20c to 800 ml of water at 80c what is the most likely final temperature of the mixture ?

Physics

2 answers:

Rom4ik [11]3 years ago

4 0

Answer: D. 50 C

Explanation:

bekas [8.4K]3 years ago

3 0

Answer:

d. 50 C

Explanation:

In this problem, we have to add 800 ml of water at 20 Celsius to 800 ml of water at 80 Celsius.

According to the 2nd law of thermodynamics, heat transfers from hot to cold temperature.

The quantity of both the different waters is equal so this makes it very easy. All we have to do is find the mean of both the temperatures:

Final temperature = (20 C + 80 C)/2

= 50 Celsius

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3 years ago

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The type of circuit in which there is only one path for an electric current to flow is known as a _____ circuit. Question 6 opti

-BARSIC- [3]

Answer:

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Explanation:

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3 years ago

A very long, solid insulating cylinder has radius R; bored along its entire length is a cylindrical hole with radius a. The axis

lawyer [7]

Answer:

The value of the electric field is $E_{net} = \dfrac{r \textbf{b}}{2\epsilon_{0}}$

Explanation:

We know that the electric field inside a solid cylinder at a distance $\textbf{r}$ from the centre is given by

$E = \dfrac{\rho \textbf{r}}{2 \epsilon_{0}}$

Let's consider the cross-section of the cylinder as shown in the figure. Let `O' be the centre of the long solid insulating cylinder having radius 'R'. Also consider that $O'$ be the cetre of the hole of radius 'a' situated at a distance 'b' from 'O'. Given, the volume charge density of the material is 'r'. So, the volume charge density inside the hole will be '-r'. Let's consider 'P' be any arbitrary point inside the hole situated at a distance 's' from $O'$ .

So, the electric field ' $E_{O}$ ' due to the long cylinder at point 'P' is given by

$E_{O} = \dfrac{r \textbf{c}}{2 \epsilon_{0}}$

and the electric field ' $E_{O'}$ 'due to the hole at point 'P' is given by

$E_{O'} = \dfrac{\rho \textbf{s}}{2 \epsilon_{0}}$

So the net electric field ( $E_{net}$ ) inside the hole is given by

$E_{net} = E_{O} - E_{O'} = \dfrac{r}{2\epsilon_{0}}(\textbf{c - s}) = \dfrac{r \textbf{b}}{2\epsilon_{0}}$

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4 years ago

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joja [24]

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3 years ago

A mail carrier leaves the post ofﬁce and drives 2.00 km to the north. He then drives in a direction 60.0° south of east for 7.00

Oxana [17]

Answer:

12.97 km

Explanation:

In order to find the resultant displacement, we have to resolve each of the 3 displacements along the x and y direction.

Taking north as positive y direction and east as positive x-direction, we have:

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$A_x = 0\\A_y = +2.00 km$