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9 months ago

Part A

Physics

1 answer:

Basile [38]9 months ago

6 0

The relationships to determine the number of calories to change 0.50 kg of 0°C ice to 0°C ice water is 1,080,000 cal.

<h3>How does heating ice that is at C affect it?</h3>

Ice melts and becomes liquid water at 0 degrees Celsius. Once all of the ice has been entirely transformed into liquid water, the temperature of the remaining ice begins to increase once more (in °C), continuing to rise until it reaches 100 °C, where it then stabilizes.

The water turns into steam when it reaches a temperature of 100 °C (D).

Water has a fusion latent heat of fusion of 80 cal/g.

Water has a 1 cal/g-C specific heat.

Water has a 540 cal/g latent heat of vaporization.

In light of this, the total amount of heat needed is 1500 g [(80 cal/g) + (1 cal/g-C)(100 - 0)C + (540 cal/g)] = 1500 g [(720 cal/g)] = 1,080,000 cal.

To learn more about Vaporization refer to:

brainly.com/question/26306578

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Answer:

$|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))$

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

$\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r$

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)$

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

$\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}$

Now, substitute this into 'dF':

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)$

Now we can integrate dF over the rod.

$\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)$

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Answer:

The capacitance of the deflecting plates is $C=1.59 pF$ .

Explanation:

The expression for the capacitance of the capacitor in terms of area and distance is as follows;

$C=\frac{\varepsilon _{0}A}{d}$

Here, C is the capacitance, A is the area, d is the distance and $\varepsilon _{0}$ is the absolute permittivity.

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s= 3.0 cm

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$A= s_^{2}$

Put s= 0.030 m.

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$C=1.59 pF$

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