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1 year ago

Part A

Physics

1 answer:

Basile [38]1 year 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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A charged object is suspended motionless in the air by the gravitational force pulling it down and an electric force pushing it

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The charge of the object must be $1.11 \times e^{-5} \text { coulomb }$

Answer: Option C

<u>Explanation:</u>

Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.

Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

$Electric field, E=\frac{\text { Force }(F)}{q}$

Here, given E = 4500 N/C and F = 0.05 N.

We need to find charge of the object (q)

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3 years ago

What will happen to the charge and potential difference if the plate area were increased while the plate separation remains unch

Alenkinab [10]

Answer:

Potential difference and charge will also increase.

Explanation:

Asking that :

What will happen to the charge and potential difference if the plate area were increased while the plate separation remains unchanged?

The charge is directly proportional to area of the plate. That is, increase in area of the plate of a capacitor will lead to the increase in the charges between the plates.

And since charge is also proportional to the magnitude of potential difference between the plates from the definition of capacitance of a capacitor which says that:

Q = CV

Therefore, increase in the area of the plate will also lead to increase in potential difference between the plates.

Therefore, if the plate area were increased while the plate separation remains unchanged, the charge and potential difference between them will also increase.

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3 years ago

Four +2 μC charges are placed at the positions (10 cm, 0 cm), (−10 cm, 0 cm), (0 cm, 10 cm), and (0 cm, −10 cm) such that they f

Rufina [12.5K]

Answer:

The force on the charge at the origin is 0 N .

Explanation:

All charges are positive. So, in x axis force exerted by the charge located in the position (10 cm, 0 cm) will be canceled with the force exerted by the charge located in the position (-10 cm, 0 cm). In the same way, in y axis the force exerted by the charge located in the position (0 cm, 10 cm) will be canceled with the force exerted by the charge located in the position (0 cm, -10 cm).

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