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3 years ago

Chemical, potential energy is stored in a battery.

Chemistry

2 answers:

Sav [38]3 years ago

5 0

Answer:

<em>If a piece of iron were placed close to this device, the iron nail would move toward the piece of iron.</em>

Explanation:

The description of the built device corresponds to a temporal electromagnet: as long as the battery is connected and supplying the potential difference (voltage) the iron nail will have magnet properties.

Since magnets attract and are attracted to iron, the iron nail and the piece of iron would feel attracted to each other.

The question is confusing because it asks what would happen to a piece of iron if it were placed close to this device, but it should be what happens to the iron nail if a piece of iron were placed close to this device.

Then, assuming the iron nail is free to move it would move toward the piece of iron due to the electromagnetic force.

Harrizon [31]3 years ago

4 0

Answer: D it moves toward piece of iron.

Explanation:

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3 years ago

If the distance between two objects is decreased to - of the original

Charra [1.4K]

Answer:

The new force will be \frac{1}{100} of the original force.

Explanation:

In the context of this problem, we're dealing with the law of gravitational attraction. The law states that the gravitational force between two object is directly proportional to the product of their masses and inversely proportional to the square of a distance between them.

That said, let's say that our equation for the initial force is:

$F = G\frac{m_1m_2}{R^2}The problem states that the distance decrease to 1/10 of the original distance, this means:[tex]R_2 = \frac{1}{10}R$

And the force at this distance would be written in terms of the same equation:

$F_2 = G\frac{m_1m_2}{R_2^2}$

Find the ratio between the final and the initial force:

$\frac{F_2}{F} = \frac{G\frac{m_1m_2}{R_2^2}}{G\frac{m_1m_2}{R^2}}$

Substitute the value for the final distance in terms of the initial distance:

$\frac{F_2}{F} = \frac{G\frac{m_1m_2}{(\frac{R}{10})^2}}{G\frac{m_1m_2}{R^2}}$

Simplify:

$\frac{F_2}{F} = \frac{\frac{1}{100R^2}}{\frac{1}{R^2}}=\frac{1}{100}$

This means the new force will be \frac{1}{100} of the original force.

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3 years ago

Predict the electronegativity of the yet undiscovered element with Z = 119.

Lera25 [3.4K]

We know that the element Z = 119 would be placed right below the Fr, in the column of the alcaline metals.

We also know that the trend in the electronegativity is to decrease when you go up-down ia group.

The known electronegativities of the elements of this group are:

Li: 0.98
Na: 0.93
K: 0.82
Rb: 0.82
Cs: 0.79
Fr: 0.70

Then the hypotetical element Z = 119 would probably have an electronegativity slightly below 0.70, for sure in the range 0.60 - 0.70.

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