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1 year ago

A copper wire carries 1 amp of electric current. what kind of charge does the electron flow create in the wire?

1 answer:

7 0

A current is just the development of electrons. Since there is no electric field from a (steady) current, this implies the wire is still electrically unbiased, ie uncharged. Accordingly the appropriate response is no charge. I trust the appropriate response will help you.

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What is the value of g on the surface of Saturn? Assume M-Saturn = 5.68×10^26 kg and R-Saturn = 5.82×10^7 m.Choose the appropria

Likurg_2 [28]

Answer:

Approximately $\rm 11.2 \; N \cdot kg^{-1}$ at that distance from the center of the planet.

Option A) The low value of $g$ near the cloud top of Saturn is possible because of the low density of the planet.

Explanation:

The value of $g$ on a planet measures the size of gravity on an object for each unit of its mass. The equation for gravity is:

$\displaystyle \frac{G \cdot M \cdot m}{R^2}$ ,

where

$G \approx 6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2$ .
$M$ is the mass of the planet, and
$m$ is the mass of the object.

To find an equation for $g$ , divide the equation for gravity by the mass of the object:

$\displaystyle g = \left.\frac{G \cdot M \cdot m}{R^2} \right/\frac{1}{m} = \frac{G \cdot M}{R^2}$ .

In this case,

$M = 5.68\times 10^{26}\; \rm kg$ , and
$R = 5.82 \times 10^7\; \rm m$ .

Calculate $g$ based on these values:

$\begin{aligned} g &= \frac{G \cdot M}{R^2}\cr &= \frac{6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2\times 5.68\times 10^{26}\; \rm kg}{\left(5.82\times 10^7\; \rm m\right)^2} \cr &\approx 11.2\; \rm N\cdot kg^{-1} \end{aligned}$ .

Saturn is a gas giant. Most of its volume was filled with gas. In comparison, the earth is a rocky planet. Most of its volume was filled with solid and molten rocks. As a result, the average density of the earth would be greater than the average density of Saturn.

Refer to the equation for $g$ :

$\displaystyle g = \frac{G \cdot M}{R^2}$ .

The mass of the planet is in the numerator. If two planets are of the same size, $g$ would be greater at the surface of the more massive planet.

On the other hand, if the mass of the planet is large while its density is small, its radius also needs to be very large. Since $R$ is in the denominator of $g$ , increasing the value of $R$ while keeping $M$ constant would reduce the value of $g$ . That explains why the value of $g$ near the "surface" (cloud tops) of Saturn is about the same as that on the surface of the earth (approximately $9.81\; \rm N \cdot kg^{-1}$ .

As a side note, $5.82\times 10^7\rm \; m$ likely refers to the distance from the center of Saturn to its cloud tops. Hence, it would be more appropriate to say that the value of $g$ near the cloud tops of Saturn is approximately $\rm 11.2 \; N \cdot kg^{-1}$ .

6 0

2 years ago

Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 20.0 m above water wit

melamori03 [73]

Answer:

Explanation:

Given that,

Height of the bridge is 20m

Initial before he throws the rock

The height is hi = 20 m

Then, final height hitting the water

hf = 0 m

Initial speed the rock is throw

Vi = 15m/s

The final speed at which the rock hits the water

Vf = 24.8 m/s

Using conservation of energy given by the question hint

Ki + Ui = Kf + Uf

Where

Ki is initial kinetic energy

Ui is initial potential energy

Kf is final kinetic energy

Uf is final potential energy

Then,

Ki + Ui = Kf + Uf

Where

Ei = Ki + Ui

Where Ei is initial energy

Ei = ½mVi² + m•g•hi

Ei = ½m × 15² + m × 9.8 × 20

Ei = 112.5m + 196m

Ei = 308.5m J

Now,

Ef = Kf + Uf

Ef = ½mVf² + m•g•hf

Ef = ½m × 24.8² + m × 9.8 × 0

Ef = 307.52m + 0

Ef = 307.52m J

Since Ef ≈ Ei, then the rock thrown from the tip of a bridge is independent of the direction of throw

7 0

1 year ago

An electric motor consumes 9.00 kj of electrical energy in 1.00 min. if one-third of this energy goes into heat and other forms

liq [111]

How do I make a question help PLZZZ

5 0

1 year ago

Calculate the weight of a 4.5 kg rabbit.

solniwko [45]

The correct answer is: 13900589.

3 0

1 year ago

Lamar has been running sprints to prepare for his next football game.He has found that he can maintain his maximum speed for 45

Svet_ta [14]

Answer:

Kindly check explanation

Explanation:

Length of race = 5km

Maximum speed = 45 yards

Converting from yards to kilometer :

1km = 1093.613 yards

x = 45 yards

(1093.613 * x) = 45

x = 45 / 1093.613

x = 0.0411480 km

Where x = maximum length for which he can maintain his maximum speed expressed in kilometers.

Therefore, with the available information, it can be concluded that Lamar cannot maintain his maximum speed for the entire 5km race and will only be able maintain his maximum speed for 0.0411 kilometers.

5 0

1 year ago