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2 years ago

Which of these are terrestrial planets? Select all that apply.

2 answers:

5 0

The answers to the question above would be letters a, b, c, and d. The terrestrial planets are those that are close to the sun, namely, Mercury, Venus, Earth, and Mars. The others are known as the gas giants since they have a thick atmosphere and are considerably cold.

lawyer [7]2 years ago

4 0

Answer:

<h2>a. Mars </h2><h2>b. Earth </h2><h2>c. Venus</h2>

Explanation:

Terrestrial planets are defined as planets which are formed by silicate. An important characteristic is that these planets have a solid surface. On the other hand, there are gaseous planets, which are bigger and more separated of the Sun.

In our Solar system, there are 4 terrestrial planets, which are Mercury, Venus, Earth, and Mars. Notice that they are the inner planets of the system.

So, in this case, the right chice is a. Mars, b. Earth and c. Venus. All three are terrestrial planets, that is, they have a rock solid surface.

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The free-electron density in a copper wire is 8.5×1028 electrons/m3. The electric field in the wire is 0.0520 N/C and the temper

meriva

Answer:

(a) 1.87 x 10⁻⁴ m/s

(b) 0.013V

Explanation:

(a) Drift speed, $v_{d}$ , is the average velocity that a charged particle can have due to an electric field. For a given current, I, the drift velocity is given by;

$v_{d}$ = $\frac{I}{qnA}$ ----------------(i)

Where;

q = amount of charge

n = free charge density

A = cross-sectional area of the wire

But current density, J, is the electric current per unit cross-section area. This is also equal to the ratio of the electric field, E, to the resistivity, p, of the material of the wire. i.e

J = $\frac{I}{A}$ = $\frac{E}{p}$

Equation (i) can then be written as follows;

$v_{d}$ = $\frac{J}{qn}$ = $\frac{E}{qnp}$

$v_{d}$ = $\frac{E}{qnp}$ ---------------------(ii)

From the question;

E = 0.0520N/C

p = 1.72 x 10⁻⁸ Ωm

n = 8.5 x 10²⁸ electrons/m³

c = charge on electron = 1.9 x 10⁻¹⁹C

Substitute these values into equation (ii) as follows;

$v_{d}$ = $\frac{0.0520}{1.9*10^{-19} * 8.5*10^{28} * 1.72*10^{-8}}$

$v_{d}$ = 1.87 x 10⁻⁴ m/s

(b) The potential difference, V, is given by the product of the electric field and the distance, d, between the two points in the wire. i.e

V = E x d [where d = 25.0cm = 0.25m]

V = 0.0520 x 0.25

V = 0.013V

4 0

2 years ago

If you calculate the thermal power radiated by typical objects at room temperature, you will find surprisingly large values, sev

OverLord2011 [107]

Answer:

best explanation of this is sentence B

Explanation:

The radiation emission of the bodies is given by the expression

P = σ A e T⁴

Where P is the power emitted in watts, σ is the Stefan-Boltzmann constant, A is the surface area of the body, e is the emissivity for black body e = 1 and T is the absolute body temperature in degrees Kelvin.

When the values are substituted the power is quite high 2.5 KW, but the medium surrounding the box also emits radiation

T box ≈ T room

P box ≈ P room

As the two powers are similar and the box can absorbed, since it has the ability to emit and absorb radiation, as the medium is also close of the temperature of the box, the amount emitted is very similar to that absorbed, so the net change in energy is very small.

In the case that the box is much hotter or colder than the surrounding medium if there is a significant net transfer.

Consequently, the best explanation of this is sentence B

5 0

2 years ago

A very long line of charge with charge per unit length +8.00 μC/m is on the x-axis and its midpoint is at x = 0. A second very l

artcher [175]

Answer:

at y=6.29 cm the charge of the two distribution will be equal.

Explanation:

Given:

linear charge density on the x-axis, $\lambda_1=8\times 10^{-6}\ C$

linear charge density of the other charge distribution, $\lambda_2=-6\times 10^{-6}\ C$

Since both the linear charges are parallel and aligned by their centers hence we get the symmetric point along the y-axis where the electric fields will be equal.

Let the neural point be at x meters from the x-axis then the distance of that point from the y-axis will be (0.11-x) meters.

<u>we know, the electric field due to linear charge is given as:</u>

$E=\frac{\lambda}{2\pi.r.\epsilon_0}$

where:

$\lambda=$ linear charge density

r = radial distance from the center of wire

$\epsilon_0=$ permittivity of free space

Therefore,

$E_1=E_2$

$\frac{\lambda_1}{2\pi.x.\epsilon_0}=\frac{\lambda_2}{2\pi.(0.11-x).\epsilon_0}$

$\frac{\lambda_1}{x} =\frac{\lambda_2}{0.11-x}$

$\frac{8\times 10^{-6}}{x} =\frac{6\times 10^{-6}}{0.11-x}$

$x=0.0629\ m$

∴at y=6.29 cm the charge of the two distribution will be equal.

9 0

2 years ago

An underground tunnel has two openings, with one opening a few meters higher than the other. If air moves past the higher openin

klasskru [66]

Answer:

There would be a pressure drop in the direction of the higher opening. This will force air to move in from the lower opening and force it to leave through the higher opening. This will create a convectional movement of air, cooling and ventilating the tunnel.

Explanation:

This is in accordance with bernoulli's law of fluid flow which states that the pressure exerted by a moving fluid is lesser than it would exert if it were at rest.

6 0

2 years ago

A solution is oversaturated with solute. Which could be done to decrease the oversaturation?

salantis [7]

Ans: Dilute the solution

Explanation:
To decrease the over-saturation, dilute the solution. Dilution<span> is the process of decreasing the solute's concentration in the </span>solution. It is<span> usually done by mixing with more solvent. In other words, to </span>dilute<span> a </span>solution<span> means to add more solvent without the addition of more solute.</span>

3 0

2 years ago

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