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2 years ago

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The sun delivers an average power of 1.499 w/m2 to the top of neptune's atmosphere. find the magnitudes of vector e max and vect

or b max for the electromagnetic waves at the top of the atmosphere.

1 answer:

sesenic [268]2 years ago

4 0

See attachment for answer:

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Select all that apply.

lianna [129]

Answer:

the earth is flat, the stars control human life, the planets revolve around the earth

Explanation:

7 0

3 years ago

Read 2 more answers

When is the pressure of the man on the ground more – lying position or standing

Wittaler [7]

Hi,

<u>The man on the ground in standing position has more pressure</u>. This is because when he stands, only his legs are in contact with the ground. While lying, his body is more in contact with the ground, therefore, he exerts less pressure.

To the point, a man standing position on the ground had more pressure.

More is the area of contact, less is the pressure efforted.

Thank you...

7 0

2 years ago

Two tiny particles having charges of +5.00 μC and +7.00 μC are placed along the x-axis. The +5.00-µC particle is at x = 0.00 cm,

Liula [17]

Answer:

The third charged particle must be placed at x = 0.458 m = 45.8 cm

Explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

$F = \frac{k*q_1*q_2}{d^2}$ Formula (1)

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁, q₂: Charges in Coulombs (C)

d: distance between the charges in meters (m)

Equivalence

1μC= 10⁻⁶C

1m = 100 cm

Data

K = 8.99 * 10⁹ N*m²/C²

q₁ = +5.00 μC = +5.00 * 10⁻⁶ C

q₂= +7.00 μC = +7.00 * 10⁻⁶ C

d₁ = x (m)

d₂ = 1-x (m)

Problem development

Look at the attached graphic.

We assume a positive charge q₃ so F₁₃ and F₂₃ are repulsive forces and must be equal so that the net force is zero:

We use formula (1) to calculate the forces F₁₃ and F₂₃

$F_{13} = \frac{k*q_1*q_3}{d_1^2}$

$F_{23} = \frac{k*q_2*q_3}{d_2^2}$

F₁₃ = F₂₃

$\frac{k*q_1*q_3}{d_1^2} = \frac{k*q_2*q_3}{d_2^2}$ We eliminate k and q₃ on both sides

$\frac{q_1}{d_1^2}= \frac{q_2}{d_2^2}$

$\frac{q_1}{x^2}=\frac{q_2}{(1-x)^2}$

$\frac{5*10^{-6}}{x^2}=\frac{7*10^{-6}}{(1-x)^2}$ We eliminate 10⁻⁶ on both sides

$(1-x)^2 = \frac{7}{5} x^2$

$1-2x+x^2=\frac{7}{5} x^2$

$5-10x+5x^2=7 x^2$

$2x^2+10x-5=0$

We solve the quadratic equation:

$x_1 = \frac{-b+\sqrt{b^2-4ac} }{2a} = \frac{-10+\sqrt{10^2-4*2*(-5)} }{2*2} = 0.458m$

$x_2 = \frac{-b-\sqrt{b^2-4ac} }{2a} = \frac{-10-\sqrt{10^2-4*2*(-5)} }{2*2} = -5.458m$

In the option x₂, F₁₃ and F₂₃ will go in the same direction and will not be canceled, therefore we take x₁ as the correct option since at that point the forces are in opposite way .

x = 0.458m = 45.8cm

8 0

3 years ago

Is gas matter? How do you know? Hint: Remember there are 2 things we need to be considered matter.

astra-53 [7]

Answer:

Gas is a state of matter that has no fixed shape and no fixed volume.

In addition to solids and liquids, gases are also a physical state in which matter can occur. All gases have weight. Unlike solids and liquids, gases will occupy the entire container that encloses them.

matter is "anything that has mass and volume (occupies space)

<em>Gases have mass. The space between gas particles is empty. Gases can be formed as products in chemical reactions. Gas particles can form bonds between them under certain conditions</em>

<em> Gases have volume which isn't fixed </em>(no fixed volume)<em> and no fixed shape. Gases expand to fill the space available. They can also be compressed into a very small space.</em>

Explanation:

7 0

2 years ago

A wattage rating of a lightbulb is the power it consumers when it is connected across a 120 V potential difference. How does the

navik [9.2K]

Answer:

The answer is C.

120 V with 60 W light bulb is 240 ohms.

120 V with 100 W light bulb is 144 ohms.

The 100 W bulb has less resistance :)

7 0

2 years ago