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3 years ago

The features of Mercury are named in honor of famous people in which fields of endeavor?

Physics

1 answer:

frozen [14]3 years ago

4 0

Answer:

Writing, Poetry, Philosophy, Painter, Lutheran pastor, Screenwriting, Music, Photography

Explanation:

The International Astronomical Union (IAU) approved the naming of 9 craters to honor artists of various fields. They are:

Alver: Betty Alver was an Estonian writer.

Donelaitis: Kristijonas Donelaitis was a Lutheran pastor

Flaiano: Ennio Flaiano was an Italian screenwriter, playwright, novelist, journalist, and drama critic.

Hurley: James Francis “Frank” Hurley was an Australian photographer and adventurer.

L’Engle: Madeleine L’Engle was an American writer

Lovecraft: Howard Phillips Lovecraft was an American author.

Petofi: Sándor Petőfi was a Hungarian poet.

Pahinui: Charles Phillip Kahahawai “Gabby” Pahinui was a Hawaiian guitar player.

Roerich: Nicholas Roerich was a Russian painter and philosopher.

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Are the statements of the Flat Earth Society an example of science or pseudoscience?

Y_Kistochka [10]

Answer:

Pseudoscience

Explanation:

Pseudoscience is defined as set of ideas that present themselves as a science, but they do not fulfill the criteria to be called science. In the same way, in the middle of 20th century some people called earth is flat. They challenged the spherical shape of earth. But their hypothesis does not fulfill the criteria of science. Therefore, the statements of flat earth society are pseudoscience.

5 0

2 years ago

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A wave with 2.0 m amplitude has a frequency of 500 Hz is travelling at a speed of 200 m/s. What is the wavelength?

Serggg [28]

Answer: 0.4m

Explanation:

Given that:

Amplitude of wave = 2.0 m

Wavelength (λ)= ?

Frequency F = 500Hz

Speed V = 200 m/s

The wavelength is measured in metres, and represented by the symbol λ.

So, apply the formula:

Wavespeed V= Frequency F xwavelength λ

200m/s = 500Hz x λ

λ = 200m/s / 500Hz

λ = 0.4m

Thus, the wavelength is 0.4 metres

3 0

3 years ago

ANSWER ASAP!!!!

trapecia [35]

As you increase the temperature, the matter begins to expand. Due to this, the distance between matter particles decreases and they are no more compact. Hence, density decreases.

3 0

3 years ago

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

2 years ago

Consider the following geometric solids.

Bad White [126]

Before answering this question, first we have to understand the effect of ratio of surface area to volume on the rate of diffusion.

The rate of diffusion for a body having larger surface area as compared to the ratio of surface area to volume will be more than a body having less surface area. Mathematically it can written as-

V∝ R [ where v is the rate of diffusion and r is the ratio of surface area to volume]

As per the question,the ratio of surface area to volume for a sphere is given $0.08m^{-1}$

The surface area to volume ratio for right circular cylinder is given $2.1m^{-1}$

Hence, it is obvious that the ratio is more for right circular cylinder.As the rate diffusion is directly proportional to the surface area to volume ratio,hence rate of diffusion will be more for right circular cylinder.

Hence the correct option is B. The rate of diffusion would be faster for the right cylinder.

5 0

2 years ago

Read 2 more answers