How is the periodic table of elements organized?

Which of the following elements is commonly found in the Earth's crust, living matter, oceans, and atmosphere? A. carbon B. argo

jekas [21]

The answer is carbon.

8 0

3 years ago

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The area in which the attraction and the repulsion of a magnet's poles are felt is a(n)

docker41 [41]

The are is called a magnetic field.

8 0

4 years ago

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a wave travels in a string at 58 m/s. a second string of 10% greater linear density has the same tension applied as in the first

ozzi

Answer:

The speed of wave in the second string is 55.3 m/s.

Explanation:

Given that,

Speed of wave in first string= 58 m/s

We need to calculate the wave speed

Using formula of speed for first string

$v_{1}=\sqrt{\dfrac{T}{\mu_{1}}}$ ...(I)

For second string

$v_{2}=\sqrt{\dfrac{T}{\mu_{2}}}$ ...(II)

Divided equation (II) by equation (I)

$\dfrac{v_{2}}{v_{1}}=\sqrt{\dfrac{\dfrac{T}{\mu_{2}}}{\dfrac{T}{\mu_{1}}}}$

Here, Tension is same in both string

So,

$\dfrac{v_{2}}{v_{1}}=\sqrt{\dfrac{\mu_{1}}{\mu_{2}}}$

The linear density of the second string

$\mu_{2}=\mu_{1}+\dfrac{10}{100}\mu_{1}$

$\mu_{2}=\dfrac{110}{100}\mu_{1}$

$\mu_{2}=1.1\mu_{1}$

Now, Put the value of linear density of second string

$\dfrac{v_{2}}{v_{1}}=\sqrt{\dfrac{\mu_{1}}{1.1\mu_{1}}}$

$v_{2}=v_{1}\times\sqrt{\dfrac{1}{1.1}}$

$v_{2}=58\times\sqrt{\dfrac{1}{1.1}}$

$v_{2}=55.3\ m/s$

Hence, The speed of wave in the second string is 55.3 m/s.

5 0

4 years ago

Two charged particles are projected into a region where a magnetic field is directed perpendicular to their velocities. if the c

goldfiish [28.3K]

The bearing could be the below:
oppositely charged, same initial direction
same charge, opposite initial direction

You can decide by utilizing your correct hand and put your fingers toward the attractive field (North to South). Thumb toward present or charged molecule. The course of your palm will demonstrate the heading of compelling set on a decidedly charged molecule and the bearing of the back of your hand will demonstrate the bearing of a contrarily charged molecule.

4 0

4 years ago

A tightrope walker is walking between two buildings holding a pole with length L = 16.0 m, and mass m p = 20.0 kg. The daredevil

Sati [7]

Answer:

F1 = 140.2[N]; F2 = 61.72[N]

Explanation:

To solve this problem, we must make a free body diagram, with the different forces and distances that apply on the Daredevil and its bar.

In the attached image is a free body diagram, with the different forces and distances that are on the equilibrium bar.

First, we performed a sum of forces at y-axis equal to zero, in this way we deduced one of the equations for the forces of the arms F1 & F2. It takes another equation, to find each of the forces.

Then we do a sum of moments equal to zero, at the point of force F1, in this way we can find an equation for Force F2. With the force F2 we replace in the first equation and we can find the force F1.

In this way the forces are:

F1 = 140.2[N]

F2 = 61.72[N]

In the attached image we can see the equations developed to find the forces F1 & F2

4 0

4 years ago