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3 years ago

This is how manganese appears in the periodic table. The arrow is pointing to the atomic of manganese.

2 answers:

5 0

The way an element is shown in the periodic table is called the element's symbol. In the case of manganese, it is Mn. Other information shown alongside the element's symbol is the atomic mass, which for manganese is 55, and the atomic number of the element, which for manganese is 25.

nalin [4]3 years ago

4 0

The answer is symbol :)

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An object that looks white when exposed to sunlight reflects all colors of light. What

Ierofanga [76]

It looks blue as it is only reflecting blue light

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3 years ago

Which of the following correctly describes the importance of demonstrations to scientific investigation?

sergey [27]

Answer:

A- A demonstration shows how something works, often including models

Explanation:

A demonstration allows, through experimentation, to show how nature works and in that way can include the explanation of scientific theories that explain the set of observed facts, that is, it serves as a demonstration of the underlying scientific principles.

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3 years ago

If the acceleration of motorboat is 4m/s^2 and the motorboat stsrtsfrol.Rest what is velocity after 6.0 s

UkoKoshka [18]

Answer:

24 m/s

Explanation:

Using v = u + at where u = initial velocity of the motorboat = 0 m/s (since the boat starts from rest), a = acceleration = 4 m/s², t = time = 6 s and v = velocity of the motorboat after 6.0 s.

Substituting the values of the variables into the equation, we have

v = u + at

= 0 m/s + 4 m/s² × 6.0 s

= 0 m/s + 24 m/s

= 24 m/s

3 0

3 years ago

A particular coaxial cable is comprised of inner and outer conductors having radii 1 mm and 3 mm respectively, separated by air.

noname [10]

Answer:

The value is $\rho_s = 4.026 *10^{-6} \ C/m^2$

Explanation:

From the question we are told that

The radius of the inner conductor is $r_1 = 1 \ mm = 0.001 \ m$

The radius of the outer conductor is $r_2 = 3 \ mm = 0.003 \ m$

The potential at the outer conductor is $V = 1.5 kV = 1.5 *10^{3} \ V$

Generally the capacitance per length of the capacitor like set up of the two conductors is

$C= \frac{2 * \pi * \epsilon_o }{ ln [\frac{r_2}{r_1} ]}$

Here $\epsilon_o$ is the permitivity of free space with value $\epsilon_o = 8.85*10^{-12} C/(V \cdot m)$

=> $C= \frac{2 * 3.142 * 8.85*10^{-12} }{ ln [\frac{0.003}{0.001} ]}$

=> $C= 50.6 *10^{-12} \ F/m$

Generally given that the potential of the outer conductor with respect to the inner conductor is positive it then mean that the outer conductor is positively charge

Generally the line charge density of the outer conductor is mathematically represented as

$\rho_l = C * V$