PLEASE HELP!! Lab: Electromagnetic Induction

I don't understand any of this please help !

OlgaM077 [116]

You can see what is the electron configuration by looking at the layout of the periodic tables. the first shell will have a max of 2 electrons on it, once the first one is filled up a second is added with a max of 8 electrons on it and so on with the 8 as a max. so He, and H will only have them on the first shell but every horizontal row is a new valence or outer shell. so lets say for carbon look at the number in the upper left corner of the box will tell you the total number of electrons you will need. so start off with the first two electrons on the first shell. now you know that carbon needs 6 electrons in total, since you can only have a max of 2 on the first shell you need a second one so on the second one you will have to have the remaining 4. now elements are most stable when they have a full valence shell becuase those are the only electrons that will react with others. so if carbon has 4 it wants to either gain or lose 4 electrons so you could say that it would bond with 4H since each H will donate 1 electron to the C valence shell making all the H and C stable. CH4(methane)

6 0

3 years ago

A 10 kg monkey climbs up a massless rope that runs over a frictionless tree limb and back down to a 15 kg package on the ground.

pshichka [43]

Answer:

A. 4,9 m/s2

B. 2,0 m/s2

C. 120 N

Explanation:

In the image, 1 is going to represent the monkey and 2 is going to be the package. Let a_mín be the minimum acceleration that the monkey should have in the upward direction, so the package is barely lifted. Apply Newton’s second law of motion:

$\sum F_y=m_1*a_m_i_n = T-m_1*g$

If the package is barely lifted, that means that T=m_2*g; then:

$\sum F_y =m_1*a_m_i_n=m_2*g-m_1*g$

Solving the equation for a_mín, we have:

$a_m_i_n=((m_2-m_1)/m_1)*g = ((15kg-10kg)/10kg)*9,8 m/s^2 =4,9 m/s^2$

Once the monkey stops its climb and holds onto the rope, we set the equation of Newton’s second law as it follows:

For the monkey: $\sum F_y = m_1*a \rightarrow T-m_1*g=m_1*a$

For the package: $\sum F_y = m_2*a \rightarrow m_2*g - T = m_2*a$

The acceleration a is the same for both monkey and package, but have opposite directions, this means that when the monkey accelerates upwards, the package does it downwards and vice versa. Therefore, the acceleration a on the equation for the package is negative; however, if we invert the signs on the sum of forces, it has the same effect. To be clearer:

For the package: $\sum F_y = -m_2*a \rightarrow T-m2*g=-m_2*a \rightarrow m_2*g -T=m_2 *a$