"The connective tissue that's found between bones."
An example would be between the vertabrae in the spine.
Like the other (inappropriate) answer said, a ligament is what attches muscles and bones.
Also, the tissue inside of bones is bone marrow.
If you know that, you can take out those two options (Option 1 and 3)
To represent in order how the animals
Answer:
![\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%7D%5B%5Cfrac%7B1%7D%7By%7D%28%5C%5Ey%29%20-%20%5Cfrac%7B1%7D%7Bx%7D%28%5C%5Ex%29%5D)
Explanation:
The electric field created by an infinitely long wire can be found by Gauss' Law.
For the electric field at point (x,y), the superposition of electric fields created by both lines should be calculated. The distance 'r' for the first wire is equal to 'y', and equal to 'x' for the second wire.
![\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) + \frac{-\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) - \frac{\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%20%3D%20%5Cvec%7BE%7D_1%20%2B%20%5Cvec%7BE%7D_2%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%20y%7D%28%5C%5Ey%29%20%2B%20%5Cfrac%7B-%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%20x%7D%28%5C%5Ex%29%5C%5C%5Cvec%7BE%7D%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%20y%7D%28%5C%5Ey%29%20-%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%20x%7D%28%5C%5Ex%29%5C%5C%5Cvec%7BE%7D%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%7D%5B%5Cfrac%7B1%7D%7By%7D%28%5C%5Ey%29%20-%20%5Cfrac%7B1%7D%7Bx%7D%28%5C%5Ex%29%5D)
Answer:
the balls reached a height of 4.9985 m
Explanation:
Given the data in the question;
mass one m = 3.8 kg
mass two M = 2.1 kg
Initial velocities
u = 22 m/s
U = { moving downward} = 12 m/s
Now, using the law conservation of linear moment;
mu + MU = v( m + M )
we solve for "v" which is the velocity of the ball s after collision;
v = (mu + MU) / ( m + M )
so we substitute our given values into the equation
v = ( ( 3.8 × 22 ) + ( 2.1 × -12) ) / ( 3.8 + 2.1 )
v = ( 83.6 - 25.2 ) / 5.9
v = 58.4 / 5.9
v = 9.898 m/s
Now, we determine required height using the following relation;
v"² - v² = 2gh
where v" is the velocity at the top which is 0 m/s and g = -9.8 m/s²
0 - v² = 2gh
v² = -2gh
so we substitute
( 9.898 )² = -2 × -9.8 × h
97.97 = 19.6 × h
h = 97.97 / 19.6
h = 4.9985 m
Therefore, the balls reached a height of 4.9985 m
Answer:
Explanation:
From the question we are told that:
Charges:
Distance 
Generally the equation for Force is mathematically given by
Therefore
