Answer:
O²⁻
Explanation:
Number of protons = 8
Number of neutrons = 9
Number of electrons = 10
What type of atom or ion is it = ?
Solution:
Protons are the positively charged particle in an atom
Neutrons do not carry any charges
Electrons are negatively charged particles
For this atom, the number of protons helps to identify what specie it is; so this is an oxygen atom.
Now,
Charge = Number of protons - Number of electrons
Charge = 8 - 10 = -2
The charge on the atom is -2 and so it is an oxygen ion with -2 charge
The ion is O²⁻
The answer is A. Bob (<span>object's length)
</span>
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
Answer:
a) (0, -33, 12)
b) area of the triangle : 17.55 units of area
Explanation:
<h2>
a) </h2>
We know that the cross product of linearly independent vectors
and
gives us a nonzero, orthogonal to both, vector. So, if we can find two linearly independent vectors on the plane through the points P, Q, and R, we can use the cross product to obtain the answer to point a.
Luckily for us, we know that vectors
and
are living in the plane through the points P, Q, and R, and are linearly independent.
We know that they are linearly independent, cause to have one, and only one, plane through points P Q and R, this points must be linearly independent (as the dimension of a plane subspace is 3).
If they weren't linearly independent, we will obtain vector zero as the result of the cross product.
So, for our problem:
![\vec{A} = \vec{P} - \vec{Q} \\\\\vec{A} = (1,0,1) - (-2,1,4)\\\\\vec{A} = (1 +2,0-1,1-4)\\\\\vec{A} = (3,-1,-3)](https://tex.z-dn.net/?f=%5Cvec%7BA%7D%20%3D%20%5Cvec%7BP%7D%20-%20%5Cvec%7BQ%7D%20%5C%5C%5C%5C%5Cvec%7BA%7D%20%3D%20%281%2C0%2C1%29%20-%20%28-2%2C1%2C4%29%5C%5C%5C%5C%5Cvec%7BA%7D%20%3D%20%281%20%2B2%2C0-1%2C1-4%29%5C%5C%5C%5C%5Cvec%7BA%7D%20%3D%20%283%2C-1%2C-3%29)
![\vec{B} = \vec{R} - \vec{Q} \\\\\vec{B} = (6,2,7) - (-2,1,4)\\\\\vec{B} = (6 +2,2-1,7-4)\\\\\vec{B} = (8,1,3)](https://tex.z-dn.net/?f=%5Cvec%7BB%7D%20%3D%20%5Cvec%7BR%7D%20-%20%5Cvec%7BQ%7D%20%5C%5C%5C%5C%5Cvec%7BB%7D%20%3D%20%286%2C2%2C7%29%20-%20%28-2%2C1%2C4%29%5C%5C%5C%5C%5Cvec%7BB%7D%20%3D%20%286%20%2B2%2C2-1%2C7-4%29%5C%5C%5C%5C%5Cvec%7BB%7D%20%3D%20%288%2C1%2C3%29)
![\vec{A} \times \vec{B} = (A_y B_z - B_y A_z) \ \hat{i} - ( A_x B_z-B_xA_z) \ \hat{j} + (A_x B_y - B_x A_y ) \ \hat{k}](https://tex.z-dn.net/?f=%5Cvec%7BA%7D%20%5Ctimes%20%20%5Cvec%7BB%7D%20%3D%20%28A_y%20B_z%20-%20B_y%20A_z%29%20%5C%20%20%5Chat%7Bi%7D%20-%20%28%20A_x%20B_z-B_xA_z%29%20%5C%20%5Chat%7Bj%7D%20%2B%20%28A_x%20B_y%20-%20B_x%20A_y%20%29%20%5C%20%5Chat%7Bk%7D)
![\vec{A} \times \vec{B} = ( (-1) * 3 - 1 * (-3) ) \ \hat{i} - ( 3 * 3 - 8 * (-3)) \ \hat{j} + (3 * 1 - 8 * (-1) ) \ \hat{k}](https://tex.z-dn.net/?f=%5Cvec%7BA%7D%20%5Ctimes%20%20%5Cvec%7BB%7D%20%3D%20%28%20%28-1%29%20%2A%203%20-%201%20%2A%20%28-3%29%20%29%20%5C%20%20%5Chat%7Bi%7D%20-%20%28%203%20%2A%203%20-%208%20%2A%20%28-3%29%29%20%5C%20%5Chat%7Bj%7D%20%2B%20%283%20%2A%201%20-%208%20%2A%20%28-1%29%20%29%20%5C%20%5Chat%7Bk%7D)
![\vec{A} \times \vec{B} = ( - 3 + 3 ) \ \hat{i} - ( 9 + 24 ) \ \hat{j} + (3 + 8 ) \ \hat{k}](https://tex.z-dn.net/?f=%5Cvec%7BA%7D%20%5Ctimes%20%20%5Cvec%7BB%7D%20%3D%20%28%20-%203%20%2B%203%20%29%20%5C%20%20%5Chat%7Bi%7D%20-%20%28%209%20%2B%2024%20%29%20%5C%20%5Chat%7Bj%7D%20%2B%20%283%20%2B%208%20%29%20%5C%20%5Chat%7Bk%7D)
![\vec{A} \times \vec{B} = 0 \ \hat{i} - 33 \ \hat{j} + 12 \ \hat{k}](https://tex.z-dn.net/?f=%5Cvec%7BA%7D%20%5Ctimes%20%20%5Cvec%7BB%7D%20%3D%200%20%5C%20%20%5Chat%7Bi%7D%20-%2033%20%5C%20%5Chat%7Bj%7D%20%2B%2012%20%5C%20%5Chat%7Bk%7D)
![\vec{A} \times \vec{B} =(0, -33, 12)](https://tex.z-dn.net/?f=%5Cvec%7BA%7D%20%5Ctimes%20%20%5Cvec%7BB%7D%20%3D%280%2C%20-33%2C%2012%29)
<h2>B)</h2>
We know that
and
are two sides of the triangle, and we also know that we can use the magnitude of the cross product to find the area of the triangle:
![|\vec{A} \times \vec{B} | = 2 * area_{triangle}](https://tex.z-dn.net/?f=%20%7C%5Cvec%7BA%7D%20%5Ctimes%20%20%5Cvec%7BB%7D%20%7C%20%3D%202%20%2A%20area_%7Btriangle%7D)
so:
![\sqrt{(-33)^2 + (12)^2} = 2 * area_{triangle}](https://tex.z-dn.net/?f=%20%5Csqrt%7B%28-33%29%5E2%20%2B%20%2812%29%5E2%7D%20%3D%202%20%2A%20area_%7Btriangle%7D)
![\sqrt{1233} = 2 * area_{triangle}](https://tex.z-dn.net/?f=%20%5Csqrt%7B1233%7D%20%3D%202%20%2A%20area_%7Btriangle%7D)
![35.114= 2 * area_{triangle}](https://tex.z-dn.net/?f=%2035.114%3D%202%20%2A%20area_%7Btriangle%7D)
![17.55 \ units \ of \ area = area_{triangle}](https://tex.z-dn.net/?f=%2017.55%20%5C%20units%20%5C%20%20of%20%5C%20area%20%3D%20%20area_%7Btriangle%7D)
Answer:
The mechanical advantage of a machine is the ratio of the load (the resistance overcome by a machine) to the effort (the force applied). For an ideal (without friction) mechanism, it is also equal to: There is no unit for mechanical advantages since the unit for both input and output forces cancel out.
Explanation: