Answer:
D. are brought from the mantle to the surface in magma that hardens into komatiite.
Explanation:
Diamond :
It is the hardest form of carbon.The atomic atoms arrange in the cubic crystal structure and this is known as diamond cubic.Another form of the diamond at room temperature is graphite.This is used for making jewelry.This is also used in the cutting process because it has high strength.
Therefore the correct option for the diamond is D.
For an uniformly accelerated motion, we can write
![2ah=v_f^2-v_0^2](https://tex.z-dn.net/?f=2ah%3Dv_f%5E2-v_0%5E2)
where
![a=g=-9.81 m/s^2](https://tex.z-dn.net/?f=a%3Dg%3D-9.81%20m%2Fs%5E2)
is the acceleration of this motion, which in this problem is the gravitational acceleration, with a negative sign because it points downward, against the direction of the motion; h=0.540 m is the distance covered by the flea, and
![v_0](https://tex.z-dn.net/?f=v_0)
is the initial velocity.
At the maximum height, the velocity is zero, so
![v_f =0](https://tex.z-dn.net/?f=v_f%20%3D0)
. Therefore we can solve to find
![v_0](https://tex.z-dn.net/?f=v_0)
:
Answer:
8050 J
Explanation:
Given:
r = 4.6 m
I = 200 kg m²
F = 26.0 N
t = 15.0 s
First, find the angular acceleration.
∑τ = Iα
Fr = Iα
α = Fr / I
α = (26.0 N) (4.6 m) / (200 kg m²)
α = 0.598 rad/s²
Now you can find the final angular velocity, then use that to find the rotational energy:
ω = αt
ω = (0.598 rad/s²) (15.0 s)
ω = 8.97 rad/s
W = ½ I ω²
W = ½ (200 kg m²) (8.97 rad/s)²
W = 8050 J
Or you can find the angular displacement and find the work done that way:
θ = θ₀ + ω₀ t + ½ αt²
θ = ½ (0.598 rad/s²) (15.0 s)²
θ = 67.3 rad
W = τθ
W = Frθ
W = (26.0 N) (4.6 m) (67.3 rad)
W = 8050 J
Answer:
a) {[1.25 1.5 1.75 2.5 2.75]
[35 30 25 20 15] }
b) {[1.5 2 40]
[1.75 3 35]
[2.25 2 25]
[2.75 4 15]}
Explanation:
Matrix H: {[1.25 1.5 1.75 2 2.25 2.5 2.75]
[1 2 3 1 2 3 4]
[45 40 35 30 25 20 15]}
Its always important to get the dimensions of your matrix right. "Roman Columns" is the mental heuristic I use since a matrix is defined by its rows first and then its column such that a 2 X 5 matrix has 2 rows and 5 columns.
Next, it helps in the beginning to think of a matrix as a grid, labeling your rows with letters (A, B, C, ...) and your columns with numbers (1, 2, 3, ...).
For question a, we just want to take the elements A1, A2, A3, A6 and A7 from matrix H and make that the first row of matrix G. And then we will take the elements B3, B4, B5, B6 and B7 from matrix H as our second row in matrix G.
For question b, we will be taking columns from matrix H and making them rows in our matrix K. The second column of H looks like this:
{[1.5]
[2]
[40]}
Transposing this column will make our first row of K look like this:
{[1.5 2 40]}
Repeating for columns 3, 5 and 7 will give us the final matrix K as seen above.
Lets suppose center of cube as origin.
All charges are equidistant from this point .
Distance = 0.5*s*√<span>(3). </span>
Total work done = 8*(9*10^9)*q/(0.5*s*√<span>(3) =
Total work </span><span> required to assemble eight identical point charges= </span>(8.31*10^10)*q J