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BlackZzzverrR [31]
4 years ago
12

__________ separating lanes mean changing lanes is illegal.

SAT
2 answers:
BartSMP [9]4 years ago
7 0
B. One solid white line

You can’t switch lanes if there is a solid white line on the road
lana [24]4 years ago
4 0

Answer:

B. One solid white line separating lanes mean changing lanes is illegal.

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Now that we have gotten to know Humbug more... what is a common character trait for him?
sukhopar [10]
He’s really selfish and he talks a lot
8 0
3 years ago
A cube of osmium metal 1. 500 cm on a side has a mass of 76. 31 g at 25 ∘c. What is its density in g/cm3 at this temperature?.
ivolga24 [154]

Answer:

22.61 g/cm³

Explanation:

The density is the mass divided by the volume

The volume of a cube is:

1.500³ = 3.375 cm³

76.31/3.375 = 22.61 g/cm³

7 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%20n%20-%20%5Csqrt%7Bc%20%2B%205%20%3D%201%7D%20" id="TexFormula1" title=" n - \sqrt{c + 5
Salsk061 [2.6K]

Answer:

c=11

Explanation:

n -  \sqrt{c  +  5}  = 1

5 -  \sqrt{c + 5}  = 1

Add sqr root of c+5 to both sides

5 = 1 +  \sqrt{c + 5}

4 =  \sqrt{c + 5}

16 = c + 5

c = 11

7 0
3 years ago
Read 2 more answers
Lammps ERROR on proc 0: Out of range atoms - cannot compute PPPM
Alexeev081 [22]

Using the knowledge in computational language in C++ it is possible to write a code that Out of range atoms:

<h3>Writting the code in C++:</h3>

<em>#include <cmath></em>

<em>#include "pppm_tip4p.h"</em>

<em>#include "atom.h"</em>

<em>#include "domain.h"</em>

<em>#include "force.h"</em>

<em>#include "memory.h"</em>

<em>#include "error.h"</em>

<em>#include "math_const.h"</em>

<em />

<em>using namespace LAMMPS_NS;</em>

<em>using namespace MathConst;</em>

<em />

<em>#define OFFSET 16384</em>

<em />

<em>#ifdef FFT_SINGLE</em>

<em>#define ZEROF 0.0f</em>

<em>#define ONEF  1.0f</em>

<em>#else</em>

<em>#define ZEROF 0.0</em>

<em>#define ONEF  1.0</em>

<em>#endif</em>

<em />

<em>void PPPMTIP4P::init()</em>

<em>{</em>

<em>  // TIP4P PPPM requires newton on, b/c it computes forces on ghost atoms</em>

<em />

<em>  if (force->newton == 0)</em>

<em>    error->all(FLERR,"Kspace style pppm/tip4p requires newton on");</em>

<em />

<em>  PPPM::init();</em>

<em>}</em>

<em />

<em>void PPPMTIP4P::particle_map()</em>

<em>{</em>

<em>  int nx,ny,nz,iH1,iH2;</em>

<em>  double *xi,xM[3];</em>

<em />

<em>  int *type = atom->type;</em>

<em>  double **x = atom->x;</em>

<em>  int nlocal = atom->nlocal;</em>

<em />

<em>  if (!std::isfinite(boxlo[0]) || !std::isfinite(boxlo[1]) || !std::isfinite(boxlo[2]))</em>

<em>    error->one(FLERR,"Non-numeric box dimensions - simulation unstable");</em>

<em />

<em>  int flag = 0;</em>

<em>  for (int i = 0; i < nlocal; i++) {</em>

<em>    if (type[i] == typeO) {</em>

<em>      find_M(i,iH1,iH2,xM);</em>

<em>      xi = xM;</em>

<em>    } else xi = x[i];</em>

<em />

<em>    nx = static_cast<int> ((xi[0]-boxlo[0])*delxinv+shift) - OFFSET;</em>

<em>    ny = static_cast<int> ((xi[1]-boxlo[1])*delyinv+shift) - OFFSET;</em>

<em>    nz = static_cast<int> ((xi[2]-boxlo[2])*delzinv+shift) - OFFSET;</em>

<em />

<em>    part2grid[i][0] = nx;</em>

<em>    part2grid[i][1] = ny;</em>

<em>    part2grid[i][2] = nz;</em>

<em />

<em>    if (nx+nlower < nxlo_out || nx+nupper > nxhi_out ||</em>

<em>        ny+nlower < nylo_out || ny+nupper > nyhi_out ||</em>

<em>        nz+nlower < nzlo_out || nz+nupper > nzhi_out) flag++;</em>

<em>  }</em>

<em />

<em>  int flag_all;</em>

<em>  MPI_Allreduce(&flag,&flag_all,1,MPI_INT,MPI_SUM,world);</em>

<em>  if (flag_all) error->all(FLERR,"Out of range atoms - cannot compute PPPM");</em>

<em>}</em>

<em>void PPPMTIP4P::make_rho()</em>

<em>{</em>

<em>  int i,l,m,n,nx,ny,nz,mx,my,mz,iH1,iH2;</em>

<em>  FFT_SCALAR dx,dy,dz,x0,y0,z0;</em>

<em>  double *xi,xM[3];</em>

<em />

<em>  FFT_SCALAR *vec = &density_brick[nzlo_out][nylo_out][nxlo_out];</em>

<em>  for (i = 0; i < ngrid; i++) vec[i] = ZEROF;</em>

<em />

<em>  int *type = atom->type;</em>

<em>  double *q = atom->q;</em>

<em>  double **x = atom->x;</em>

<em>  int nlocal = atom->nlocal;</em>

<em />

<em>  for (int i = 0; i < nlocal; i++) {</em>

<em>    if (type[i] == typeO) {</em>

<em>      find_M(i,iH1,iH2,xM);</em>

<em>      xi = xM;</em>

<em>    } else xi = x[i];</em>

<em />

<em>    nx = part2grid[i][0];</em>

<em>    ny = part2grid[i][1];</em>

<em>    nz = part2grid[i][2];</em>

<em>    dx = nx+shiftone - (xi[0]-boxlo[0])*delxinv;</em>

<em>    dy = ny+shiftone - (xi[1]-boxlo[1])*delyinv;</em>

<em>    dz = nz+shiftone - (xi[2]-boxlo[2])*delzinv;</em>

<em />

<em>    compute_rho1d(dx,dy,dz);</em>

<em />

<em>    z0 = delvolinv * q[i];</em>

<em>    for (n = nlower; n <= nupper; n++) {</em>

<em>      mz = n+nz;</em>

<em>      y0 = z0*rho1d[2][n];</em>

<em>      for (m = nlower; m <= nupper; m++) {</em>

<em>        my = m+ny;</em>

<em>        x0 = y0*rho1d[1][m];</em>

<em>        for (l = nlower; l <= nupper; l++) {</em>

<em>          mx = l+nx;</em>

<em>          density_brick[mz][my][mx] += x0*rho1d[0][l];</em>

<em>        }</em>

<em>      }</em>

<em>    }</em>

<em>  }</em>

<em>}</em>

See more about C++ at brainly.com/question/19705654

#SPJ1

6 0
2 years ago
The writer is working on a revision of the passage and has written the following thesis statement for it. Based on the evidence
hram777 [196]

Writers often write stories or fictions from any perspective of their choice. The writer's main claim or thesis in the new version of the passage is Da Vinci was a great artist because he had exotropia.

Exotropia is a medical condition where there is an outward turning of the eyes.

It is known to be a very common type of strabismus often holding up to 25 percent of all ocular misalignment in early childhood.

People with this condition often notice their vision then becomes blurry or they have double vision when their eye points outward .

The options to the question is below

Which of the following is most likely the main claim that the writer is planning to make in the new version of the passage?

A The only way for an artist to be successful is to have exotropia.

B. Da Vinci was more successful as an artist than others who also had exotropia.

C. Studying just a few works of art can provide compelling information about an artist’s physical conditions.

D. The only way to fully understand an artist’s work is to study the artist’s physical conditions.

E. Da Vinci was a great artist because he had exotropia

Learn more from

brainly.com/question/25222246

4 0
3 years ago
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