What can we say about the rock as it falls to earth

A probe orbiting Venus exerts a gravitational force of 2. 58 × 103 N on Venus. Venus has a mass of 4. 87 × 1024 kg. The mass of

andriy [413]

Answer:

9.08

Explanation:

To solve for the distance between the probe and the center of Venus, use the formula:

$\mathrm{F = \frac{G\: m1 \:m2}{r^2}}$

Plugin the values:

G =

$6.67\times10^{-11} \mathrm{\frac{m3}{kg\: s2}}$

$\mathrm{F} =2.58 \times 10^3\mathrm{N}$

$\mathrm{m1 = 4.87 \times 10^{24}\:kg}$

$\mathrm{m2 = 655\: kg}$

Hence:

$\mathrm{F = \frac{G\: m1 \:m2}{r^2}}$

$2.58 \times 10^{23} = 6.67 \times 10^{-11} (4.87 \times 10^{24})$

$\frac{(655)}{r^2}$

$\mathrm{r = 9.08 \times 10^6 \:km}$

4 0

3 years ago

What do you think of when you hear the word education?

kakasveta [241]

Being overworked!! :D

5 0

3 years ago

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>

#include <cmath>

#include "pppm_tip4p.h"

#include "atom.h"

#include "domain.h"

#include "force.h"

#include "memory.h"

#include "error.h"

#include "math_const.h"

using namespace LAMMPS_NS;

using namespace MathConst;

#define OFFSET 16384

#ifdef FFT_SINGLE

#define ZEROF 0.0f

#define ONEF 1.0f

#else

#define ZEROF 0.0

#define ONEF 1.0

#endif

void PPPMTIP4P::init()

{

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

if (force->newton == 0)

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

PPPM::init();

}

void PPPMTIP4P::particle_map()

{

int nx,ny,nz,iH1,iH2;

double *xi,xM[3];

int *type = atom->type;

double **x = atom->x;

int nlocal = atom->nlocal;

if (!std::isfinite(boxlo[0]) || !std::isfinite(boxlo[1]) || !std::isfinite(boxlo[2]))

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

int flag = 0;

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

if (type[i] == typeO) {

find_M(i,iH1,iH2,xM);

xi = xM;

} else xi = x[i];

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

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

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

part2grid[i][0] = nx;

part2grid[i][1] = ny;

part2grid[i][2] = nz;

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

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

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

}

int flag_all;

MPI_Allreduce(&flag,&flag_all,1,MPI_INT,MPI_SUM,world);

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

}

void PPPMTIP4P::make_rho()

{

int i,l,m,n,nx,ny,nz,mx,my,mz,iH1,iH2;

FFT_SCALAR dx,dy,dz,x0,y0,z0;

double *xi,xM[3];

FFT_SCALAR *vec = &density_brick[nzlo_out][nylo_out][nxlo_out];

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

int *type = atom->type;

double *q = atom->q;

double **x = atom->x;

int nlocal = atom->nlocal;

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

if (type[i] == typeO) {

find_M(i,iH1,iH2,xM);

xi = xM;

} else xi = x[i];

nx = part2grid[i][0];

ny = part2grid[i][1];

nz = part2grid[i][2];

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

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

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

compute_rho1d(dx,dy,dz);

z0 = delvolinv * q[i];

for (n = nlower; n <= nupper; n++) {

mz = n+nz;

y0 = z0*rho1d[2][n];

for (m = nlower; m <= nupper; m++) {

my = m+ny;

x0 = y0*rho1d[1][m];

for (l = nlower; l <= nupper; l++) {

mx = l+nx;

density_brick[mz][my][mx] += x0*rho1d[0][l];

}

}

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

#SPJ1

6 0

2 years ago

What is the melting point of a substance?

Veronika [31]

The correct answer should be C. The temperature at which the solid state turns into a liquid state.
A doesn’t make sense (a solid needs to melt, not a liquid)
B is evaporation I would think.

3 0

3 years ago

Which statement describes the 3d, 4s, and 4p orbitals of arsenic (as) based on its electronic configuration and position in the

professor190 [17]

The statement that describes the 3d, 4s, and 4p orbitals of arsenic (as) based on its electronic configuration and position in the periodic table is; The 3d and 4s orbitals are completely filled, and the 4p orbital is partially filled.

The electronic configuration of arsenic is given as; 1s² 2s² 2p^(6) 3s² 3p^(6) 4s² 3d^(10) 4p³.

Now we can see that in that electronic configuration the orbitals are, s, p and D.

Now, the maximum electrons for each of the 3 orbitals are;

Maximum electrons for d orbital = 10 electrons

Maximum electrons for s orbital = 2 electrons

Maximum electrons for p orbital = 6 electrons.

Now, when we compare the maximum number of electrons for each orbital to the ones in electron configuration of arsenic, it is clear from the last 3 orbitals that the 3d and 4s orbitals are completely filled while the 4p orbital is half filled or partially filled.

Read more about electronic configuration at: brainly.com/question/24258336

6 0

3 years ago