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Archy [21]
3 years ago
5

schools often use concrete rewards to increase adaptive behaviors. how might this be a disadvantage for students intrinsically m

otivated to learn?
SAT
1 answer:
OleMash [197]3 years ago
6 0

Answer:

Intrinsic motivation is a type of motivation that an individual has that comes internally, that is, the individual does a specific task or activity mainly because he or she enjoys it or wants to learn from it. In this scenario, a student studies or behaves properly because he or she is motivated to become better instead of doing it because he or she will receive a reward or a good appraisal from the teacher or parents.

Explanation:

You might be interested in
After the torque has been applied for 0.3 s, what are the magnitude and direction of the angular momentum about the center of th
vredina [299]

The magnitude and direction of the angular momentum about the center of the disk is 0.996 kgm²/s.

<h3>Angular acceleration of the disk</h3>

The angular acceleration of the disk is determined by applying the principle of conservation of angular momentum as shown below;

Fr = Iα

α = Fr/I

Where;

  • I is the moment of inertia of the solid disk

I = ¹/₂MR²

I = ¹/₂ x (0.5) x (0.08)²

I = 1.6 x 10⁻³ kgm²

\alpha = \frac{14 \times 0.22}{1.6\times 10^{-3}}\\\\\alpha = 1,925 \ rad/s^2

<h3>Final angular velocity of the disk</h3>

ωf = ωi + αt

ωf = 45 + (1925 x 0.3)

ωf = 622.5 rad/s

<h3>Angular momentum</h3>

L = Iω

L = (1.6 x 10⁻³) x (622.5)

L = 0.996 kgm²/s

Thus, the magnitude and direction of the angular momentum about the center of the disk is 0.996 kgm²/s.

The complete question is below:

A uniform solid disk with radius 8 cm has mass 0.5 kg (moment of inertia I = (1/2)MR²). A constant force 14 N is applied as shown. At the instant shown, the amount of string that has wound off the disk is d = 22 cm, and the angular velocity of the disk is 45 radians/s into the page.

Learn more about angular momentum here: brainly.com/question/7538238

4 0
2 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>

<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
A disk-shaped platform has a known rotational inertia. the platform is mounted
notka56 [123]

Based on the scenario of things, No additional data are necessary, due to the fact that the rotational inertia of the disk and its initial angular velocity are known.

<h3>What is  angular impulse?</h3>

The term angular impulse is known to be the output of a torque and the time it took to be equal to the change found in angular momentum of a body that is said to be free to move.

Note that in the above situation, what should be used to determine the angular impulse on the disk is already given and as such, No additional data are necessary, because the rotational inertia of the disk and its initial angular velocity are known.

See full question below

What additional data, if any, should a student collect to determine the angular impulse on the disk?

A

The time interval in which the net torque is applied, because a net torque is not exerted on the disk at a single instant in time.

B

The net torque exerted on the disk, because a net torque is responsible for an angular impulse.

C

The force of friction exerted on the disk, because a force component perpendicular to the line connecting the axis of rotation and the point of application of the force results in a torque about that axis.

D

No additional data are necessary, because the rotational inertia of the disk and its initial angular velocity are known.

Learn more about platform from

brainly.com/question/13792010

3 0
2 years ago
How are refraction and diffraction similar? how are they different? check all that apply
FinnZ [79.3K]

Explanation:

they are diffrent cause refraction is solid and diffrection is liquid

6 0
3 years ago
Of segments and , which of the segments has a greater area based on the given information? justify with your work. circle has ra
WITCHER [35]

Of segments and, which of the segments has a greater area based on the given information circle has a radius 12m has the greater base.

<h3>What is the area of base?</h3>

Substitute the fee of "r" into the equation for the place of a circle: place = πr^2. Note that π is the image for pi, which is about three.14. For example, a circle with a radius of three cm could yield an equation like this: place = π3^2.

  1. Once you discover the place of the world, all you need to do is get the place of the triangle after which subtract the place of the triangle from the world and to be able to let you know the place of CFD and CED and can help you realize which one is more.
  2. For instance, Lets discover A, area place = 90/360 x πx10 ^ 2 = 25pi
  3. Now we could get the place of the triangle, that is place triangle = 1/2x bh xsin C
  4. Now we could discover B of the triangle, that is place angle = 1/2x bhx sin place triangle = 1/2 x 12 x12sin 60 = 36 sqrt ( three CFD=sec place - place of gle = 24 pi-36 sqrt three)=13.04m^ .
  5. So A has the more area base.

Read more about the area:

brainly.com/question/25689052

#SPJ1

8 0
2 years ago
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