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3 years ago

Which Of The Following Research Methods Are Widely Used By Psychologists?

Physics

1 answer:

goldenfox [79]3 years ago

4 0

Answer: As you said D.

Explanation:

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An electron, moving south, enters a magnetic field of certain strength. Because of this field the electron curves upward. What i

e-lub [12.9K]

Answer: The magnetic field points to the west.

Explanation:

If we take the plane as:

North = positive y-axis

East = positive x-axis.

We have that the electron is moving south, so the velocity of the electron can be written in vector form as:

V = (0, -v, 0)

Now, when the electron interacts with the magnetic field, the electron moves upwards.

We know that the interaction between an electron and a magnetic field is:

F = q*VxB

So we have that this force acts on the z-direction.

Now, to solve this we can use the righ hand rule.

First, we point wit our hand to the direction of the velocity (negative y-axis) now, we want that our thumb points up (the direction of the force) so the side where our palm faces is the direction of the field B.

But, remember that an electron has a negative charge, so the actual equation is:

F = -q*VxB

So the magnetic field actually points in the opposite direction of our palm, to the west.

Now, we can also solve it mathematically as:

F = (0, 0, f) = -q*(0, -v,0)x(a, b, c)

where B = (a, b, c) is the vector of the magnetic field.

(0, 0, f) = -q*(-v*b -0, -0 + 0, 0 -(- a*v)) = -q*(-v*b, 0, a*v)

then we have that b must be equal to zero, and that:

f = -q*a*v

and f is positive, then we have:

a = -(f/q*v)

then the vector of the magnetic field is:

B = (-(f/q*v), 0, 0)

so it points in the negative x-axis, that is the West, as we found earlier.

4 0

3 years ago

Let surface S be the boundary of the solid object enclosed by x^2+z^2=4, x+y=6, x=0, y=0, and z=0. and, let f(x,y,z)=(3x)i+(x+y+

babunello [35]

a. I've attached a plot of the surface. Each face is parameterized by

• $\mathbf s_1(x,y)=x\,\mathbf i+y\,\mathbf j$ with $0\le x\le2$ and $0\le y\le6-x$ ;

• $\mathbf s_2(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf k$ with $0\le u\le2$ and $0\le v\le\frac\pi2$ ;

• $\mathbf s_3(y,z)=y\,\mathbf j+z\,\mathbf k$ with $0\le y\le 6$ and $0\le z\le2$ ;

• $\mathbf s_4(u,v)=u\cos v\,\mathbf i+(6-u\cos v)\,\mathbf j+u\sin v\,\mathbf k$ with $0\le u\le2$ and $0\le v\le\frac\pi2$ ; and

• $\mathbf s_5(u,y)=2\cos u\,\mathbf i+y\,\mathbf j+2\sin u\,\mathbf k$ with $0\le u\le\frac\pi2$ and $0\le y\le6-2\cos u$ .

b. Assuming you want outward flux, first compute the outward-facing normal vectors for each face.

$\mathbf n_1=\dfrac{\partial\mathbf s_1}{\partial y}\times\dfrac{\partial\mathbf s_1}{\partial x}=-\mathbf k$

$\mathbf n_2=\dfrac{\partial\mathbf s_2}{\partial u}\times\dfrac{\partial\mathbf s_2}{\partial v}=-u\,\mathbf j$

$\mathbf n_3=\dfrac{\partial\mathbf s_3}{\partial z}\times\dfrac{\partial\mathbf s_3}{\partial y}=-\mathbf i$

$\mathbf n_4=\dfrac{\partial\mathbf s_4}{\partial v}\times\dfrac{\partial\mathbf s_4}{\partial u}=u\,\mathbf i+u\,\mathbf j$

$\mathbf n_5=\dfrac{\partial\mathbf s_5}{\partial y}\times\dfrac{\partial\mathbf s_5}{\partial u}=2\cos u\,\mathbf i+2\sin u\,\mathbf k$

Then integrate the dot product of f with each normal vector over the corresponding face.

$\displaystyle\iint_{S_1}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{6-x}f(x,y,0)\cdot\mathbf n_1\,\mathrm dy\,\mathrm dx$

$=\displaystyle\int_0^2\int_0^{6-x}0\,\mathrm dy\,\mathrm dx=0$

$\displaystyle\iint_{S_2}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,0,u\sin v)\cdot\mathbf n_2\,\mathrm dv\,\mathrm du$

$\displaystyle=\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=-8$

$\displaystyle\iint_{S_3}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^6\mathbf f(0,y,z)\cdot\mathbf n_3\,\mathrm dy\,\mathrm dz$

$=\displaystyle\int_0^2\int_0^60\,\mathrm dy\,\mathrm dz=0$

$\displaystyle\iint_{S_4}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,6-u\cos v,u\sin v)\cdot\mathbf n_4\,\mathrm dv\,\mathrm du$

$=\displaystyle\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=\frac{40}3+6\pi$

$\displaystyle\iint_{S_5}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^{\frac\pi2}\int_0^{6-2\cos u}\mathbf f(2\cos u,y,2\sin u)\cdot\mathbf n_5\,\mathrm dy\,\mathrm du$

$=\displaystyle\int_0^{\frac\pi2}\int_0^{6-2\cos u}12\,\mathrm dy\,\mathrm du=36\pi-24$

c. You can get the total flux by summing all the fluxes found in part b; you end up with 42π - 56/3.

Alternatively, since S is closed, we can find the total flux by applying the divergence theorem.

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\iiint_R\mathrm{div}\mathbf f(x,y,z)\,\mathrm dV$

where R is the interior of S. We have

$\mathrm{div}\mathbf f(x,y,z)=\dfrac{\partial(3x)}{\partial x}+\dfrac{\partial(x+y+2z)}{\partial y}+\dfrac{\partial(3z)}{\partial z}=7$

The integral is easily computed in cylindrical coordinates:

$\begin{cases}x(r,t)=r\cos t\\y(r,t)=6-r\cos t\\z(r,t)=r\sin t\end{cases},0\le r\le 2,0\le t\le\dfrac\pi2$

$\displaystyle\int_0^2\int_0^{\frac\pi2}\int_0^{6-r\cos t}7r\,\mathrm dy\,\mathrm dt\,\mathrm dr=42\pi-\frac{56}3$

as expected.

4 0

3 years ago

Suppose that you are swimming in a river while a friend watches from the shore. In calm water, you swim at a speed of 1.25 m/s .

aliya0001 [1]

Answer: The observing friend will the swimmer moving at a speed of 0.25 m/s.

Explanation:

Let S be the speed of the swimmer, given as 1.25 m/s

Let $S_{0}$ be the speed of the river's current given as 1.00 m/s.

Note that this speed is the magnitude of the velocity which is a vector quantity.

The direction of the swimmer is upstream.

Hence the resultant velocity is given as, $S_{R}$ = S — S 0 $S_{0}$

$S_{R}$ = 1.25 — 1

$S_{R}$ = 0.25 m/s.

Therefore, the observing friend will see the swimmer moving at a speed of 0.25 m/s due to resistance produced by the current of the river.

6 0

3 years ago

What is the correct name for this formula BaCI 2?

KIM [24]

BaCI2 stands for Barium Chloride.

3 0

3 years ago

Read 2 more answers

3. A pendulum with a 1.0-kg weight is set in motion from a position 0.04 m above the lowest point on the path of the weight.

gavmur [86]

Answer: K.E = 0.4 J

Explanation:

Given that:

M = 1.0 kg

h = 0.04 m

K.E = ?

According to conservative of energy

K.E = P.E

K.E = mgh

K.E = 1 × 9.81 × 0.04

K.E = 0.3924 Joule

The kinetic energy of the pendulum at the lowest point is 0.39 Joule

6 0

3 years ago