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3 years ago

SOHCAHTOA MATH REVIEW Find the angle x from the diagram above

Physics

1 answer:

mestny [16]3 years ago

5 0

Explanation:

Hey there!

Here,

It is a Right angled triangle.

Now, Taking angle x as a reference angle.

We get,

perpendicular (p) = 24.

Hypotenuse (h) = 29.

Using sin ratio we get.

$\sin( \alpha ) = \frac{p }{h}$

Now, keep values and simplify them.

$\sin(x) = \frac{24}{29}$

sinx = 0.8275.

$x = { \sin }^{ - 1} (0.8275)$

Therefore the value of x is 55.842°. Or 55°.

Hope it helps....

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Two pulses are moving along a string. One pulse is

raketka [301]

Answer:

The answer is the 3rd option!

6 0

2 years ago

A toroid having a square cross section, 5.49 cm on a side, and an inner radius of 18.1 cm has 450 turns and carries a current of

joja [24]

Answer:

(a) 4.27 x 10^-4 Telsa

(b) 3.28 x 10^-4 Telsa

Explanation:

side of square, a = 5.49 cm

inner radius, r = 18.1 cm = 0.181 m

number of turns,N = 450

current, i = 0.859 A

(a)

The magnetic field due to a solenoid due to inner radius is

$B = \frac{\mu_{0}Ni}{2\pi r_{inner}}$

$B = \frac{4\pi\times 10^{-7}\times 450\times 0.859}{2\pi \times 0.181}$

B = 4.27 x 10^-4 Telsa

(b)

The outer radius is R = 18.1 + 5.49 = 23.59 cm = 0.236 m

The magnetic field due to the outer radius is

$B = \frac{\mu_{0}Ni}{2\pi r_{outer}}$

$B = \frac{4\pi\times 10^{-7}\times 450\times 0.859}{2\pi \times 0.236}$

B = 3.28 x 10^-4 Tesla

7 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

2 years ago

6.) What is

Feliz [49]

Answer:

100 V

Explanation:

Hi there!

Ohm's law states that $V=IR$ where V is the voltage, I is the current and R is the resistance.

Plug the given information into Ohm's law (R=50, I=A)

$V=IR\\V=(2)(50)\\V=100$

Therefore, the voltage across this current is 100 V.

I hope this helps!

6 0

2 years ago

Calculate the velocity of a wave that has a frequency of 60 Hz and wavelength of 2.0 m/s

mote1985 [20]

Answer:We have , a relation in frequency f and wavelength λ of a wave having the velocity v as ,

v=fλ ,

given f=60Hz , λ=20m ,

therefore velocity of wave , v=60×20=1200m/s

3 0

3 years ago