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Mars2501 [29]
3 years ago
10

Plz helpI need it fast​

Physics
2 answers:
Rufina [12.5K]3 years ago
7 0

Answer:

perpendicular to

Explanation:

it means perpendicular to .....should u come across something like this / / , this one means parallel to .....

katovenus [111]3 years ago
3 0

Answer:

perpendicular

Explanation:

Some of the most popular symbols are:

Heart symbol: this represents love, compassion and health.

Dove symbol: this represents peace, love, and calm.

Raven symbol: this represents death and doom.

Tree symbol: this represents growth, nature, stability, and eternal life.

Owl symbol: this represents wisdom and intelligence.

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x rays with wavelength of 2.0nm scatter from a nacl crystal with plane spacing 0.281 nm find the scattering
ValentinkaMS [17]

Explanation:

It is given that,

Wavelength of x-rays = 2 nm

Plane spacing, d = 0.281 nm

It is assumed to find the scattering angle for second order maxima.

For 2nd order, Bragg's law is given by :

2d\sin\theta=n\lambda

For second order, n = 2

\sin\theta=\dfrac{n\lambda}{2d}\\\\\sin\theta=\dfrac{2\times 2\ nm}{2\times 0.28\ nm}\\\\\theta=\sin^{-1}(7.14)

Here, θ is not defined. Also, the wavelength of x-rays is more than the plane spacing. It means that it cannot produce any diffraction maximum.

4 0
3 years ago
A table is pushed it 15 metre across a room with the force of 50N.what is the work done? ​
kiruha [24]

Answer:

W=f*d

= 50 * 15

=750 J

Therefore, the work done is 750 joule( J)

6 0
3 years ago
An electron is accelerated through 2.40×10³V from rest and then enters a uniform 1.70-T magnetic field. What are(b) the minimum
Korolek [52]

The maximum value of Force is 7.88x10^{-12}

The minimum value of magnetic force is 0

Potential V=2.40x10^{3}.

Magnetic field = B = 1.70 T.

we need to find

a) The maximum value of Force

Since the particle is moving from rest, the potential energy change is 2.40x10^{3}?

Potential Energy= vq= 2,40x10^{3}x (-1.6 x 10x10^{-19}%).

= -8.84x10^{-16}

Now this will be equal to 1/2 mv²

=  1/2 mv²= 8.84x 10^{-16}.

v=2.903x10^{7}m/s :

Wow Fmax = qvb

=1.6x10^{-19}x2.903x10^{7}x1.70 .

=7.88x10^{-12}

b) The minimum value of magnetic force

The minimum Value of magnetic force will be

when sinθ=0

The minimum value is 0

To know more about magnetic field refer to brainly.com/question/13091447

#SPJ4

5 0
2 years ago
If the frequencies of two component waves are 24 Hz and 20 Hz, they should produce _______ beats per second
dimulka [17.4K]
They should produce 44 beats per second 
3 0
3 years ago
Read 2 more answers
Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.60 m away from the slits. (As
Naddik [55]

Answer:

a

   \lambda = 1.667 nm

b

     \theta  =  0.8681^o

Explanation:

From the question we are told that

   The distance of separation is d  =  0.220 \ mm  =  0.00022 \ m

    The  is distance of the screen from the slit is  D   =  2.60 \ m

    The distance between the central bright fringe and either of the adjacent bright   y  =  1.97 cm  =  1.97 *10^{-2}\ m

Generally  the condition for constructive interference is  

      d sin \tha(\theta ) =  n \lambda

From the question we are told that small-angle approximation is valid here.

So    sin (\theta ) = \theta

=>        d \theta  =  n \lambda

=>        \theta =  \frac{n *  \lambda }{d }

Here n is the order of maxima and the value is  n =  1 because we are considering the central bright fringe and either of the adjacent bright fringes

Generally the distance between the central bright fringe and either of the adjacent bright  is mathematically represented as

         y  =  D * sin (\theta )

From the question we are told that small-angle approximation is valid here.

So

       y  =  D * \theta

=>   \theta  =  \frac{ y}{D}

So

     \frac{n *  \lambda }{d } = \frac{y}{D}

     \lambda =\frac{d * y }{n * D}

substituting values

       \lambda =  \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }

        \lambda = 1.667 *10^{-6}

        \lambda = 1.667 nm

In the b part of the question we are considering the next set of bright fringe so  n=  2

    Hence

     dsin (\theta ) =  n \lambda

    \theta  =  sin^{-1}[\frac{ n  *  \lambda }{d} ]

    \theta  =  sin^{-1}[\frac{ 2  *  1667 *10^{-9}}{ 0.00022} ]

    \theta  =  0.8681^o

7 0
4 years ago
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