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stepan [7]
3 years ago
5

What needs to be the relation between the two forces to cause the upper magnet to float without moving?

Physics
1 answer:
OleMash [197]3 years ago
7 0

Answer:

If it points the other way, the fields subtract, for a lower energy, and so the magnet prefers to turn to point in this way. Magnets in uniform fields feel torques which make them turn around if they are not pointing in the right direction, but there is no net force making the magnet want to levitate.

Explanation:

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Light passes through a 0.15 mm-wide slit and forms a diffraction pattern on a screen 1.25 m behind the slit. The width of the ce
Elena L [17]

Answer:

The value is  \lambda =  900 \ nm

Explanation:

From the question we are told that

   The width of the slit is  a =  0.15 \ mm = 0.00015 \  m

     The distance of the screen from the slit is D = 1.25 m

      The width of the central maximum is y =  0.75 \ cm  = 0.0075 \ m

Generally the width of the central maximum is mathematically represented as

         y   =  \frac{m *  D  *  \lambda}{a}

Here  m is the order of the fringe and given that we are considering the central maximum, the order will be  m =  1  because the with of the central maximum separate's the and first maxima

So

        \lambda     =     \frac{a y}{ m *  D }

=>     \lambda     =     \frac{ 0.000015 *  0.0075}{ 1  *  1.2 }

=>     \lambda     =   900 *10^{-9} \  m

=>      \lambda =  900 \ nm

6 0
3 years ago
FIRST PERSON WILL BE MARKED BRAINLIEST, THANKED, AND RATED A 5!
svetoff [14.1K]
A is the answer to what your asking
4 0
3 years ago
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An alpha particle has a charge of +2e and a mass of 6.64 x 10-27 kg. It is accelerated from rest through a potential difference
kondor19780726 [428]

Answer:

a) v = 1.075*10^7 m/s

b) FB = 7.57*10^-12 N

c) r = 10.1 cm

Explanation:

(a) To find the speed of the alpha particle you use the following formula for the kinetic energy:

K=qV          (1)

q: charge of the particle = 2e = 2(1.6*10^-19 C) = 3.2*10^-19 C

V: potential difference = 1.2*10^6 V

You replace the values of the parameters in the equation (1):

K=(3.2*10^{-19}C)(1.2*10^6V)=3.84*10^{-13}J

The kinetic energy of the particle is also:

K=\frac{1}{2}mv^2       (2)

m: mass of the particle = 6.64*10^⁻27 kg

You solve the last equation for v:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(3.84*10^{-13}J)}{6.64*10^{-27}kg}}\\\\v=1.075*10^7\frac{m}{s}

the sped of the alpha particle is 1.075*10^6 m/s

b) The magnetic force on the particle is given by:

|F_B|=qvBsin(\theta)

B: magnitude of the magnetic field = 2.2 T

The direction of the motion of the particle is perpendicular to the direction of the magnetic field. Then sinθ = 1

|F_B|=(3.2*10^{-19}C)(1.075*10^6m/s)(2.2T)=7.57*10^{-12}N

the force exerted by the magnetic field on the particle is 7.57*10^-12 N

c) The particle describes a circumference with a radius given by:

r=\frac{mv}{qB}=\frac{(6.64*10^{-27}kg)(1.075*10^7m/s)}{(3.2*10^{-19}C)(2.2T)}\\\\r=0.101m=10.1cm

the radius of the trajectory of the electron is 10.1 cm

6 0
4 years ago
SOMEONE PLEASE HELP ME ASAP PLEASE!!!!!​
Zina [86]

Answer:

x₁ = 58.09 m

Explanation:

Displacement is calculated by finding the final distance away from a point then subtracting the initial distance.

Given that initial position of object is x=25.89 m

Displacement Δx=32.2 m

Final position x₁=?

Δx = x₁-x

32.2= x₁-25.89

32.2+25.89 = x₁

58.09 m =x₁

5 0
3 years ago
Find the conductivity of a conduit with a cross-sectional area of 0.60 cm2 and a length of 15 cm, given that its conductance G i
marta [7]
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