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3 years ago

12

A vertical wire carries current in the upward direction. an electron is traveling parallel to the wire. what is the angle α betw

een the velocity of the electron and the magnetic field of the wire?

1 answer:

Luden [163]3 years ago

7 0

Angle α between the velocity of theelectron and the magnetic field of the wire will be 90 degree

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How do nerves tie the other parts of the nervous system together?

Svetradugi [14.3K]

They do it by followinng the centeral nervous system

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3 years ago

A layer of oil (n = 1.38) floats on an unknown liquid. A ray of light originates in the oil and passes into the unknown liquid.

Vinil7 [7]

Answer:

Refractive index of unknown liquid = 1.56

Explanation:

Using Snell's law as:

$n_i\times {sin\theta_i}={n_r}\times{sin\theta_r}$

Where,

${\theta_i}$ is the angle of incidence ( 65.0° )

${\theta_r}$ is the angle of refraction ( 53.0° )

${n_r}$ is the refractive index of the refraction medium (unknown liquid, n=?)

${n_i}$ is the refractive index of the incidence medium (oil, n=1.38)

Hence,

$1.38\times {sin65.0^0}={n_r}\times{sin53.0^0}$

Solving for ${n_r}$ ,

Refractive index of unknown liquid = 1.56

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3 years ago

You have two identical boxes. You place skittles in one and the snicker bars in the other. Which box can hold more pieces of can

vichka [17]

The skittles because they are small so you can fit more in

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3 years ago

If the magnitude of the magnetic field is 6.50 mT at a distance of 12.8 cm from a long straight current carrying wire, what is t

Lunna [17]

Answer: magnitude of the magnetic field at a distance of 19.4 cm from the wire=4.29mT

Explanation:

According to Biot-Savart law, A magnetic field generated by a current carrying wire at a distance is represented as

B=μ₀I/ 2πr

B = magnetic field intensity 1000 mT =1T, 6.50mT = 6.50 X 10^-3T

μ₀ =permeability of free space 4π × 10−7 H/m

I = current intensity

r = radius, 100cm = 1m, 12.8 cm= 12.8 x 10^-2m

6.50 X 10^-3 = μ₀ x I/ 2 π X 12.8 X 10^-2

I =6.50 X 10 ^-3 X 2π X X 12.8 X 10^-2/ 4π × 10−7 H/m

I= 4160 A

when the magnetic field is at 19.4 cm from the wire

B=μ₀I/ 2πr

= 4π × 10−7 H/m x4160/ 2π x 19.4 x 10^-2

=0.004288

= 4.29x 10 ^-3T

= 4.29mT

7 0

3 years ago

A stone is thrown horizontally at 60.0 m/sm/s from the top of a very tall cliff. Calculate its horizontal position and vertical

svp [43]

Answer:

X-Positions: Y-Positions

x(0) = 0 y(0) = 0

x(2) = 120 m y(2) = 19.6 m

x(4) = 240 m y(4) = 78.4 m

x(6) = 360 m y(6) = 176.4 m

x(8) = 480 m y(8) = 313 m

x(10) = 600m y (10) = 490 m

Explanation:

X-Positions

First, we choose to take the horizontal direction as our x-axis, and the positive x-axis as positive.
After being thrown, in the horizontal direction, no external influence acts on the stone, so it will continue in the same direction at the same initial speed of 60. 0 m/s
So, in order to know the horizontal position at any time t, we can apply the definition of average velocity, rearranging terms, as follows:

$x = v_{ox} * t = 60.0 m/s * t(s)$

It can be seen that after 2 s, the displacement will be 120 m, and each 2 seconds, as the speed is constant, the displacement will increase in the same 120 m each time.

Y-Positions

We choose to take the vertical direction as our y-axis, taking the downward direction as our positive axis.
As both axes are perpendicular each other, both movements are independent each other also, so, in the vertical direction, the stone starts from rest.
At any moment, it is subject to the acceleration of gravity, g.
As the acceleration is constant, we can find the vertical displacement (taking the height of the cliff as the initial reference level), using the following kinematic equation:

$y = \frac{1}{2} * g* t^{2} = \frac{1}{2} * 9.8 m/s2 * t(s)^{2}$

Replacing by the values of t, we get the following vertical positions, from the height of the cliff as y = 0:
y(2) = 2* 9.8 m/s2 = 19.6 m
y(4) = 8* 9.8 m/s2 = 78.4 m
y(6) = 18*9.8 m/s2 = 176.4 m
y(8) = 32*9.8 m/s2 = 313.6 m
y(10)= 50 * 9.8 m/s2 = 490.0 m

5 0

3 years ago