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Lisa [10]
2 years ago
6

A very long half-cylinder of radius r carries a total current is uniformly distributed across the circumference. Determine the m

agnetic field along the central axis of the cylinder.
Physics
1 answer:
Rufina [12.5K]2 years ago
8 0

The magnetic field along the central axis of the cylinder will be <em>determined </em>using the formula given that all parameters are known

dBz=\frac{uoR^2di}{2r^3}

<h3>Magnetic field </h3>

Generally, A magnetic field is also referred to as a vector field that shows the magnetic presence on electric charges and magnetic materials.

The formula for magnetic field  is

Where

u=4\pi*10^{-7H/M}

Therefore, the magnetic field along the central axis of the cylinder will be determined using the formula given that all parameters are known

For more information on Magnets

brainly.com/question/7802337

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A ball thrown into the air has its greatest kinetic energy as it reaches the highest point in its path. please select the best a
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F - False.

Its greatest kinetic energy is at the point of release.

It has the least kinetic energy, zero, at its highest point in its path.
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A woman can row her canoe at 2.5 m/s. If she faces an opposing current of 3.0 m/s, how fast will she go forward?
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Answer:

V = - 0.5 [m/s]

Explanation:

In order to solve this problem, we must use the principle of relative speeds. This is for an observer who is on the edge of the river he can see how the river moves to the left and the woman tries to move to the right but can not since:

V_{total}=-3+2.5\\V_{total}=-0.5 [m/s]

That is, the person sees how the woman moves to the left but with avelocity of 0.5 [m/s] to the left

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3 years ago
A student pushes against a wall with a force of 30N. The wall does not move. What amount of force does the wall exert on the stu
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C

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they both have to be the same for both to not move

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2 years ago
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Suppose a rocket ship in deep space moves with constant acceleration equal to 9.80 m/s2, which gives the illusion of normal grav
DochEvi [55]

Answer:

a) 3673469.39 seconds

b) 6.61×10¹⁴ m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0.12×3×10⁸ m/s

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

Equation of motion

v=u+at\\\Rightarrow 0.12\times 3\times 10^8=0+9.8t\\\Rightarrow t=\frac{0.12\times 3\times 10^8}{9.8}=3673469.39\ s

Time taken to reach 12% of light speed is 3673469.39 seconds

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{(0.12\times 3\times 10^8)^2-0^2}{2\times 9.8}\\\Rightarrow s=6.61 \times 10^{14}\ m

The distance it would have to travel is 6.61×10¹⁴ m

7 0
3 years ago
SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the
Firlakuza [10]

Answer:

The length of his shadow is decreasing at a rate of 1.13 m/s

Explanation:

The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.

Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.

By similar triangles,

H/D = h/L

L = hD/H

Also, when the man is 4 m from the building, the length of his shadow is L = D - 4

So, D - 4 = hD/H

H(D - 4) = hD

H = hD/(D - 4)

Since h = 2 m and D = 12 m,

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Since L = hD/H

and h and H are constant, differentiating L with respect to time, we have

dL/dt = d(hD/H)/dt

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Since h = 2 m and H = 3 m,

dL/dt = h(dD/dt)/H

dL/dt = 2 m(-1.7 m/s)/3 m

dL/dt = -3.4/3 m/s

dL/dt = -1.13 m/s

So, the length of his shadow is decreasing at a rate of 1.13 m/s

5 0
3 years ago
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