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Neporo4naja [7]
3 years ago
15

What is the direction of the field halfway between two horizontal parallel wires if the top wire has a current of 4 A to the lef

t and the bottom wire has a current of 2 A to the right?
A) Out of the page
B) into the page
C) Left
D) The magnetic fields cancel out at that position
Physics
1 answer:
Alex777 [14]3 years ago
4 0

Answer:

A) Out of the page.

Explanation:

Right-hand rule points the direction of the magnetic field at any point.

<u>Top wire</u>: Current is to the left. Point your thumb to the left and curl your other fingers around the wire. The tips of the four fingers points the direction of the field at that point. In this case, out of the page.

<u>Bottom wire</u>: Current is to the right. Point your thumb to the right and curl your other fingers around the wire. The tips of the four finger points out of the page again.

So, the total field produced by both wires is directed out of the page.

Another method to figure out the direction is the mathematical method.

Use the B-field formula:

d\vec{B} = \frac{\mu_0}{4\pi}\frac{Id\vec{l}\times \^r}{r^2}

The cross product between the direction of the current and the target position gives the direction of the B-field. If the left is -x direction and downwards is the -y direction, then

(-\^x) \times (-\^y) = +\^z for the top wire.

(+\^x) \times (+\^y) = +\^z for the bottom wire.

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A wooden rod of negligible mass and length 80.0cm is pivoted about a horizontal axis through its center. A white rat with mass 0
algol13

Answer:

The speed of the animals is 1.64m/s.

Explanation:

Let us work with variables and call the mass of the two rats m_1 and m_2,  and the length of the rod L.

Using the law of conservation of energy, which says the potential energies of the rats must equal their kinetic energies, we know that when the rod swings to the vertical position,

$m_1\frac{L}{2}g -m_2\frac{L}{2}g = \frac{1}{2}m_1v^2+\frac{1}{2}m_2v^2$

$(m_1 -m_2)g\frac{L}{2} = \frac{1}{2}(m_1+m_2)v^2$,

solving for v, we get:

$\boxed{v = \sqrt{\frac{(m_1 -m_2)gL}{(m_1+m_2)}} }$

Putting in the values for m_1, m_2, g, and L we get:

$v = \sqrt{\frac{(0.450kg -0.220kg)(9.8m/s^2)(0.8m)}{(0.450kg+0.220kg)}} $

\boxed{ v= 1.64m/s}

Therefore, as the rod swings through the vertical position , the speed of the rats is 1.64 m/s.

7 0
2 years ago
In order to determine the mass moment of inertia of a flywheel of radius 600 mm, a 12-kg block is attached to a wire that is wra
shtirl [24]

Answer:

Explanation:

Given that,

When Mass of block is 12kg

M = 12kg

Block falls 3m in 4.6 seconds

When the mass of block is 24kg

M = 24kg

Block falls 3m in 3.1 seconds

The radius of the wheel is 600mm

R = 600mm = 0.6m

We want to find the moment of inertia of the flywheel

Taking moment about point G.

Then,

Clockwise moment = Anticlockwise moment

ΣM_G = Σ(M_G)_eff

M•g•R - Mf = I•α + M•a•R

Relationship between angular acceleration and linear acceleration

a = αR

α = a / R

M•g•R - Mf = I•a / R + M•a•R

Case 1, when y = 3 t = 4.6s

M = 12kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 4.6²

3 × 2 = 4.6²a

a = 6 / 4.6²

a = 0.284 m/s²

M•g•R - Mf = I•a / R + M•a•R

12 × 9.81 × 0.6 - Mf = I × 0.284/0.6 + 12 × 0.284 × 0.6

70.632 - Mf = 0.4726•I + 2.0448

Re arrange

0.4726•I + Mf = 70.632-2.0448

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Second case

Case 2, when y = 3 t = 3.1s

M= 24kg

Using equation of motion

y = ut + ½at², where u = 0m/s

3 = ½a × 3.1²

3 × 2 = 3.1²a

a = 6 / 3.1²

a = 0.6243 m/s²

M•g•R - Mf = I•a / R + M•a•R

24 × 9.81 × 0.6 - Mf = I × 0.6243/0.6 + 24 × 0.6243 × 0.6

141.264 - Mf = 1.0406•I + 8.99

Re arrange

1.0406•I + Mf = 141.264 - 8.99

1.0406•I + Mf = 132.274 equation 2

Solving equation 1 and 2 simultaneously

Subtract equation 1 from 2,

Then, we have

1.0406•I - 0.4726•I = 132.274 - 68.5832

0.568•I = 63.6908

I = 63.6908 / 0.568

I = 112.13 kgm²

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