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3 years ago

Why were you able to see the effects of the magnetic fields using iron filings

2 answers:

6 0

The iron filings aligned themselves in a pattern, along with the magnet connecting the north and south poles. This creates the “magnetic field”. Interacting with a metal object in between the magnetic properties of the metal interact with the magnet, expressing signs of magnetic force acting on it and the metal, pulled together letting you see the see the effects of the magnetic fields using iron filings.

Talja [164]3 years ago

6 0

Iron filing are soft magnetic materials. The are highly attracted by magnets.
They are used in the study of magnetic fields.
Another reason why they are used because they are light and and they can aline themselves in the paths of the magnetic fields. This way the fields patterns can be seen clearly.

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which actions most likely cause the domains within a material to lose their alignment and become more randomized

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answer:

heating the material

placing the material in a magnetic field of opposite polarity

hitting the material

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3 years ago

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Which of the following is not a primary cause of eroding rock

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3 years ago

What will change the velocity of a periodic wave?

Gwar [14]

The one that will change the velocity of a periodic wave is :
B. Changing the medium of the wave
Waves is always determined by the properties of the medium, which means that changing the medium will change the velocity of the wave

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3 years ago

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For a moon orbiting its planet, rp is the shortest distance between the moon and its planet andra is the longest distance betwee

Natasha2012 [34]

Answer: D. 0.57

Explanation:

The formula to calculate the eccentricity $e$ of an ellipse is (assuming the moon's orbit in the shape of an ellipse):

$e=\frac{r_{a}-r_{p}}{r_{a}+r_{p}}$

Where:

$r_{a}$ is the apoapsis (the longest distance between the moon and its planet)

$r_{p}=0.27 r_{a}$ is the periapsis (the shortest distance between the moon and its planet)

Then:

$e=\frac{r_{a}-0.27 r_{a}}{r_{a}+0.27 r_{a}}$

$e=\frac{0.73 r_{a}}{1.27 r_{a}}$

$e=0.57$ This is the moon's orbital eccentricity

3 0

3 years ago

5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switc

ludmilkaskok [199]

Answer:

Δθ₁ = 172.5 rev

Δθ₁h = 43.1 rev

Δθ₂ = 920 rev

Δθ₂h = 690 rev

Explanation:

Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:

$\Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2} (1)$

Now, we need first to find the value of the angular acceleration, that we can get from the following expression:

$\omega_{f1} = \omega_{o} + \alpha * \Delta t (2)$

Since the machine starts from rest, ω₀ = 0.
We know the value of ωf₁ (the operating speed) in rev/min.
Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:

$3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)$

Replacing by the givens in (2):

$57.5 rev/sec = 0 + \alpha * 6 s (4)$

Solving for α:

$\alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)$

Replacing (5) and Δt in (1), we get:

$\Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev (6)$

in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:

$\Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev (7)$

In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:

$\Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2} (8)$

First of all, we need to find the value of the angular acceleration during the second period.
We can use again (2) replacing by the givens:
ωf =0 (the machine finally comes to an stop)
ω₀ = ωf₁ = 57.5 rev/sec
Δt = 32 s

$0 = 57.5 rev/sec + \alpha * 32 s (9)$

Solving for α in (9), we get:

$\alpha_{2} =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)$

Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:

$\Delta \theta_{2} = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)$

In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:
$\Delta \theta_{2h} = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)$

7 0

3 years ago