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ipn [44]
3 years ago
13

How does the junkyard magnet let go of metal?

Physics
1 answer:
klemol [59]3 years ago
7 0

Answer:

Wrecking yards employ extremely powerful electromagnets to move heavy pieces of scrap metal or even entire cars from one place to another. ... This creates a magnetic field around the coiled wire, magnetizing the metal as if it were a permanent magnet

Explanation:

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With the advent of the plastic balloon and the beginning of the unmanned ... That would lead to a more sophisticated ballast system that uses fine steel or iron
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What is the stretch when you pull with a force of 25 N on a spring with a spring constant of 8 N/m? *
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Hooke's Law

\tt F=k.\Delta x

k = spring constant

x = stretch

F = force

Input the value

\tt \Delta x=\dfrac{F}{k}=\dfrac{25}{8}=3.125\rightarrow 3.13\:m

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For a charged particle, a constant magnetic field can be used to change
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The image shows positivle and negative charged particles bouncing around. Which state of matter is most likely represented in th
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An electron moves with a constant horizontal velocity of 3.0 × 106 m/s and no initial vertical velocity as it enters a deflector
Ghella [55]

Answer:

a = 5.05 x 10¹⁴ m/s²

Explanation:

Consider the motion along the horizontal direction

v_{x} = velocity along the horizontal direction = 3.0 x 10⁶ m/s

t = time of travel

X = horizontal distance traveled = 11 cm = 0.11 m

Time of travel can be given as

t = \frac{X}{v_{x}}

inserting the values

t = 0.11/(3.0 x 10⁶)

t = 3.67 x 10⁻⁸ sec

Consider the motion along the vertical direction

Y = vertical distance traveled = 34 cm = 0.34 m

a = acceleration = ?

t = time of travel  = 3.67 x 10⁻⁸ sec

v_{y} = initial velocity along the vertical direction = 0 m/s

Using the kinematics equation

Y = v_{y} t + (0.5) a t²

0.34 = (0) (3.67 x 10⁻⁸) + (0.5) a (3.67 x 10⁻⁸)²

a = 5.05 x 10¹⁴ m/s²

7 0
3 years ago
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