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EastWind [94]
3 years ago
14

If a permanent magnet is dropped or struck by a hammer, it may loose it's magnetism. Explain why.

Physics
1 answer:
Rainbow [258]3 years ago
8 0
This is because strong vibrations will affect the alignment of the magnetic domains within the magnetic material. The domains would change the alignment thus affecting the strength of the magnets which solely depends on the alignment of the domains. The strength of a magnet is the sum of the magnetic fields of all the domains in the magnet. Therefore, the net result will be a weaker magnet or the magnet loosing its magnetism.
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A hydraulic system for a dentist's chair is designed to be able to lift 3,112 newtons. The surface area over which this force is
In-s [12.5K]

Answer:

13.8 N

Explanation:

Pressure on the one end of the hydraulic system = Pressure on the other end

Pressure = Force / Area where Force is in Newton, area is in m²

so Force of one end (F1) / area of that end = force of the other end (F2) / area of that end

3112 / ( 707 /10000) in m² = F2 / ( 3.14 / 10000) in m²

cross multiply

44016.97 × 0.000314 = 13.82 N

5 0
3 years ago
Which of the following is a conversion of potential to kinetic energy?
ruslelena [56]
Its either releasing a stretch rubber band and a ball rolling down a hill because rubber have a potential to move when its stretch and its an eleastic potential energy also a ball rolling down a hill is a gravitational to kinetic energy because the ball started at rest which is art the top of the hill and it has a potential to fall. Once its started rolling its energy transfer into kinetic energy.
Hope this helps
4 0
3 years ago
A ball is thrown vertically upwards with a velocity v and an initial kinetic energy Ek. When half way to the top of its flight,
Mama L [17]

Answer:

Option A

\frac{V}{\sqrt{2} } ,\frac{E}{2}

Explanation:

When the ball its thrown up, at half way of its flight it means half of its vertical height which is \frac{h}{2}.  

potential energy = mgh

since it moved half way of height

P.E = \frac{mgh}{2}

This means for the body to have gained half of its P.E, it will loose half of its kinetic energy.

Final kinetic energy(E_{1}) = E/2

kinetic energy = \frac{1}{2}mv^{2}

let the final velocity at halfway flight be v1

E_{1} = E/2

\frac{1}{2}mv_{1}^{2} =\frac{\frac{1}{2}mv^{2}}{2}

cross multiply we have

mv_{1}^{2} =\frac{1}{2}mv^{2}

cancel m from both sides

v_{1}^{2} = \frac{1}{2}v^{2}

take the square root of both sides,

v_{1} =\sqrt{\frac{v^{2} }{2} }

v_{1} =\frac{v}{\sqrt{2} }

Thus our final velocities will be E/2 and \frac{v}{\sqrt{2} }

5 0
3 years ago
Frank’s automobile engine runs at 100∘C. One day, when the outside temperature is 15∘C, he turns off the ignition and notes that
lapo4ka [179]

Answer:

the engine cool to 40^{o}C at 14.07 minutes

Explanation:

Given information

T(5) = 70^{o}C

T_{0} = 100^{o}C

C = 15^{o}C

Newton's law of cooling :

T(t) = C + (T_{0} - C) e^{-kt}

where

T(t) = temperature at any given time

C = surrounding temperature

T_{0} = initial temperature of heated object

k = cooling constant

to find the the time when the engine will be cooled down to 40^{o}C, we first need to find the cooling constant, k

when t = 5, T(5) = 70^{o}C

so,

T(t) = C + (T_{0} - C) e^{-kt}

T(5) = 15 + (100 - 15) e^{-5k}

70 = 15 + (85) e^{-5k}

e^{-5k} = (70 - 15) / 85

-5k = ln (55/85)

k = - ln (55/85) / 5

k = 0.087

thus, we have the eqaution

T(t) = 15 + (85) e^{-0.087t}

now we can determine the time when T(t) = 40^{o}C

40 = 15 + (85) e^{-0.087t}

e^{-0.087t} = (40-15)/85

-0.0087t = ln (25/85)

t = - ln (25/85)/0.087

t = 14.07 minutes

8 0
3 years ago
If the absolute temperature of a gas is 600 K, the temperature in degrees Celsius is
devlian [24]
326.85 degrees Celsius
7 0
4 years ago
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