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allochka39001 [22]
3 years ago
8

Describe two methods by a magnet can be demagnetized​

Physics
2 answers:
Mekhanik [1.2K]3 years ago
8 0

Answer:

Demagnetizing a magnet :

  1. By hitting with a hammer
  2. By heating with a high temperature
  3. By dropping the magnet
leva [86]3 years ago
5 0
Being hit with a hammer
Being dropped
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Choose two forces and compare and contrast these forces. These must be different forces than used in the prior question. Provide
marin [14]

Answer:

Explanation:

There are 4 forces. These are 1) Gravity, 2) Weak Nuclear Force, 3) Electromagnetism, and 4) Strong Nuclear Force. 

Order of strength from weakest to strongest: Gravity, Weak Nuclear Force, Electromagnetism, Strong Nuclear Force

Type of Range:

Gravity - Unlimited range

Weak Nuclear Force - Limited range

Electromagnetism - Infinite range

Strong Nuclear Force - Limited Range

Found in:

Gravity - Exists between all objects with mass

Weak Nuclear Force - Governs over beta decays like the emission of electron or positron

Electromagnetism - the attraction found between particles that are electrically charged

Strong Nuclear Force - Found in atoms and subatomic particles. It is responsible for holding the atoms' nucleus together.

6 0
3 years ago
What is the kinetic energy f a 25kg object movingat a velocity of 10m/s?
alex41 [277]
Using K.E=1/2MV^2
answer is 125joules
5 0
3 years ago
A centrifuge rotor rotating at 10,000 rpm is shut off and is eventually brought to rest by a frictional force of 1.20m n. if the
Pani-rosa [81]
<span>Answer: The moments of inertia are listed on p. 223, and a uniform cylinder through its center is: I = 1/2mr2 so I = 1/2(4.80 kg)(.0710 m)2 = 0.0120984 kgm2 Since there is a frictional torque of 1.20 Nm, we can use the angular equivalent of F = ma to find the angular deceleration: t = Ia -1.20 Nm = (0.0120984 kgm2)a a = -99.19 rad/s/s Now we have a kinematics question to solve: wo = (10,000 Revolutions/Minute)(2p radians/revolution)(1 minute/60 sec) = 1047.2 rad/s w = 0 a = -99.19 rad/s/s Let's find the time first: w = wo + at : wo = 1047.2 rad/s; w = 0 rad/s; a = -99.19 rad/s/s t = 10.558 s = 10.6 s And the displacement (Angular) Now the formula I want to use is only in the formula packet in its linear form, but it works just as well in angular form s = (u+v)t/2 Which is q = (wo+w)t/2 : wo = 1047.2 rad/s; w = 0 rad/s; t = 10.558 s q = (125.7 rad/s+418.9 rad/s)(3.5 s)/2 = 952.9 radians But the problem wanted revolutions, so let's change the units: q = (5528.075087 radians)(revolution/2p radians) = 880. revolutions</span>
6 0
3 years ago
Suppose you are pushing a 3 kg box with a force of 25 N (directed parallel to the ground) over a distance of 15 m. Afterward, th
alex41 [277]

Answer: 321 J

Explanation:

Given

Mass of the box m=3\ kg

Force applied is F=25\ N

Displacement of the box is s=15\ m

Velocity acquired by the box is v=6\ m/s

acceleration associated with it is a=\dfrac{F}{m}

\Rightarrow a=\dfrac{25}{3}\ m/s^2

Work done by force is W=F\cdot s

W=25\times 15\\W=375\ J

change in kinetic energy is \Delta K

\Rightarrow \Delta K=\dfrac{1}{2}m(v^2-0)\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 6^2\\\\\Rightarrow \Delta K=\dfrac{1}{2}\times 3\times 36\\\\\Rightarrow \Delta K=54\ J

According to work-energy theorem, work done by all the forces is equal to the change in the kinetic energy

\Rightarrow W+W_f=\Delta K\quad [W_f=\text{Work done by friction}]\\\\\Rightarrow 375+W_f=54\\\Rightarrow W_f=-321\ J

Therefore, the magnitude of work done by friction is 321\ J

3 0
3 years ago
Little Billie has a mass of 25 kg. The acceleration of gravity on the moon is one-sixth of the value on Earth. What is Little Bi
sineoko [7]

Answer:

Mass is constant everywhere,

But weight is different,

If earth g = 10 then moon's is 1.6666667

Now billie's weight in moon is 41.6667

6 0
3 years ago
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