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Nataly [62]
2 years ago
10

You are given a material which produces no initial magnetic field when in free space. when it is placed in a region of uniform m

agnetic field, the material produces an additional internal magnetic field parallel to the original field. however, this induced magnetic field disappears when the external field is removed. what type of magnetism does this material exhibit? you are given a material which produces no initial magnetic field when in free space. when it is placed in a region of uniform magnetic field, the material produces an additional internal magnetic field parallel to the original field. however, this induced magnetic field disappears when the external field is removed. what type of magnetism does this material exhibit? diamagnetism paramagnetism ferromagnetism
Physics
1 answer:
mamaluj [8]2 years ago
7 0
The correct answer is:
<span>paramagnetism 

In fact, paramagnetic materials, when they are placed in a magnetic field, they form an internal magnetic field parallel to the external one and in the same direction. However, unlike ferromagnetic materials, they do not retain their magnetization, so when the external magnetic field is removed, their internal induced magnetic field disappears.</span>
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cold air is more dense than warm water so it sinks to the bottom of the pool

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You are called as an expert witness to analyze the following auto accident: Car B, of mass 2000 kg, was stopped at a red light w
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Answer:

a). va=17.23 \frac{m}{s} or 38.54 mph

b). v=38.54 mph and limit is 35 mph

c). Completely inelastic

d). Eka=192.967 kJ

Ekt=76.071 kJ

Explanation:

m_{a}=1300kg\\m_{b}=2000kg\\x_{f}=7.25m\\u_{k}=0.65

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m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}

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F-F_{uk} =0\\F=F_{uk}\\F_{uk}=u_{k}*m*g\\F=m*a\\a=\frac{F}{m}\\ a=\frac{F_{uk}}{m}\\a=\frac{u_{k}*m*g}{m}\\a=u_{k}*g\\a=0.65*9.8\frac{m}{s^{2}} \\a=6.39\frac{m}{s^{2}}

Now with the acceleration can find the time and the velocity final that make the distance 7.25m being united

x_{f}=x_{o}+v_{o}*t+2*a*t^{2}\\x_{o}=0\\v_{o}=0\\x_{f}=2*a*t^{2}\\t^{2}=\frac{x_{f}}{2*a}\\t=\sqrt{\frac{7.25m}{6.37\frac{m}{s^{2} } } } \\t=1.06s

So the velocity final can be find using this time

v_{f}=v_{o}+a*t\\v_{o}=0\\v_{f}=6.37\frac{m}{s^{2} } *1.06s\\v_{f}=6.79 \frac{m}{s}

a).

Replacing in the first equation the final velocity can find the initial velocity

m_{a}*v_{a}+m_{b}*v_{b}=(m_{a}+m_{b})*v_{f}

v_{b}=0

v_{a}= \frac{(m_{a}+m_{b)*v_{f}}}{m_{a}}\\v_{a}= \frac{(1300+2000)*6.37}{1300}\\v_{a}=17.23 \frac{m}{s}

b).

35mph*\frac{1m}{0.000621371mi} *\frac{1h}{3600s}=15.646\frac{m}{s}

Velocity limit in m/s is 15.646 m/s and the initial velocity is 17.23 m/s

so is exceeding the speed limit in about 1.58 m/s

or in miles per hour

3.5 mph

c).

The collision is complete inelastic because any mass can be returned to the original mass, so even they are no the same mass however in the moment they move the distance 7.25m as a same mass the motion is considered completely inelastic

d).

Ek=\frac{1}{2}*m*(v)^{2}\\  Eka=\frac{1}{2}*1300kg*(17.23\frac{m}{s})^{2}\\Eka=192.967 kJ\\Ekt=\frac{1}{2}*m*(v)^{2}\\Ekt=\frac{1}{2}*3300kg*(6.79\frac{m}{s})^{2}\\Ekt=76.071 kJ

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Answer:

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d. 7500

Explanation:

Please see attachment

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