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3 years ago

What are 2 ways electromagnets are used

1 answer:

lisabon 2012 [21]3 years ago

4 0

Electromagnets are used for various purposes but I fathom in this instance, the questioner is asking about how electromagnetics can be used to attraction or repulsion.
Example, electromagnets are used for attraction in cranes which attach them to containers in order to lift them.
Meanwhile, Maglev trains use electromagnets repulsive properties.

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Elements with the largest atomic radius are found in the lower left hand of the periodic table.

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2 years ago

Use the crisscross method to find the chemical formula for the ionic compound formed by strontium (Sr) and bromine (Br).

nignag [31]

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Explanation:

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3 years ago

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Fantom [35]

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3 years ago

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The gravitational acceleration on Mars is 3.71 m/s2. If the pendulums were set in motion on the red planet, how would that affec

Elanso [62]

On Earth, the period of a pendulum is given by:
$T_{earth}=2\pi \sqrt{ \frac{L}{g_{earth} }$
where L is the length of the pendulum and $g_{earth}=9.81~m/s^2$ is the gravitational acceleration on Earth.
Similarly, the period of the same pendulum on Mars will be
$T_{mars}=2\pi \sqrt{ \frac{L}{g_{mars} }$
where $g_{mars}=3.71~m/s^2$ is the gravitational acceleration on Mars.
Therefore, if we want to see how does the period of the pendulum on Mars change compared to the one on Earth, we can do the ratio between the two of them:
$\frac{T_{mars}}{T_{earth}}= \sqrt{ \frac{g_{earth}}{g_{mars}} } = \sqrt{ \frac{9.81~m/s^s}{3.71~m/s^2} }=1.63$
Therefore, the period of the pendulum on Mars will be 1.63 times the period on Earth.

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3 years ago

A spring of negligible mass has force constant of 1600 Newtons per meter. (a) How far must the spring be compressed for 3.20 Jou

frez [133]

Answer:

a) x=63.0 $x10^{-3} m$ or 6.3cm

b) x=116.0 $x10^{-3} m$ or 11.6cm

Explanation:

a).

The elastic potential energy is modeling by equation :

$U1=\frac{1}{2}*k*x^{2} \\K=1600 \frac{N}{m}\\ U=3.2J\\m=1.2kg\\x^{2}=\frac{2*U}{k}\\x=\sqrt{\frac{2*3.2J}{1600 \frac{N}{m}}} \\x=\sqrt{4x10^{-3} m^{2}}\\ x=0.06324m$

b).

The work energy theorem explain which work is done in this case. the motion began from the rest so K1=K2 equal zero, Ug1 is no yet done and U2is also zero because is the potential energy

$Ug=K2\\mgy=\frac{1}{2}*k*x^{2} \\but \\y=h+x=0.80m+x\\m*g*(0.80m+x)=\frac{1}{2}*k*x^{2}$

Solving for x

$2*(m*g(h+x))=k*x^{2} \\k*x^{2}-(2*m*g*x)-(2*m*g*h)=0\\1600x^{2} -2*1.2kg*9.8\frac{m}{s^{2}}*x-2*1.2kg*0.80m=0\\1600x^{2} -23.52x-18.816m=0$

$x=-\frac{b+/-\sqrt{b^{2}-4*a*c } }{2*a} \\x=-\frac{-23.52+/-\sqrt{-23.52^{2}-4*1600*-18.816 } }{2*a} \\x=11.79 +/- 173.9\\x1=-0.10 m\\x2=0.116$

The negative is discard so

x=0.116m

7 0

3 years ago