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BabaBlast [244]
3 years ago
13

Why might an electromagnet be used to pick up old cars in junkyards?

Physics
2 answers:
siniylev [52]3 years ago
7 0
B) Hope it helps ,Have a nice day :)
Law Incorporation [45]3 years ago
6 0
<h3><u>Answer;</u></h3>

B) Electromagnets are powerful and can be turned on and off easily.

<h3><u>Explanation;</u></h3>
  • <em><u>Electromagnets are made by winding a wire through a magnetic material such as soft iron core making a coil, and then passing electric current through the coil thus making the soft iron core magnetic.</u></em>
  • The strength of the electromagnet depends on the number of coils and also the size of the electric current through the coil.
  • <em><u>Its advantageous to use electromagnets as compared to using permanent magnets since they are powerful and can be turned on and off easily and also the strength of the electromagnet can be easily adjusted by regulating the amount of current flowing through the coils. </u></em>



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Spiderman, whose mass is 70.0 kg, is dangling on the free end of a 12.2-m-long rope, the other end of which is fixed to a tree l
Lana71 [14]

Answer:

U = -3978.8 J

Explanation:

The work of the gravitational force U just depends of the heigth and is calculated as:

U = -mgh

Where m is the mass, g is the gravitational acceleration and h the alture.

for calculate the alture we will use the following equation:

h = L-Lcos(θ)

Where L is the large of the rope and θ is the angle.

Replacing data:

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h = 5.8 m

Finally U is equal to:

U = -70(9.8)(5.8)

U = -3,978.8 J

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3 years ago
Calculate the pressure exerted by a 4000N camel on the sand. The camel’s feet have a
Harrizon [31]

Answer:

The pressure exerted by camel feet is <u>2000 N/m²</u>.

Step-by-step explanation:

<h3><u>Solution</u> :</h3>

Here, we have given that ;

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  • Total area of camel feet = 2 m²

We need to find the pressure exerted by camel feet.

As we know that :

{\longrightarrow{\pmb{\sf{Pressure= \dfrac{Area}{Force}}}}}

Substituting all the given values in the formula to find the pressure exerted by camel feet.

\begin{gathered} \begin{array}{l} {\longrightarrow{\sf{Pressure= \dfrac{Area}{Force}}}} \\  \\ {\longrightarrow{\sf{Pressure= \dfrac{4000}{2}}}}  \\  \\ {\longrightarrow{\sf{Pressure= \cancel{\dfrac{4000}{2}}}}} \\  \\ {\longrightarrow{\sf{Pressure= 2000 \: N/{m}^{2}}}} \\  \\\star \:  \small\underline{\boxed{\sf{\purple{Pressure= 2000 \: N/{m}^{2}}}}} \end{array}\end{gathered}

Hence, the pressure exerted by camel feet is 2000 N/m².

\rule{300}{2.5}

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and since we're find when it's half we could either change it to 1/2 or 2, but 2 is easier to use
m=1
v=2
r=1
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So the velocity in the 1st part is half the velocity in the 2nd part and the centripetal force is 4× less
The answer is the centripetal force is 1/4 as big the second time around
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