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mario62 [17]
3 years ago
8

Research to find out 5 uses of magnets in our daily life in medical contexts, route finding, security, sorting steel from other

materials for recycling, etc.
Chemistry
1 answer:
bearhunter [10]3 years ago
4 0

Answer:

Medical: Magnets are used for the RMN, which uses waves and magnets to create images of the body, this process is really complex because is deeply related to quantum mechanics, so i will not dive in the topic.

Security: You can attach a magneto to a sensor, and when the magnet detects a metal, a current flows to the sensor and it can make a noise. This system is used to detect if someone has something of metal (like a knife for example) and a similar system is used to detect land mines or metals under the ground, which can be related to "route finding".

Sorting steel: When you have a mix of different metals, some of them will be attracted to the magnet and others not. An example where this is used is when you have a piece of steel with impurities, you can melt it and use a magnet to only grab the steel and leave most of the impurities.

Recycling. Big magnets are used to manage large pieces of steel that may be difficult or dangerous to handle by hand, for example. in order to move old cars and objects of similar size of steel, a big magnet is used.

Music: Modern instruments like electric guitars or electric basses, use magnets in the pickups: The magnet detects the movement of the strings, and this creates a current in a copper wire that is winded around the magnets, and that current then is transformed into sound by the amplifier.

You might be interested in
Mechanical twinning occurs in metals having which type(s) of crystal structure(s)?
kari74 [83]
Mechanical twinning occurs in metals having body center cubic and hexagonal closed packed structures. Twinning is said to occur when a portion of a crystal takes up an orientation that is related to the orientation of the untwinned lattice in a definite symmetrical manner. 
5 0
3 years ago
Steve walks from his house 5 I'm south then turns east and walks 2 km. Then he walks 9 in North to his older sister's house. She
IgorLugansk [536]

Answer:

18 km

Explanation:

To know the distance, you need to be clear in the concept of distance.

According to the physics, the distance is a measurement that refers to the fact of how much an object has move from one point to another.

In this case, we have a person named Steve and he's walking through several points.

To calculate the distance that he moved, you just need to sum all the values of distance. In this case, it's not neccesary to do calculations regarding the components of the x or y axis, because we are not talking about displacement, where it actually counts. But in this case, we just need all the covered distance that Steven walked since the beggining.

So, no more talk, let's do calculations:

D = 5 + 2 + 9 + 2 = 18 km

So the final answer would be:

<h2>Distance = 18 km</h2>

Hope this helps

5 0
2 years ago
H2O2 → H2O + O2
vladimir2022 [97]

Answer:

75.47

Explanation:

AS WE CANT WEIGHT AIR THE ANSWER IS CORRECT

8 0
2 years ago
What is the hydrogen ion and hydroxide ion concentration for a pOH of 3.557? Is the solution an acid or a base?​
UkoKoshka [18]

Answer:

Hydrogen ion concentration is 2.773E-4Mole per DM3

Hydroxide ion concentration is 4.093E-11

It is an acid

Explanation:

PH(H3O)=-LogH3O

PH=3.557

H3O=antilog of -3.557

antilog of -3.557=antilog (0.443 -4)

antilog0.443 ×antilog -4=2.773E-4mole per DM3.

(H3O)(OH)=10^-14

H30=2.773×10^-14

OH= 1.00E-14/2.773E-4

=4.093E-11

it is an acid since the concentration of hydrogen ion is more than the hydroxide ion

4 0
3 years ago
Read 2 more answers
Student conducts an experiment to determine the enthalpy of solution for lithium chloride dissolved in water. the student combin
nirvana33 [79]

Answer:-  \Delta H_s_o_l_n=-37.2\frac{kJ}{mol}

Solution:- First of all we calculate the heat absorbed or released when the solute is added to the solvent. Here the solute is LiCl and the solvent is water.

To calculate the heat absorbed or released we use the formula:

q=ms\Delta T

q = heat absorbed or released

m = mass of solution

s = specific heat capacity

and \Delta T = change in temperature

mass of solution = mass of solute + mass of solvent

mass of solution = 5.00 g + 100.0 g = 105.0 g

(note:- density of pure water is 1 g per mL so the mass is same as its volume)

\Delta T = 33.0 - 23.0 = 10.0 degree C

s = 4.18\frac{J}{g.^0C}

Let's plug in the values in the formula and calculate q.

q = 105.0g(4.18\frac{J}{g.^0C})(10.0^0C)

q = 4389 J

To calculate the enthalpy of solution that is \Delta H_s_o_l_n we convert q to kJ and divide by the moles of solute.

moles of solute = 5.00g(\frac{1mol}{42.39g})

= 0.118 moles

q = 4389J(\frac{1kJ}{1000J}) = 4.389 kJ

\Delta H_s_o_l_n=\frac{4.389kJ}{0.118mol}

\Delta H_s_o_l_n = 37.2\frac{kJ}{mol}

Since the heat is released which is also clear from the rise in temperature of the solution, the sign of enthapy of solution will be negative.

So,  \Delta H_s_o_l_n=-37.2\frac{kJ}{mol}


4 0
3 years ago
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