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2 years ago

12

What is one way you can destroy a magnet’s magnetism? a. by putting it in water b. by cooling it c. by heating it d. by breaking

it into pieces

2 answers:

elixir [45]2 years ago

8 0

By heating it up is the answer

telo118 [61]2 years ago

5 0

the answer is C LOLOLO

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A demographic tool used to predict a population's future growth is

azamat

C. Age structure. Hope I helped :)

7 0

2 years ago

A student constructed a series circuit consisting of a 12.0-volt battery, a 10.0-ohm lamp, and a

Zolol [24]

Considering the unknown resistence as R and using the Ohm's First Law, we have:

$i= \frac{V}{R_{eq}} \\ 0.5= \frac{12}{R+10} \\ R+10=24 \\ R=14-Ohm$

The equivalent resistence is given by the resistor series with the lamp resistence.

$R_{eq}=R+10 \\ R_{eq}=14+10 \\ \boxed {R_{eq}=24-Ohm}$

If you notice any mistake in my english, please let me know, because i am not native.

6 0

2 years ago

Below is a circuit schematic of sources and resistors (Figure 3). VS = 10V , R1 = 100Ω, R2 = 50Ω, R3 = 25Ω, IS = 2A. Calculate t

lorasvet [3.4K]

Answer:

$V_3\approx 4.28\,\,V$

$I_1=0.0572\,\,amps$

$I_3\approx 0.171\,\,amps$

Explanation:

Notice that this is a circuit with resistors R1 and R2 in parallel, connected to resistor R3 in series. It is what is called a parallel-series combination.

So we first find the equivalent resistance for the two resistors in parallel:

$\frac{1}{Re}= \frac{1}{R1}+\frac{1}{R2}\\\frac{1}{Re}= \frac{1}{100}+\frac{1}{50}\\\frac{1}{Re}= \frac{3}{100}\\Re=\frac{100}{3} \,\,\Omega$

By knowing this, we can estimate the total current through the circuit,:

$Vs=I\,*\,(\frac{100}{3} +25)\\10=I\,*\,\frac{175}{3} \\I=\frac{30}{175} \,amps$

So approximately 0.17 amps

and therefore, we can estimate the voltage drop (V3) in R3 uisng Ohm's law:

$V_3=\frac{30}{175} *\,25=\frac{30}{7} \approx 4.28\,\,V$

So now we know that the potential drop across the parellel resistors must be:

10 V - 4.28 V = 5.72 V

and with this info, we can calculate the current through R1 using Ohm's Law:

$I_1=\frac{V_1}{R_1} =\frac{5.72}{100} =0.0572\,\,amps$

4 0

2 years ago

Which of the following statements about Gaussʹs law are correct? (There may be more than one correct choice.) Question 3 options

irina1246 [14]

Answer:

If a Gaussian surface is completely inside an electrostatic conductor, the electric field must always be zero at all points on that surface.

Explanation:

Option A is incorrect because, given this case, it is easier to calculate the field.

Option B is incorrect because, in a situation where the surface is placed inside a uniform field, option B is violated

Option C is also incorrect because it is possible to be a field from outside charges, but there will be an absence of net flux through the surface from these.

Hence, option D is the correct answer. "If a Gaussian surface is completely inside an electrostatic conductor, the electric field must always be zero at all points on that surface."

3 0

2 years ago

If a clock expends 2 W of power from a 1.5 V battery what amount of current is supplying the clock?

wolverine [178]

1.3 A
If a clock expends 2 W of power from a 1.5 V battery, what amount of current is supplying
the clock?
solution
as we know
p=vi
i=p/v
=2/1.5
=1.3A

7 0

2 years ago