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2 years ago

When asked how to create an electromagnet, the best answer would be, "You can create an electromagnet by

2 answers:

5 0

The correct answer is D. wrapping an insulated wire around a metal with ferromagnetic properties and applying an electric current.

Explanation

An electromagnet is a non permanent magnet that is created with an electric current .When current flows through a wire it induces a magnetic field around the wire. When the wire is wrapped around in a coil, the magnetic field that is induced is concentrated in the center of the coil, creating a north and south pole. The field is that of a solenoid.

Electromagnets are usually made by wrapping insulated coil of wire around a ferromagnetic metal. This focuses the electromagnetic field and creates a magnet as long as there is current running trough the insulated wire.

The correct answer is D. wrapping an insulated wire around a metal with ferromagnetic properties and applying an electric current.

Norma-Jean [14]2 years ago

4 0

Choice-D will do the job nicely.

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a car is running with the velocity of 72 km/h . what will be it's velocity after 5 seconds if it's acceleration -2 m/s² ?

sladkih [1.3K]

Answer:

Explanation:

79 second

8 0

2 years ago

Two identical circular, wire loops 35.0 cm in diameter each carry a current of 2.80 A in the same direction. These loops are par

defon

Answer:

The answer is " $4659.2 \times 10^{-24} \ N$ "

Explanation:

The magnetic field at ehe mid point of the coils is,

$\to B=\frac{\mu_0 i R^2}{(R^2+x^2)^{\frac{3}{2}}}\\\\$

Here, i is the current through the loop, R is the radius of the loop and x is the distance of the midpoint from the loop.

$\to B=\frac{(4\pi\times 10^{-7})(2.80\ A) (\frac{0.35}{2})^2}{( (\frac{0.35}{2})^2+ (\frac{0.24}{2})^2)^{\frac{3}{2}}}\\\\$

$=\frac{(12.56 \times 10^{-7})(2.80\ A) \times 0.030625}{( 0.030625+ 0.0144)^{\frac{3}{2}}}\\\\=\frac{ 1.07702 \times 10^{-7} }{0.0095538976}\\\\=112.730955 \times 10^{-7}\\\\=1.12\times 10^{-5}\ \ T\\$

Calculating the force experienced through the protons:

$F=qvB=(1.6 \times 10^{-19}) (2600)(1.12 \times 10^{-5})= 4659.2 \times 10^{-24}\ N$

7 0

2 years ago

Which of the following lists the typical steps of the scientific method in the

irinina [24]

B. Ask a question - Hypothesize - Experiment - Analyze - Draw a conclusion

Explanation:

The scientific method is a systematic approach in which investigations are conducted in science.

It follows systematic manner;

An observation in nature is made using the senses.
Questions are then asked about such observations. This is the inquiry phase.
A hypothesis is then formula to predict what might cause the observation.
Experiments are then conducted
The results are analyzed and a conclusion is drawn

learn more:

Experiments brainly.com/question/5096428

#learnwithBrainly

4 0

2 years ago

A car moving with an initial speed v collides with a second stationary car that is one-half as massive. After the collision the

Mashutka [201]

Answer:

4v/3

Explanation:

Assume elastic collision by the law of momentum conservation:

$m_1v = m_1v_1 + m_2v_2$

where v is the original speed of car 1, v1 is the final speed of car 1 and v2 is final speed of car 2. m1 and m2 are masses of car 1 and car 2, respectively

Substitute $m_2 = m_1/2 \& v_1 = v/3$

$m_1v = \frac{m_1v}{3} + \frac{m_1v_2}{2}$

Divide both side by $m_1$ , then multiply by 6 we have

$6v = 2v + 3v_2$

$3v_2 = 4v$

$v_2 = \frac{4v}{3}$

So the final speed of the second car is 4/3 of the first car original speed

5 0

2 years ago

Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ

bija089 [108]

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; $H_0 = 51km/s/Mly$

Age of the universe; $t = \ ?$

We know that, the reciprocal of the Hubble's constant ( $H_0$ ) gives an estimate of the age of the universe ( $t$ ). It is expressed as:

$Age\ of\ Universe; t = \frac{1}{H_0}$

Now,

Hubble's constant; $H_0 = 51km/s/Mly$

We know that;

$1\ light\ years = 9.46*10^{15}m$

$1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m$

Therefore;

$H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\$

Now, we input this Hubble's constant value into our equation;

$Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years$

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0

2 years ago