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laila [671]
3 years ago
4

An electromagnet is created using a battery, an insulated copper wire, and an iron nail. The wire is wrapped around the nail 20

times and attached to the battery. Which pair of changes will definitely increase the strength of the electromagnet?
Physics
2 answers:
Veseljchak [2.6K]3 years ago
4 0
Think of each turn as a tiny electromagnet, and they all add up. So increasing the number of turns will increase the magnet strength. And if the current increases then the magnetic field strength increases. Current = Voltage/Resistance. So increasing the voltage will increase the current.
so if there is "Increase the voltage of the battery and increases the number of turns of the coil" then that option should be the answer 
Roman55 [17]3 years ago
4 0

^The answer is C if doing USA test prep !!!


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Consider the following neutral electron configurations in which n has a constant value. Which configuration would belong to the
grandymaker [24]

Answer:

he configuration with the highest electronic affinity is 2s2 2p5

Explanation:

Electronic affinity is the variation of energy when we add an electron to a neutral atom to form an ion

When an electron is added, it must occupy a space is the sub-level of the atom, giving more stability when it approaches the configuration of a complete shell with eight electrons (noble gas), so the affinity must increase when moving in a period Group VIII noble gases)

Let's examine the given settings

In this case, when adding an electron, 2s2 is very far from a complete level configuration, so its affinity must be small.

2s2 2p2 when adding an electro the one has a little more affinity, but is still a long way from a full shell, it would be missing 3 electrons

2s2 2sp5 this is the atom with the highest electronic affinity, since i = that when adding an electron the ion has the configuration of a noble gas. This is the most stable on the list

2s2 2p6 already has a full shell making it very difficult to insert an electron into this atom.

In summary, the configuration with the highest electronic affinity is 2s2 2p5

3 0
3 years ago
Read 2 more answers
(b) Un bloque rectangular de madera se mide con dimensiones de 11.2 cm x 3.4 cm x
vlabodo [156]

Answer:

V = 156.13 [cm³]

Explanation:

El volumen de un solido con forma de paralepipedo se puede calcular por medio de la siguiente formula:

V = ancho*largo*alto

donde:

V = volumen [cm³]

ancho = 3.4 [cm]

largo = 11.2 [cm]

alto = 4.1 [cm]

Ahora reemplazando.

V = 3.4*11.2*4.1\\V = 156.13 [cm^{3}]

5 0
3 years ago
The loaded cab of an elevator has a mass of 3.0 x 10 3 kg and moves 200 m up the shaft in 20 s at constant speed. At what averag
Alexeev081 [22]

The average rate at which the cable does work is 294,000 J/s.

The given parameters:

  • <em>mass, m = 3000 kg</em>
  • <em>height, h = 200 m</em>
  • <em>time of motion, t = 20 s</em>

The average rate at which the cable does work is calculated as follows;

P = \frac{E}{t} \\\\P = \frac{mgh}{t} \\\\P = \frac{3000 \times 9.8 \times 200}{20} \\\\P = 294,000 \ J/s

Thus, the average rate at which the cable does work is 294,000 J/s.

Learn more about energy and power here: brainly.com/question/13387946

4 0
3 years ago
A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from
Ulleksa [173]

Answer:

1270.64\ \text{J}

Explanation:

m = Mass of object = \dfrac{mg}{g}

mg = Weight of object = 20 N

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

\theta = Angle of slope = 30^{\circ}

f = Force of friction

fd = Work done against friction

The force balance of the system is

\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}

The work done against friction is 1270.64\ \text{J}.

8 0
3 years ago
Derive an expression for the gravitational potential energy U(r) of the object-earth system as a function of the object's distan
Drupady [299]

Answer:

U(r)=-\frac{Gm_Emr^2}{2R^3_E}

Explanation:

We are given that

Gravitational force=F_g=\frac{Gm_Emr}{R^3_E}

r=0,U(0)=0

We know that

Gravitational potential energy=-\int F_gdr

U(r)=-\int\frac{Gm_Emr}{R^3_E}dr

U(r)=-\frac{Gm_Em}{R^3_E}\times \frac{r^2}{2}+C

Substitute r=0 ,U(0)=0

0=0+C

C=0

Substitute the value

U(r)=-\frac{Gm_Emr^2}{2R^3_E}

4 0
4 years ago
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