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3 years ago

SOMEONE HELP

Physics

2 answers:

tatyana61 [14]3 years ago

8 0

Answer:

The answer is solenoid and metal

Explanation:

Nimfa-mama [501]3 years ago

6 0

The answer is: An electromagnet is a (Solenoid) with a (metal) core.

Plz mark brainliest and be safe out there

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According to the Law of Reflection, the angle of incidence

Elan Coil [88]

Answer:

Explanation:

this is because the law of reflection states that the angle of incidence equals to the angle of the reflection.

hope this helps, if not please report it

someone else can try it

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2 years ago

The density of a certain type of plastic is 0.75 g/cm3 . If a sheet of this plastic is 10.0 m long, 1.0 m wide, and 1 cm thick,

Oksanka [162]

Answer:

7.5 g

Explanation:

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2 years ago

The electric potential difference across the membrane of a body cell is 0.070 V (higher on the outside than on the inside). The

12345 [234]

Answer:

The electric field is $8.75 \times 10^{6}~v~m^{-1}$ and the ditection is from outer to inner side of the membrane.

Explanation:

We know the electric field ( $\vec{E}$ ) is given by $\vec{E} = - \nabla V$ , 'V' being the potential.

In 1-D, it can be written as

$E=\dfrac{V}{d}$

where 'd' is the separation of space in between the potential difference is created.

Given, $V = 0.070~V~$ and the thickness of the cell membrane is $d = 8.0 \times 10^{-9}~m$ .

Therefore the created electric field through the cell membrane is

$E = \dfrac{0.07~V}{8 \times 10^{-9}~m} = 8.75 \times 10^{6}~m~s^{-1}$

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2 years ago

In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance

maw [93]

Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = $X_{L} = j\omega L = 2\pi fL$

where

R = resistance

$X_{L} = Inductive Reactance$

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

$C = \frac{\epsilon_{o}A}{x}$

where

x = separation between the parallel plates

Thus

C ∝ $\frac{1}{x}$

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$

Also,

Z ∝ I

Therefore,

$\frac{Z}{I} = \frac{Z'}{I'}$

$\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}$

${R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})$

${R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})$

Solving the above eqn:

$X_{C} = 4R$

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2 years ago

Which of the following gases are typically used for colorful lighting when an electric current is apllied

Leona [35]

Well i really need to see the choices but i think you mean neon

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2 years ago

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