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3 years ago

Which statement describes the magnetic field within a bar magnet?

Physics

1 answer:

Aleksandr [31]3 years ago

6 0

Since you did not put the choices of statements, here is what describes magnetic field within a bar magnet:

- from a bar magnet, magnetic field lines are forming closed lines.

- at either pole of the magnet, the magnetic field of a bar magnet is strongest. It is equally compared to the north pole with the south pole

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How does the area of 0.3 m2 of iron change when heated from 20 ° C to 60 ° C?

DerKrebs [107]

Answer:

The quantitative relationship between heat transfer and temperature change contains all three factors: Q = mcΔT, where Q is the symbol for heat transfer, m is the mass of the substance, and ΔT is the change in temperature. The symbol c stands for specific heat and depends on the material and phase. The specific heat is the amount of heat necessary to change the temperature of 1.00 kg of mass by 1.00ºC. The specific heat c is a property of the substance; its SI unit is J/(kg ⋅ K) or J/(kg ⋅ ºC). Recall that the temperature change (ΔT) is the same in units of kelvin and degrees Celsius. If heat transfer is measured in kilocalories, then the unit of specific heat is kcal/(kg ⋅ ºC).

Explanation:

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3 years ago

Which of the following statements is NOT a correct assumption of the classicalmodel of an ideal gas?A. The molecules are in rand

Elena-2011 [213]

The answer is D, because the collision's between molecules are elastic, not inelastic.

6 0

3 years ago

Read 2 more answers

A point charge q is located at the center of a spherical shell of radius a that has a charge −q uniformly distributed on its sur

muminat

Answer:

a) E = 0

b) $E = \dfrac{k_e \cdot q}{ r^2 }$

Explanation:

The electric field for all points outside the spherical shell is given as follows;

a) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

From which we have;

$E \cdot A = \dfrac{{\Sigma Q}}{\varepsilon _{0}} = \dfrac{+q + (-q)}{\varepsilon _{0}} = \dfrac{0}{\varepsilon _{0}} = 0$

E = 0/A = 0

E = 0

b) $\phi_E = \oint E \cdot dA = \dfrac{\Sigma q_{enclosed}}{\varepsilon _{0}}$

$E \cdot A = \dfrac{+q }{\varepsilon _{0}}$

$E = \dfrac{+q }{\varepsilon _{0} \cdot A} = \dfrac{+q }{\varepsilon _{0} \cdot 4 \cdot \pi \cdot r^2}$

By Gauss theorem, we have;

$E\oint dS = \dfrac{q}{\varepsilon _{0}}$

Therefore, we get;

$E \cdot (4 \cdot \pi \cdot r^2) = \dfrac{q}{\varepsilon _{0}}$

The electrical field outside the spherical shell

$E = \dfrac{q}{\varepsilon _{0} \cdot (4 \cdot \pi \cdot r^2) }= \dfrac{q}{4 \cdot \pi \cdot \varepsilon _{0} \cdot r^2 }= \dfrac{q}{(4 \cdot \pi \cdot \varepsilon _{0} )\cdot r^2 }$

$k_e= \dfrac{1}{(4 \cdot \pi \cdot \varepsilon _{0} ) }$

Therefore, we have;

$E = \dfrac{k_e \cdot q}{ r^2 }$

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3 years ago

A runner covers 40 m in 4.55 seconds ,what is their average velocity?

goldenfox [79]

Velocity = displacement/time

I hope i helped you <33

3 0

3 years ago

When a cube is inscribed in a sphere of radius r, the length Lof a side of the cube is . If a positive point charge Qis placed a

Nana76 [90]

Answer:

Ф_cube /Ф_sphere = 3 /π

Explanation:

The electrical flow is

Ф = E A

where E is the electric field and A is the surface area

Let's shut down the electric field with Gauss's law

Фi = ∫ E .dA = $q_{int}$ / ε₀

the Gaussian surface is a sphere so its area is

A = 4 π r²

the charge inside is

q_{int} = Q

we substitute

E 4π r² = Q /ε₀

E = 1 / 4πε₀ Q / r²

To calculate the flow on the two surfaces

* Sphere

Ф = E A

Ф = 1 / 4πε₀ Q / r² (4π r²)

Ф_sphere = Q /ε₀

* Cube

Let's find the side value of the cube inscribed inside the sphere.

In this case the radius of the sphere is half the diagonal of the cube

r = d / 2

We look for the diagonal with the Pythagorean theorem

d² = L² + L² = 2 L²

d = √2 L

we substitute

r = √2 / 2 L

r = L / √2

L = √2 r

now we can calculate the area of the cube that has 6 faces

A = 6 L²

A = 6 (√2 r)²

A = 12 r²

the flow is

Ф = E A

Ф = 1 / 4πε₀ Q/r² (12r²)

Ф_cubo = 3 /πε₀ Q

the relationship of these two flows is

Ф_cube /Ф_sphere = 3 /π

8 0

4 years ago