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4 years ago

Please help on this one?

Physics

1 answer:

melomori [17]4 years ago

6 0

Answer:

B, on the right end

Explanation:

The north pole of a bar magnet is indicated by the direction of the arrows. It always point out of North and towards South. Therefore it is on the right end.

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An experiment on the earth's magnetic field is being carried out 1.00 m from an electric cable. What is the maximum allowable cu

Whitepunk [10]

Answer:

The allowed current in the cable is 1.15 A.

Explanation:

Given that,

Distance = 1.00 m

Suppose the magnetic field is $2.3\times10^{-5}\ T$ and if the experiment is to be accurate to 1.0 %

We need to calculate the current

Using formula of magnetic field

$B=\dfrac{\mu_{0}I}{2\pi r}$

$I=\dfrac{B\times2\pi r}{\mu_{0}}$

Put the value into the formula

$I=\dfrac{2.3\times10^{-5}\times2\times\pi\times1.00}{4\pi\times10^{-7}}$

$I=115\ A$

If the experiment is to be accurate to 1.0%

Then,

We need to calculate the allowed current in the cable

$I'=(1.0%)\times I$

$I'=\dfrac{1.0}{100}\times115$

$I'=1.15\ A$

Hence, The allowed current in the cable is 1.15 A.

4 0

3 years ago

Calculate the volume of this regular solid.

fenix001 [56]

Answer:

$2122.64 cm^3$

Explanation:

Some data in the problem are missing.

Missing values:

Radius of the cylinder: 13 cm

Height of the cylinder: 4 cm

The volume of a cylinder is given by

$V=\pi r^2 h$

where

r is the radius of the base of the cylinder

h is the height of the cylinder

In this problem, we have:

r = 13 cm (radius)

h = 4 cm (height)

Therefore, the volume of the cylinder is:

$V=\pi (13)^2(4)=2122.64 cm^3$

6 0

3 years ago

Cho lực F ⃗=6x^3 i ⃗-4yj ⃗ tác dụng lên vật làm vật chuyển động từ A(-2,5) đến B(4,7). Vậy công của lực là:

Natasha2012 [34]

The work done by $\vec F$ along the given path C from A to B is given by the line integral,

$\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r$

I assume the path itself is a line segment, which can be parameterized by

$\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath$

with 0 ≤ t ≤ 1. Then the work performed by F along C is

$\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}$