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3 years ago

Could you pick up a nail using the curved part of the horseshoe magnet farthest from the poles?

Physics

1 answer:

nikitadnepr [17]3 years ago

5 0

Yes

Explanation:

One can pick up a nail using the curved part of the horseshoe magnet farthest from the poles.

A horse shoe magnet has magnetic fields all around it.

In the presence of magnetic force fields, any magnetic object will be attracted to it.
This is the case with any magnet.
A magnet is any object with magnetic fields all around it.
This field causes attraction to any magnetic objects especially metals.

learn more:

Electromagnet brainly.com/question/2191993

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You might be interested in

When the termination is a terminal block, care must be taken to ensure a good electrical connection without damaging the conduct

Tamiku [17]

Answer:

When the termination is a terminal block, care must be taken to ensure a good electrical connection without damaging the conductor. Terminals should not be used for more than one

Explanation:

The Terminal block being a modular block, having insulated frame, which can secure more than two wires in it. It has a conducting strip in it. These terminal clocks helps in making the connection safer as well as organised. These terminal blocks are used for power distribution in safer way. Its potential is it can distribute power from single to multiple output. The conductor is used for making it proper contact.

8 0

3 years ago

In a compound microscope, the objective has a focal length of 1.0 cm, the eyepiece has a focal length of 2.0 cm, and the tube le

expeople1 [14]

Answer:

The value is $m \approx 310$

Explanation:

From the question we are told that

The focal length of the objective is $f_o = 1.0 \ cm$

The focal length of the eyepiece is $f_e = 2.0 \ cm$

The tube length is $L = 25 \ cm$

Generally the magnitude of the overall magnification is mathematically represented as

$m = m_o * m_e$

Where $m_o$ is the objective magnification which is mathematically represented as

$m_o = \frac{L}{f_o }$

=> $m_o = \frac{25}{1 }$

=> $m_o = 25$

$m_e$ is the eyepiece magnification which is mathematically evaluated as

$m_e = \frac{L }{f_e }$

$m_e = \frac{25 }{ 2}$

$m_e = 12.5 \ cm$

$m = 25 * 12.5$

$m \approx 310$

6 0

3 years ago

A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (

SIZIF [17.4K]

The solution to the questions are given as

$t=40.39 \mathrm{sec}$
$\varepsilon &=(0.12v)e^{0.057t}$
the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}$

$\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$$

$\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}$

$\varepsilon &=(0.12v)e^{0.057t}$

(b) $Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$$

$\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}$

In conclusion, the direction of the induced current will be Counterclockwise.