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2 years ago

The magnet below is cut in half. what will be the result?

Physics

2 answers:

ozzi2 years ago

3 0

<h2>Answer:</h2>

<h2>Explanation:</h2>

This is what we call a permanent magnet. By the way, the magnetic phenomena were first observed about 2500 years ago near the ancient city of Magnesia, what is today Manisa, located in western Turkey, when people saw fragments of magnetized iron. So <em>what happens if you cut a magnet in half? </em>Well, a magnet has two ends, the first one is called a north pole or N pole while the other end is a south pole or S pole, so if you break a bar magnet, each piece has a north and south pole, no matter the size of each new bar although the smaller the piece, the weaker its magnetism. This is true because unlike electric charges, you always find magnetic poles in pairs, that is, ¡they can't be isolated! The option is C. because in the great bar the north pole is to the left while the south pole is to right.

DedPeter [7]2 years ago

3 0

Answer: C

Explanation:

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in a certain pinhole camera, the screen is 10 cm from the pinhole. When the pinhole is placed 6 cm away from a tree ,a sharp ima

vagabundo [1.1K]

Answer:

Height of the tree is <u>9.6 cm</u>

Explanation:

We know that, Magnification of an image is written as follows.

$\left(\frac{H_{i}}{H_{o}}\right)=\left(\frac{D_{i}}{D_{o}}\right)$

Where,

$\begin{array}{l}{\mathrm{H}_{0}=\text { height of the object }} \\ {\mathrm{H}_{\mathrm{i}}=\text { height of the image }} \\ {\mathrm{D}_{0}=\text { distance of the object }} \\ {\mathrm{D}_{\mathrm{i}}=\text { distance of the image }}\end{array}$

As per given question,

$\begin{array}{l}{\mathrm{H}_{1}=\text { height of the image }=\text { height of the image of the tree on screen }=16 \mathrm{cm}} \\ {\mathrm{D}_{0}=\text { distance of the object }=\text { distance of the tree from the pinhole }=6 \mathrm{cm}} \\ {D_{1}=\text { distance of the image }=\text { distance of the image from the pinhole }=10 \mathrm{cm}} \\ {\mathrm{H}_{0}=\text { height of the object }=\text { height of the tree }}\end{array}$

Substitute the values in the above formula,

$\begin{array}{l}{\left(\frac{H_{i}}{H_{o}}\right)=\left(\frac{D_{i}}{D_{o}}\right)} \\ {\left(\frac{16}{H_{o}}\right)=\left(\frac{10}{6}\right)} \\ {\mathrm{H}_{\mathrm{o}}=\left(\frac{16 \times 6}{10}\right)} \\ {\mathrm{H}_{\mathrm{o}}=9.6 \mathrm{cm}}\end{array}$

Height of the tree is 9.6 cm.

8 0

3 years ago

What type of front occurs when cold air and warm air are next to each other but are at a standstill

Korvikt [17]

Answer:

A stationary front.

(Hope this helped! c:)

5 0

3 years ago

Can someone help with these

yaroslaw [1]

Givens

delta B = 0.20 T/s
A = 0.07 m^2
R = 3.5 ohms

Formula

Φ = ΔB*A

e = Φ

Solution (first part)

e = 0.2 * 0.07

e = 0.014 emf

Solution (second part)

i = e/R

i = 0.014 / 3.5

i = 4 * 10^-3

i = 4 ma

Answer

Givens

N = 200 turns

Φ = 30 degrees

Delta B = 0.45 T/s

phi = 30 degrees

r = 0.06 meters

Formula

e = -N * delta B * A * Cos(phi)

Solution

e = -200 * 0.45 (pi r^2) * Cos(30)

e = - 200 * 0.45 * (3.14 * 0.06^2) * cos(30)

e = 0.881 emf

Answer

4 0

3 years ago

A group of hikers hear an echo 2.6 s after they shout. The temperature is 20 ◦C. How far away is the mountain that reflected the

Vadim26 [7]

Answer:

$x = 445.9\,m$

Explanation:

The velocity of sound in air at 20 °C is 343 meters per second. The distance of mountain can be derived by the following expression:

$2\cdot x = v_{air}\cdot \Delta t$

$x = \frac{v_{air}\cdot \Delta t}{2}$

$x = \frac{(343\,\frac{m}{s} )\cdot (2.6\,s)}{2}$

$x = 445.9\,m$

4 0

3 years ago

A cricketer can throw a ball to a maximum horizontal distance of 100m.How much high above the ground can he throw same ball

rosijanka [135]

<span>The range of an object thrown at a velocity v with an angle of elevation 45 degrees is S = v^2/g 100 = v^2/g => v^2 = 100*g In the same throw, the maximum height of the ball can be derived from the equation v^2 - u^2 = 2*g*h It is is h = = = 50 metres</span>

5 0

3 years ago