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3 years ago

From which end (north or south) of a bar magnet do magnetic field lines emerge?

1 answer:

5 0

They emerge from the north end. In modern magnetics we don't use the idea of north and south anymore but instead use a vector called the magnetization. It points toward what is traditionally called "north".

You might be interested in

A stationary rock on a hill has

Mekhanik [1.2K]

Answer: potential energy but no kinetic energy

Explanation:

Since the rock is stationary, velocity is zero, therefore no kinetic energy,but there's potential energy because the rock is at rest,

5 0

2 years ago

Read 2 more answers

How many butter, which has a usable energy content of 6.0 Cal/g (= 6000 cal/g), would be equivalent to the change in gravitation

Mazyrski [523]

Answer:

175.96 g

Explanation:

Potential energy required for the man to climb 7.07 km = m g h.

= 64 x 9.8 x 7070

= 4.434 x 10⁶ J

= 4.434 X 10⁶ / 4.2 cals

= 1.0557 x 10⁶ cals

= 1.0557 x 10⁶ / 6000 g of butter

= 175.96 g of butter.

3 0

2 years ago

In a fireworks display, a rocket is launched from the ground with a speed of 18.0 m/s and a direction of 51.0° above the horizon

Masja [62]

Answer:38.66 m

Explanation:

Given

launch angle $\theta =51^{\circ}$

launch velocity $u=18 m/s$

center of mass continue to travel its original Path so it center of mass will be at a distance of

$R=\frac{u^2\sin 2\theta }{g}$

$R=\frac{18^2\sin 102}{9.8}$

$R=32.33 m$

Center of mass will be at $x=32.33 m$

(b)if one of the piece will be at x=26 m then other will be at

$x_{com}=\frac{m_1x_1+m_2x_2}{m_1+m_2}$

$x_{com}=\frac{\frac{m}{2}\cdot 26+\frac{m}{2}\cdot x_0}{\frac{m}{2}+\frac{m}{2}}$

$32.33=\frac{26+x_0}{2}$

$x_0=38.66 m$

8 0

3 years ago

Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are o

polet [3.4K]

Answer:

a. charge C experiences the greatest net force, and charge B receives the smallest net force

b. ratio=9

Explanation:

<u>Electrostatic Force</u>

Two point-charges $q_1$ and $q_2$ separated a distance d will exert a force on each other of a magnitude given by the Coulomb's formula

$\displaystyle F=\frac{k\ q_1\ q_2}{r^2}$

Where k is the proportional constant of value

$k=9*10^9\ N.m^2/c^2$

The diagram provided in the question shows four identical charges (let's assume their value is Q) separated by identical distance (of value d). The force between the charges next to others is

$\displaystyle F_1=\frac{k\ Q\ Q}{d^2}$

$\displaystyle F_1=\frac{k\ Q^2}{d^2}$

The force between charges separated 2d is

$\displaystyle F_2=\frac{k\ Q^2}{(2d)^2}$

$\displaystyle F_2=\frac{k\ Q^2}{4d^2}$

And the force between the charges A and D is

$\displaystyle F_3=\frac{k\ Q^2}{(3d)^2}$

$\displaystyle F_3=\frac{k\ Q^2}{9d^2}$

Now, let's analyze each charge and the force applied to them by the others

Let's recall equally signed charges repel each other and differently signed charges attrach each other

Charge A. It receives force to the left from B and C and to the right from D

$\displaystyle F_A=-F_1-F_2+F_3=-\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{9d^2}$

$\displaystyle F_A=\frac{k\ Q^2}{d^2}(-1-\frac{1}{4}+\frac{1}{9})$

$\displaystyle F_A=-\frac{41}{36}F_1$

Charge B. It receives force to the right from A and D and to the left from C

$\displaystyle F_B=F_1-F_1+F_2=\frac{k\ Q^2}{d^2}-\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{4d^2}$

$\displaystyle F_B=\frac{1}{4}F_1$

Charge C. It receives forces to the right from all charges.

$\displaystyle F_C=F_2+F_1+F_1=\frac{k\ Q^2}{4d^2}+\frac{k\ Q^2}{d^2}+\frac{k\ Q^2}{d^2}$

$\displaystyle F_C=\frac{9}{4}F_1$

Charge D. It receives forces to the left from all charges

$\displaystyle F_D=-F_3-F_2-F_1=-\frac{k\ Q^2}{9d^2}-\frac{k\ Q^2}{4d^2}-\frac{k\ Q^2}{d^2}$

$\displaystyle F_D=-\frac{49}{36}F_1$

Comparing the magnitudes of each force is just a matter of computing the fractions

$\displaystyle \frac{41}{36}=1.13,\ \frac{1}{4}=0.25,\ \frac{9}{4}=2.25,\ \frac{49}{36}=1.36$

We can see the charge C experiences the greatest net force, and charge B receives the smallest net force

The ratio of the greatest to the smallest net force is

$\displaystyle \frac{\frac{9}{4}}{\frac{1}{4}}=9$

The greatest force is 9 times the smallest net force

7 0

3 years ago

How does an increase in temperature generally affect the rate of a reaction?

Zigmanuir [339]

Most reactions are exothemic. If the forward reaction of an equilibrium reaction is exothemic then the reverse reaction must be endothermic.

If a system in equilibrium is heated, it will move in exothermic direction to give out heat energy.

7 0

3 years ago