All categories

3 years ago

The separation between two magnets is reduced by a factor of 5. How does the force between them change?

Physics

1 answer:

gtnhenbr [62]3 years ago

7 0

Answer:Increase by a factor of 25

Explanation:

We know that magnetic force is proportional to $\frac{1}{r^2}$

where r is the distance between them

$F\propto B\propto \frac{1}{r^2}$

Thus $F\propto \frac{1}{r^2}$

Where $B$ =magnetic field

If distance is reduced by a factor of $5$ then force must increase by a factor of $25$ because

$r\rightarrow \frac{r}{5}$

so $F\propto \dfrac{1}{(\frac{r}{5})^2}$

$F\propto \frac{25}{r^2}$

You might be interested in

What are the three rules for writing the electron configuration of elements

astraxan [27]

Answer:

The aufbau principle

The Pauli exclusion principle

Hund's rule of maximum multiplicity

Explanation:

The aufbau principle:

The fundamental electronic configuration is achieved by placing the electrons one by one in the different orbitals available for the atom, which are arranged in increasing order of energy.

The Pauli exclusion principle:

Two electrons of the same atom cannot have their four equal quantum numbers. Because each orbital is defined by the quantum numbers n, l, and m, there are only two possibilities ms = -1/2 and ms = +1/2, which physically reflects that each orbital can contain a maximum of two electrons, having opposite spins

Hund's rule of maximum multiplicity:

This rule says that when there are several electrons occupying degenerate orbitals, of equal energy, they will do so in different orbitals and with parallel spins, whenever this is possible. Because electrons repel each other, the minimum energy configuration is one that has electrons as far away as possible from each other, and that is why they are distributed separately before two electrons occupy the same orbital.

4 0

3 years ago

In the figure below the pulley is a solid disk of mass M and radius R with rotational inertia MR 2/2. Two blocks one of mass m a

matrenka [14]

Assuming you are looking for the acceleration a:

1. $m_1a = T_1 -m_1g$
2. $m_2a = m_2g - T_2$
where T is the tension and a is the acceleration of the blocks. The acceleration of the two blocks and the acceleration of the pulley must be equal.

The torque on the pulley is given by:
3. $\tau = \overrightarrow r \times \overrightarrow F = (T_2 - T_1)R = I\alpha = \frac{1}{2} MR^2 \frac{a}{R}$
where $I = \frac{1}{2} mR^2$ and $a = \alpha R$ .

Combining the three equations:
$T_2 - T_1 = \frac{1}{2} Ma \\ m_2g - m_2a -m_1g - m_1a = (m_2-m_1)g - (m_1 + m_2)a = \frac{1}2}Ma \\ \\ a = \frac{(m_2 - m_1)g}{m_1 + m_2 + \frac{1}{2}M }$

6 0

3 years ago

A 1.5 kg cart is attached to a spring with spring constant of 5 N/m. The cart & spring is pulled to stretch the spring by 3

vaieri [72.5K]

22.5 J

Explanation:

Given:

x = 3 m

$k = 5\:\text{N/m}$