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Fofino [41]
3 years ago
13

A bar magnet is immersed in a heap of iron fillings and pulled out . The amount of iron fillings clinging to the ?

Physics
1 answer:
grin007 [14]3 years ago
8 0

Answer:

  North Pole is almost equal to the South Pole.

Explanation:

  The magnetic fields are weaker in the center of a magnet than at the poles.  Poles have the maximum magnetic strength of equal intensity. When dipped in a heap of iron fillings, the fillings gets attracted to the poles than other parts of magnet. Therefore, the amount of iron fillings attracted to both the poles will be almost equal.

<em>Hope this helps</em>

<em>:)</em>

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your answer for this question is B

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How does depth of liquid affect pressure in solids
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The higher the depth, the higher the mass of the liquid that creates pressure
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Which type of seismic wave can only pass through the earths mantle?
mezya [45]
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6 0
4 years ago
Suppose you toss a tennis ball upward.a) Does the kinetic energy of the ball increase or decrease as it moves higher?b) What hap
Lilit [14]

Answer:

a) No. The kinetic energy of the ball decreases.

b) The potential energy of the ball increases.

c) The ball would go half of the original distance.

Explanation:

a) The kinetic energy would be converted to potential energy as the ball goes higher. Since the total mechanical energy is conserved, the kinetic energy would decrease.

b) The potential energy of the ball would increase. Since the total mechanical energy of the ball is conserved, the ball would lose speed, and therefore kinetic energy. In order to compensate the loss of kinetic energy, the ball would gain potential energy as it goes higher.

c) The relation of the energy and mass is as follows:

K = \frac{1}{2}mv^2\\U = mgh

According to the energy conservation

K_1 + U_1 = K_2 + U_2\\\frac{1}{2}mv^2 + 0 = 0 + mgh\\\frac{1}{2}v^2 = gh

The maximum height that the ball reaches is proportional to the initial velocity. If the ball would be imparted with the same amount of energy, its final potential energy would be the same. However, in order to have the same potential energy (mgh), its height would be half of the original case.

mgh = (2m)g(\frac{h}{2})

7 0
3 years ago
Rank the deformations of the following rods in terms of the magnitude of the average normal strain: (a) The length of a 1-m-long
kipiarov [429]

To solve this problem we will consider the concepts related to the normal deformation on a surface, generated when the change in length is taken per unit of established length, that is, the division between the longitudinal fraction gained or lost, over the initial length. In general mode this normal deformation can be defined as

\epsilon = \frac{\delta}{l} = \frac{l_0-l}{l}

Here,

\delta= Change in final length (l_0) and the initial length l

PART A)

\epsilon = \frac{\delta_1}{l}

\epsilon = \frac{l_0-l}{l_0}

\epsilon = \frac{1.02-1}{1}

\epsilon = 0.01961

PART B)

\epsilon = \frac{\delta_1}{l}

\epsilon = \frac{l_0-l}{l_0}

\epsilon = \frac{2-1.05}{2}

\epsilon = 0.475

PART C)

\epsilon = \frac{\delta_1}{l}

\epsilon = \frac{l_0-l}{l_0}

\epsilon = \frac{3.07-3}{3}

\epsilon = 0.0233

Therefore the rank of this deformation would be  B>C>A

7 0
3 years ago
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