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3 years ago

9

If you place a magnet under a clear dish, and sprinkle iron fillings over it, above what part of the magnet will most of the fil

lings gather

2 answers:

choli [55]3 years ago

8 0

Mostly all around the center the highest magnetic force and also around it because of the magnets magnetic field

xxTIMURxx [149]3 years ago

5 0

round the corners of the magnet that where it is stronger

hope this helps :)

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Mengapa indra peraba tidak dapat digunakan untuk membandingkan suhu dengan tepat?

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3 years ago

. A huge pile of leaves was wrapped in a tarp in the middle of a lawn. The wrapped leaves weigh 580 newtons. The coefficient of

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The force required is 319 N

Explanation:

The force of static friction is a force that acts an object on a surface, when this object is pushed by another force to put it in motion. The direction of the force of friction is opposite to the direction of the force of push, and its value increases as the force of push increases, up to a maximum value given by:

$F_f = \mu W$

where

$\mu$ is the coefficient of friction

W is the weight of the object

Therefore, in order to put the object in motion, the force applied must be greater than this value.

For the pile of leaves in this problem, we have:

$\mu = 0.55$ (coefficient of friction)

$W=580 N$ (weight of the leaves)

Substituting, we find:

$F=(0.55)(580)=319 N$

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3 years ago

In preparation for an electrophoresis procedure, enzymes are added to DNA in order to

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3 years ago

The gold has a density of 19300 kg/m3 calculate the mass of one gold bar 1= 2.54cm

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3 years ago

A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.

iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

$F = \frac{kq_1q_2}{d^2}$

Here

k = Coulomb's Constant

$q_{1,2} =$ Charge of each object

d = Distance

Our values are given as,

$q_1 = 1 \mu C$

$q_2 = 6 \mu C$

d = 1 m

$k = 9*10^9 Nm^2/C^2$

a) The electric force on charge $q_2$ is

$F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}$

$F_{12} = 54 mN$

Force is positive i.e. repulsive

b) As the force exerted on $q_2$ will be equal to that act on $q_1$ ,

$F_{21} = F_{12}$

$F_{21} = 54 mN$

Force is positive i.e. repulsive

c) If $q_2 = -6 \mu C$ , a negative sign will be introduced into the expression above i.e.

$F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}$

$F_{12} = F_{21} = -54 mN$

Force is negative i.e. attractive

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3 years ago