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3 years ago

How would you go about separating the components of a mixture of sand, gravel, salt and iron filings

Physics

1 answer:

larisa86 [58]3 years ago

4 0

Pour the entire components into water.

First the iron filings can be separated using a magnet as iron is a magnetic element.

By pouring the mixture into water, it allows the salt to dissolve in the water, while the sand will not.

Next we can run this solution through a filtration device to separate the sand from the water.

Now all that's left is to let the water evaporate so that the salt will be exposed that dissolved into the water.

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Do rolling down a grassy hill has kinetic energy or potential energy

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If you're moving, then you have kinetic energy.

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With fuel prices for combustible engine automobiles increasing, researchers and manufacturers have given more attention to the c

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Answer:

Then the difference of weight between the two cars are:

Δw = 14210 - 5292 = 8918 N

Explanation:

An object's weigh due to the gravitational attraction force of the earth is:

w = mg

Where: m is the object's mass

g is the gravitational acceleration in the surface earth

g = 9.8 m/s2

The the ultralight car's weight is:

$w_{uc} = (540)(9.8)$

$w_{uc} = 5292 N$

And the Honda Accord's weight is:

$w_{HA} = (1450)(9.8)$

$w_{HA} = 14210 N$

Then the difference of weight between the two cars are:

Δw = 14210 - 5292 = 8918 N

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3 years ago

How do I solve this

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multiply grav pull by mass of astro maybe with a calculator

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3 years ago

Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el

rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

$q_1 = 21.0\mu C$

$q_1 = 47.0\mu C$

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

$E = \frac{kq}{x}^2$

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

$\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2$

solving for x we get

$\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}$

x = 0.200 m

b)electric field at half way mean x =0.25

$E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}$

E = 3.84*10^{-4} N/C

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2 years ago

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