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2 years ago

14

How is specific gravity or density used in identifying minerals

1 answer:

Elena L [17]2 years ago

7 0

<span>In summary, density and specific gravity are properties used to help identify minerals. Density is a measure of the mass of a certain volume of the sample. Specific gravity is a unitless measure, and it is the ratio of the mass of a substance to the mass of an equal volume of water.</span>

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9. A 50 kg physics i student trudges from the 1st floor to the 3rd floor, going up a total height of 13 m.

SVETLANKA909090 [29]

Answer:

6370 J

Explanation:

By the law of energy conservation, the work done by the student would be the change in potential enegy from 1st floor to 3rd floor, or a change of 13 m

$W = E_p = mgh$

where m = 50kg is the mass of the student, g = 9.8 m/s2 is the gravitational constant and h = 13 m is the height difference

$W = 50*9.8*13 = 6370 J$

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2 years ago

Which of these observations would most likely be seen in stage N

Free_Kalibri [48]

'N' is labelled 'runoff'. That's when the water is down. Streets are wet, worms come out of the ground, and water flows down the street to the sewer.

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2 years ago

When light propagates from a material with a given index of refraction into a material with a smaller index of refraction, the s

Ket [755]

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2 years ago

A circular loop of wire with a diameter of 0.626 m is rotated in a uniform electric field to a position where the electric flux

Mila [183]

Answer:

C) 2.44 × 106 N/C

Explanation:

The electric flux through a circular loop of wire is given by

$\Phi = EA cos \theta$

where

E is the electric field

A is the cross-sectional area

$\theta$ is the angle between the direction of the electric field and the normal to A

The flux is maximum when $\theta=0^{\circ}$ , so we are in this situation and therefore $cos \theta =1$ , so we can write

$\Phi = EA$

Here we have:

$\Phi = 7.50\cdot 10^5 N/m^2 C$ is the flux

d = 0.626 m is the diameter of the coil, so the radius is

r = 0.313 m

and so the area is

$A=\pi r^2 = \pi (0.313 m)^2=0.308 m^2$

And so, we can find the magnitude of the electric field:

$E=\frac{\Phi}{A}=\frac{7.50\cdot 10^5 Nm^2/C}{0.308 m^2}=2.44\cdot 10^6 N/C$

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3 years ago

What kind of energy is stored when you squeeze a mattress

KengaRu [80]

Answer:

elastic energy

Explanation:

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1 year ago

Read 2 more answers