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4 years ago

What is the five part definition for an object to be considered a mineral?

1 answer:

7 0

Definition<span> of a </span>Mineral<span>. A </span>mineral<span> is a naturally-occurring, homogeneous solid with a definite, but generally not fixed, chemical composition and an ordered atomic arrangement. It is usually formed by inorganic processes (got from google</span>

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Please help these with these two questions

Gennadij [26K]

<h2>If the distance between Earth and moon is increased , the gravitational force will decrease by a proportional amount .</h2><h2>The iron filings sprinkle around the magnet are arranged in regular pattern . The reason for this is that magnet can attract magnetic substances through various materials .</h2>

Explanation:

If the distance between Earth and moon is increased , the gravitational force will decrease by a proportional amount .

The iron filings sprinkle around the magnet are arranged in regular pattern . The reason for this is that magnet can attract magnetic substances through various materials .

8 0

3 years ago

A circular loop of flexible iron wire has an initial circumference of 164cm , but its circumference is decreasing at a constant

Travka [436]

Answer:

emf = 0.02525 V

induced current with a counterclockwise direction

Explanation:

The emf is given by the following formula:

$emf=-\frac{\Delta \Phi_B}{\Delta t}=-B\frac{\Delta A}{\Delta t}$ $\ \ =-B\frac{A_2-A_1}{t_2-t_1}$ (1)

ФB: magnetic flux = BA

B: magnitude of the magnetic field = 1.00T

A2: final area of the loop; A1: initial area

t2: final time, t1: initial time

You first calculate the final A2, by taking into account that the circumference of loop decreases at 11.0cm/s.

In t = 4 s the final circumference will be:

$c_2=c_1-(11.0cm/s)t=164cm-(11.0cm/s)(4s)=120cm$

To find the areas A1 and A2 you calculate the radius:

$r_1=\frac{164cm}{2\pi}=26.101cm\\\\r_2=\frac{120cm}{2\pi}=19.098cm$

r1 = 0.261 m

r2 = 0.190 m

Then, the areas A1 and A2 are:

$A_1=\pi r_1^2=\pi (0.261m)^2=0.214m^2\\\\A_2=\pi r_2^2=\pi (0.190m)^2=0.113m^2$

Finally, the emf induced, by using the equation (1), is:

$emf=-(1.00T)\frac{(0.113m^2)-(0.214m^2)}{4s-0s}=0.0252V=25.25mV$

The induced current has counterclockwise direction, because the induced magneitc field generated by the induced current must be opposite to the constant magnetic field B.

4 0

3 years ago

15 pointsss please help meeeeeeeeeee

AveGali [126]

A. Acceleration! ㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤㅤ

5 0

3 years ago

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A physics student mounts two thin lenses along a single optical axis (the lenses are at right angles to the line connecting them

Tatiana [17]

Answer:

A) q₂ = 75.98 cm, B) q₂' = 115.38 cm, C)

Explanation:

A) This is an exercise in geometric optics, as the two lenses are separated by a greater distance than their focal lengths from each lens, they must be worked as independent lenses.

Lens 1. More to the left

let's use the constructor equation

$\frac{1}{f} = \frac{1}{p} + \frac{1}{q}$

where f is the focal length, p and q are the distance to the object and the image, respectively,

We must assume a distance to the object to perform the calculation, suppose that the object is 50 cm from lens 1 that is further to the left of the system.

$\frac{1}{q_1} = \frac{1}{f} - \frac{1}{p}$

$\frac{1}{q_1} = \frac{1}{14.8} - \frac{1}{50}$

1 / q₁ = 0.04756

q₁ = 21.0227 cm

this image is the object for the second lens that has f₂ = 14.8 cm

the distance must be measured from the second lens

p₂ = 39.4 -q₁

p₂ = 39.4 -21.0227

p₂ = 18.38 cm

let's use the constructor equation

1 / q₂ = 1 / f - 1 / p2

$\frac{1}{q_2} = \frac{1}{14.8} - \frac{1}{18.38}$

$\frac{1}{q_2}$ = 0.01316

q₂ = 75.98 cm

measured from the second lens

B) the position of the final image with respect to the first lens is

q₂’= q₂ + 39.4

q₂'= 75.98 +39.4

q₂' = 115.38 cm

C) the magnification of a lens is

m = - q / p

in this case the image measured from lens 2 is q2 = 75.98 cm

the distance to the object from the first lens is p1 = 50cm

m = - 75.98 / 50

m = -1.5 X

the negative sign indicates that the image is inverted

4 0

3 years ago

A golf ball is hit from the ground with an initial velocity of 208 feet per second. Assume the starting height of the ball is 0

kirill115 [55]

It will take 13 seconds for the golf ball to hit the ground. The correct answer between all the choices given is the last choice or letter D. I am hoping that this answer has satisfied your query and it will be able to help you in your endeavor, and if you would like, feel free to ask another question.

7 0

3 years ago

Read 2 more answers