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loris [4]
3 years ago
6

Your classmate states that only precious minerals, such as diamonds, are valuable. Based on your lesson on the rock cycle, you _

____.
A.agree with her; pound for pound, diamonds cost the most
B.agree with her; diamonds are valuable, but there are other minerals that are used in industry that have a high value, as well
C.disagree with her; the cost of diamonds has dropped sharply over the past ten years
D.disagree with her; diamonds have no true value
Physics
2 answers:
butalik [34]3 years ago
8 0

The correct answer is B.agree with her; diamonds are valuable, but there are other minerals that are used in industry that have a high value, as well

Diamonds are valuable but there are also many minerals that are used in Industry and are valuable too. For example Gold is a valuable element that is used in the production of Micro-processors. Platinum is another mineral that is used in the production of Catalytic converters and nitric acid. It is also very expensive.

FinnZ [79.3K]3 years ago
4 0
B, agree with her but there are other minerals that are used that have a high value as well

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What is meant by the term fitness​
Artemon [7]

Answer:

An organisms ability to survive and reproduce in a particular environment.

4 0
3 years ago
The drawing shows a bicycle wheel resting against a small step whose height is h = 0.110 m. The weight and radius of the wheel a
Pavel [41]

Answer:

F=27.39N

Explanation:

Take sum of torques at the point the step touches the wheel, that eliminates two torques

ΣT=T_{N}+T_{f}+T_{W}

Since we are looking for when the wheel just starts to rise up N-> 0 so no torque due to normal force

T_{N}=0

The perpendicular lever arm for the F force is R-h

T_{f}=F*(r-h)

And the T of gravity according to the image

T_{W}=W*(\sqrt{r^2-(r-h)^2}

ΣT=0

T_{N}+T_{f}+T_{W}=0

F*(r-h)+W*(\sqrt{r^2-(r-h)^2}=0

F=\frac{W*(\sqrt{r^2-(r-h)^2}}{r-h}

F=\frac{24.9 N*(\sqrt{0.336^2-(0.336-0.110)^2}}{(0.336-0.11)}

F=27.39N

4 0
3 years ago
If 2 objects are moved by the same force (F):
kipiarov [429]

Answer:y=mx+b 58+5

Explanation:

8 0
2 years ago
Read 2 more answers
Un objeto de 200 gramos está amarrado del extremo de una cuerda y gira describiendo un círculo horizontal de 1.20 m de radio a r
vesna_86 [32]

Answer:

La tensión es 85.3 N.

Explanation:

Cuando el objeto gira en dirección horizontal, la sumatoria de fuerzas se puede calcular usando la segunda ley de Newton:  

\Sigma F_{x} =ma_{c}

T = ma_{c}  

Dado que el movimiento es horizontal, el peso (que está en el eje y) no contribuye en la sumatoria de fuerzas en el eje x. Por lo que la única fuerza actuando sobre el objeto en la dirección del movimiento es la tensión.  

En donde:                                          

m: es la masa del objeto = 200 g = 0.200 kg

a_{c}: es la aceleración centrípeta

La aceleración centrípeta viene dada por:  

a_{c} = \omega^{2} r

En donde:    

ω: es la velocidad angular del objeto = 3 rev/s

r: es el radio = 1.20 m

Entonces, la tensión es:

T = m\omega^{2} r = 0.200 kg(3\frac{rev}{s}*\frac{2\pi rad}{1 rev})^{2}*1.20 m = 85.3 N

   

Por lo tanto, la tensión es 85.3 N.  

Espero que te sea de utilidad!                                                                          

5 0
2 years ago
A car travels at a constant speed up a ramp making an angle of 28 degrees with the horizontal component of velocity is 40 kmh^-1
Hoochie [10]

Answer:

Vx = 35.31 [km/h]

Vy = 18.77 [km/h]

Explanation:

In order to solve this problem, we must decompose the velocity component by means of the angle of 28° using the cosine function of the angle.

v_{x} = 40*cos(28)\\V_{x} = 35.31 [km/h]

In order to find the vertical component, we must use the sine function of the angle.

V_{y}=40*sin(28)\\V_{y} = 18.77 [km/h]

7 0
3 years ago
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