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Ulleksa [173]
3 years ago
14

What elements are most common in minerals?

Physics
2 answers:
Leokris [45]3 years ago
4 0
<span>These elements are what make up minerals. The two most common elements are Oxygen (O) at 47% and Silicon (Si) at 28%. Other common elements are Aluminium (Al) at 8%, Iron (Fe) at 5%, Calcium (Ca) at 3.5%, Sodium (Na) at 3%, Potassium (K) at 2.5% and Magnesium (Mg) at just 2%.
 The information I have provided to you is from google </span>
sasho [114]3 years ago
3 0
OXYGEN & SILICON are most common in minerals......
You might be interested in
1. Faça as transformações:
zhannawk [14.2K]

Answer:

seconds (s) = hours (h) *3,600 ; h = \frac{s}{3,600} \\g = kg * 1,000; kg = \frac{g}{1,000} \\cm = \frac{m}{100};m = cm * 100

1. a) 0.5 h = 1,800 s

  h) 20 cm = 0.2 m

  b) 2.0 h = 7,200 s

   i) 5.0 kg = 5,000 g

  c) 3.5 h = 12,600 s

  j) 1.5 kg = 1,500 g

  d) 1/4 h = 900 s

  k) 450.0 g = 0.45 kg

  e) 3.0 m = 300 cm

   l) 20.0 g = 0.02 kg

  f) 2.5 m = 250 cm

  m) 500.0 g = 0.5 kg

   g) 0.5 m = 500 mm

   n) 1000.0 g = 1 kg

3 0
3 years ago
A circuit is made of a battery, a light bulb, and a 2 resistor. The battery has a voltage of 3 volts. When connected, the ammete
Monica [59]

Answer:

3ohms

Explanation:

From Ohm's Law

V = IR

V is that voltage = 3volts

I = current = 1amp

R = resistance in ohms

Putting those values into the above formula.

3volts = 1amp×R

Making R the subject

R = 3/1

R = 3ohms

The resistance of the light bulb is 3ohms.

6 0
4 years ago
An instrument is thrown upward with a speed of 15 m/s on the surface of planet X where the acceleration due to gravity is 2.5 m/
Katen [24]
<h2>Answer: 12 s</h2>

Explanation:

The situation described here is parabolic movement. However, as we are told <u>the instrument is thrown upward</u> from the surface, we will only use the equations related to the Y axis.

In this sense, the main movement equation in the Y axis is:

y-y_{o}=V_{o}.t-\frac{1}{2}g.t^{2}    (1)

Where:

y  is the instrument's final position  

y_{o}=0  is the instrument's initial position

V_{o}=15m/s is the instrument's initial velocity

t is the time the parabolic movement lasts

g=2.5\frac{m}{s^{2}}  is the acceleration due to gravity at the surface of planet X.

As we know y_{o}=0  and y=0 when the object hits the ground, equation (1) is rewritten as:

0=V_{o}.t-\frac{1}{2}g.t^{2}    (2)

Finding t:

0=t(V_{o}-\frac{1}{2}g.t^{2})   (3)

t=\frac{2V_{o}}{g}   (4)

t=\frac{2(15m/s)}{2.5\frac{m}{s^{2}}}   (5)

Finally:

t=12s

3 0
3 years ago
The acclaimed environmental book, Silent Spring, was written by this woman in 1962.
olga_2 [115]
Rachel Carson was the author of the acclaimed environmental book, Silent Spring. The book was published in 1962.
It documented the negative effect that synthetic pesticides have on the environment, specifically on birds.
This book laid bare to the American public what chemical companies indirectly contributed to the environment. 
The public furor this book caused led to the reversal of the national pesticide policy, a nationwide ban on DDT for agricultural uses, and inspired the creation of the U.S. Environmental Protection Agency.
7 0
3 years ago
A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1
bija089 [108]

1.

Answer:

Part a)

\rho = 1.35 \times 10^{-5}

Part b)

\alpha = 1.12 \times 10^{-3}

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

i = 18.7 A

V = 17.0 volts

now we will have

V = I R

17.0 = 18.7 R

R = 0.91 ohm

now we know that

R = \rho \frac{L}{A}

0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho = 1.35 \times 10^{-5}

Part b)

Now at higher temperature we have

V = I R

17.0 = 17.3 R

R = 0.98 ohm

now we know that

R = \rho \frac{L}{A}

0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}

\rho' = 1.46 \times 10^{-5}

so we will have

\rho' = \rho(1 + \alpha \Delta T)

1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))

\alpha = 1.12 \times 10^{-3}

2.

Answer:

Part a)

i = 1.55 A

Part b)

v_d = 1.4 \times 10^{-4} m/s

Explanation:

Part a)

As we know that current density is defined as

j = \frac{i}{A}

now we have

i = jA

Now we have

j = 1.90 \times 10^6 A/m^2

A = \pi(\frac{1.02 \times 10^{-3}}{2})^2

so we will have

i = 1.55 A

Part b)

now we have

j = nev_d

so we have

n = 8.5 \times 10^{28}

e = 1.6 \times 10^{-19} C

so we have

1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d

v_d = 1.4 \times 10^{-4} m/s

8 0
4 years ago
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