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Lynna [10]
3 years ago
13

Which of these are minerals? Select all that apply A) mica B) quartz C) steam D) copper

Physics
2 answers:
posledela3 years ago
6 0
<span>A) mica B) quartz C) steam D) copper is allminerals</span>
Molodets [167]3 years ago
5 0
Your answer would be:
Copper
Quartz

I hope that this helps. :)
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1. A fixed pulley is a machine that increases the effort force.
IRISSAK [1]

For the specific questions given, these would be the answers:

1) False

2) False

3) True

4) True

5) True

I am hoping that these answers have satisfied your queries and it will be able to help you, thank you and have a nice day.

6 0
3 years ago
Which of the following is NOT correct? . . a A changing electric field can produce a changing magnetic field . . b A steady magn
postnew [5]
The correct answer of the given question above would be option B. The statement that is not correct is that, a steady magnetic field produces a steady current. The rest of the statements are all correct. <span>An unchanging/static magnetic field (relative to a wire/circuit) induces zero current.</span>
4 0
3 years ago
Read 2 more answers
3. A 5 gm/100 ml solution of drug X is stored in a closed test tube
JulijaS [17]

Answer:

See explanation

Explanation:

The degradation of the drug is a first order process;

Hence;

ln[A] = ln[A]o - kt

Where;

ln[A] = final concentration of the drug

ln[A]o= initial concentration of the drug = 5 gm/100

k= degradation constant = 0.05 day-1

t= time taken

When [A] =[ A]o - 0.5[A]o = 0.5[A]o

ln2.5 = ln5 - 0.05t

ln2.5- ln5 = - 0.05t

t= ln2.5- ln5/-0.05

t= 0.9162 - 1.6094/-0.05

t= 14 days

b) when [A] = [A]o - 0.9[A]o = 0.1[A]o

ln0.5 = ln5 -0.05t

t= ln0.5 - ln5/0.05

t= -0.693 - 1.6094/-0.05

t= 46 days

5 0
3 years ago
The Kelvin scale is the most common temperature scale used in what
dalvyx [7]
It's mostly used in CHEMICAL PROCESSES.
7 0
3 years ago
Read 2 more answers
I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a
attashe74 [19]
<span>The angular momentum of a particle in orbit is 

l = m v r 

Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2" 

m_1 v_1 r_1 = m_2 v_2 r_2 

Assuming that the mass did not change, conservation of angular momentum demands that 

v_1 r_1 = v_2 r_2 

or 

v1 = v_2 (r_2/r_1) 

Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have 

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>
3 0
3 years ago
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