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3 years ago

13

Which mineral identification property deals with how a mineral splits along geometric planes determined by its crystal structure

?
A. tenacity B. fracture C. cleavage D. hardness

1 answer:

S_A_V [24]3 years ago

5 0

║∦Cleavage. <span>The tendency of a </span>mineral<span> to break along these flat parallel surfaces is known as cleavage. </span>

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please answer 1-3 and use a separate piece of paper thank you i've been trying those problems for 3 day.... ikr Sad

11Alexandr11 [23.1K]

$\frac{total distance}{ total time}$

$\frac{72m}{36s} = 2m/s$

Same thing for the rest
2) 5km/h
3)360km/h

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3 years ago

In an effort to decrease the mAs of an exposure, the 15% rule of kVp change may be considered. Changing the original kVp of 84 u

Roman55 [17]

Answer:

Will result in :Greater Compton scatter interaction

Explanation

This is because The 15% rule states that changing the kVp by 15% has the same effect as doubling the mAs, or reducing the mAs by 50%; that is A 15% increase in kVp has the same effect as doubling the mAs and vice versa but in this case reducing mas causes more Compton scatter because Compton scattering occurs at a lower kvp

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3 years ago

Name 2 common methods of polymerization.

navik [9.2K]

Answer: Addition polymerization & Condensation polymerization

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3 years ago

A coil with an inductance of 2.3 H and a resistance of 14 Ω is suddenly connected to an ideal battery with ε = 100 V. At 0.13 s

klemol [59]

Given Information:

Resistance = R = 14 Ω

Inductance = L = 2.3 H

voltage = V = 100 V

time = t = 0.13 s

Required Information:

(a) energy is being stored in the magnetic field

(b) thermal energy is appearing in the resistance

(c) energy is being delivered by the battery?

Answer:

(a) energy is being stored in the magnetic field ≈ 219 watts

(b) thermal energy is appearing in the resistance ≈ 267 watts

(c) energy is being delivered by the battery ≈ 481 watts

Explanation:

The energy stored in the inductor is given by

$U = \frac{1}{2} Li^{2}$

The rate at which the energy is being stored in the inductor is given by

$\frac{dU}{dt} = Li\frac{di}{dt} \: \: \: \: eq. 1$

The current through the RL circuit is given by

$i = \frac{V}{R} (1-e^{-\frac{t}{ \tau} })$

Where τ is the the time constant and is given by

$\tau = \frac{L}{R}\\ \tau = \frac{2.3}{14}\\ \tau = 0.16$

$i = \frac{110}{14} (1-e^{-\frac{t}{ 0.16} })\\i = 7.86(1-e^{-6.25t})\\\frac{di}{dt} = 49.125e^{-6.25t}$

Therefore, eq. 1 becomes

$\frac{dU}{dt} = (2.3)(7.86(1-e^{-6.25t}))(49.125e^{-6.25t})$

At t = 0.13 seconds

$\frac{dU}{dt} = (2.3) (4.37) (21.8)\\\frac{dU}{dt} = 219.11 \: watts$

(b) thermal energy is appearing in the resistance

The thermal energy is given by

$P = i^{2}R\\P = (7.86(1-e^{-6.25t}))^{2} \cdot 14\\P = (4.37)^{2}\cdot 14\\P = 267.35 \: watts$

(c) energy is being delivered by the battery?

The energy delivered by battery is

$P = Vi\\P = 110\cdot 4.37\\P = 481 \: watts$

4 0

3 years ago

Argon makes up 0.93% by volume of air. Calculate its solubility (in mol/L) in water at 20°C and 1.0 atm. The Henry's law constan

vladimir2022 [97]

Answer:

S= 1.40x10⁻⁵mol/L

Explanation:

The Henry's Law is given by the next expression:

$S = k_{H} \cdot p$ (1)

<em>where S: is the solubility or concentration of Ar in water, $k_{H}$ : is Henry's law constant and p: is the pressure of the Ar </em>

<u>Since the argon is 0.93%, we need to multiply the equation (1) by this percent:</u>

$S = 1.5 \cdot 10^{-3} \frac{mol}{L\cdot atm} \cdot 1.0atm \cdot \frac{0.93}{100} = 1.40 \cdot 10^{-5} \frac{mol}{L}$

Therefore, the argon solubility in water is 1.40x10⁻⁵mol/L.

Have a nice day!

8 0

3 years ago