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Varvara68 [4.7K]
4 years ago
9

Which of the following are effective treatments for some musculoskeletal conditions? Select all that apply. PLEASE ANSWER

Chemistry
1 answer:
Dimas [21]4 years ago
6 0

Answer:

1,3, 4

Explanation:

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A tank of gas has partial pressures of nitrogen and oxygen equal to 1.61 × 10^4 kPa and 4.34 × 10^3 kPa, respectively. What is t
goldfiish [28.3K]

Answer:

answer is 2.04*10^4 kPa

Explanation:

Dalton's Law of Partial Pressures.

it is because that when the volume and the temperature of a gaseous mixture are kept constant, the total pressure of the mixture is equal to the sum of the partial pressures of its gaseous components .

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Help please because why not
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B because you can feel the current
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Occasionally, deposition causes the main channel of a stream to divide into several smaller channels called _____.
timurjin [86]

ANSWER: D) distributaries

EXPLANATION: When a deposition causes the main channel of a stream to divide into several smaller channel it is called a distributary. They are a common feature of river deltas and is also known as river bifurcation. A distributary generally occurs near a lake or ocean and can also occur near a confluence of large stream. This is popularly also known as arm or channel.

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In science What are lithium properties
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Answer:

Lithium has a melting point of 180.54 C, a boiling point of 1342 C, a specific gravity of 0.534 20 C

Explanation:

6 0
2 years ago
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Suppose you store a 15.5 g piece of aluminum (Cp of Al = 0.902 J/g⁰C) in the refrigerator at 2.3⁰C and then drop it into your co
Scorpion4ik [409]

Answer:

605.4 J

Explanation:

When a certain substance absorbs a certain amount of energy, its temperature increases according to  the equation:

Q=mC\Delta T

where

Q is the heat absorbed

m is the mass of the substance

C is the specific heat capacity of the substance

\Delta T is the change in temperature

In this problem, we have:

m = 15.5 g is the mass of the piece of aluminium

C = 0.902 J/g⁰C is the specific heat of the aluminium

\Delta T = 45.6-2.3=43.3^{\circ}C is the change in temperature of the aluminium (in fact, at thermal equilibrium, the block of aluminium reaches the same final temperature as the coffee)

Therefore, the energy absorbed is

Q=(15.5)(0.902)(43.3)=605.4 J

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3 years ago
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