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maxonik [38]
3 years ago
10

A scientist completely dissolves gaseous oxygen into a container of liquid water. Which of the following best describes the cont

ents of the container after the oxygen has been dissolved?
A.
an element
B.
a solution
C.
a compound
D.
a substance

STUDY ISLAND
Physics
2 answers:
soldier1979 [14.2K]3 years ago
8 0

Answer:

c

Explanation:

sdas [7]3 years ago
5 0

Answer:

Forgive me if I'm wrong but I think the answer is C.

Explanation:

I say this because liquid water is H20 (a combination of two elements) and Oxygen is a singular element. Compound is defined as a thing that is composed of two or more separate elements and thus, I think it's C.

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<span>when it returns to its original level after encountering air resistance, its kinetic energy is decreased. 
In fact, part of the energy has been dissipated due to the air resistance.

The mechanical energy of the ball as it starts the motion is:
</span>E=K = 100 J
<span>where K is the kinetic energy, and where there is no potential energy since we use the initial height of the ball as reference level.
If there is no air resistance, this total energy is conserved, therefore when the ball returns to its original height, the kinetic energy will still be 100 J. However, because of the presence of the air resistance, the total mechanical energy is not conserved, and part of the total energy of the ball has been dissipated through the air. Therefore, when the ball returns to its original level, the kinetic energy will be less than 100 J.</span>
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3 years ago
Which factors describe a gas
Alekssandra [29.7K]

Answer:

B. changing shape and changing volume

Explanation:

*no definite shape (takes the shape of its container)

*no definite volume

7 0
3 years ago
A car with a mass of 2.0×10^3 kg is traveling at 15m/s .what is the momentum of the car ?
Maurinko [17]
Hello,

<span>A car with a mass of 2.0×10^3 kg is traveling at 15m/s. We need to find the momentum of the car. To do so, follow this formula:

p=mv

Where,

p = momentum 
m = mass
v = </span>velocity

The cars mass is 2.0E3 and its velocity is 15m/s. Therefore:

p=2.0  x 10^3 *15 or 2000(15)

p=30000

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3 years ago
Urgent!!!!!! 20 points!!!!!!<br> how are the speed, wavelength and frequency of a wave related
zheka24 [161]

Answer:

The higher the frequency, the shorter the wavelength. Because all light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength. Speed shows how long it takes for wavelengths to travel.

6 0
3 years ago
A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri
Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = \frac{I}{nqA}

Where;

n = number of free electrons per cubic meter

q =  electron charge

A =  cross-sectional area of the wire

<em>First let's calculate the number of free electrons per cubic meter (n)</em>

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M)          -------------(ii)

<em>Substitute the values of Nₐ, ρ and M into equation (ii) as follows;</em>

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

<em>Now let's calculate the drift electron</em>

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

<em>Substitute these values into equation (i) as follows;</em>

v = \frac{I}{nqA}

v = \frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

6 0
3 years ago
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