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hram777 [196]
3 years ago
5

What does it mean to say the mass is conserved during a physical change ?

Physics
2 answers:
vesna_86 [32]3 years ago
5 0

The total mass of the substance remains the same from beginning to end. The physical properties of the substance, such as size and shape, may change. The amount of matter in the substance does not change.

Olin [163]3 years ago
4 0
Mater doesn't just appear or disappeared. Chemical elements are still there just the connections and how it combines changes.

So what goes into your chemical eqation must still exist after the change.
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a

Explanation:

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3 years ago
Using calcium’s atomic structure, as shown in the image, what is this element’s atomic number?
ElenaW [278]

Answer:

Atomic number of calcium is 20.

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2 years ago
An object is dropped and is in free fall. Each second, the position of the object is marked. The distance between each mark is m
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C. hope this helps :)
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3 years ago
g stAn experienced spear fisherman sees a small fish swimming in a tidal pool. If the fisherman sees the fish at approximately a
Luba_88 [7]

Answer:

  θ = 28.9

Explanation:

For this exercise let's use the law of refraction

          n₁ sin θ₁ = n₂ sin θ₂

where we use index 1 for air and index 2 for water where the fish is

        sin θ₂ = n₁ / n₂ sin θ₁

in this case the air repair index is 1 and the water 1.33

we substitute

        sin θ₂ = 1 / 1.33 sin t 40

        sin θ = 0.4833

         

        θ = sin⁻¹ 0.4833

        θ = 28.9

5 0
3 years ago
The volume electric charge density of a solid sphere is given by the following equation: The variable r denotes the distance fro
qwelly [4]

Answer:

62.8 μC

Explanation:

Here is the complete question

The volume electric charge density of a solid sphere is given by the following equation: ρ = (0.2 mC/m⁵)r²The variable r denotes the distance from the center of the sphere, in spherical coordinates. What is the net electric charge (in μC) of the sphere if the radius of the sphere is 0.5 m?

Solution

The total charge on the sphere Q = ∫∫∫ρdV where ρ = volume charge density = 0.2r² and dV = volume element in spherical coordinates = r²sinθdθdrdΦ

So,  Q =  ∫∫∫ρdV

Q =  ∫∫∫ρr²sinθdθdrdΦ

Q =  ∫∫∫(0.2r²)r²sinθdθdrdΦ

Q =  ∫∫∫0.2r⁴sinθdθdrdΦ

We integrate from r = 0 to r = 0.5 m, θ = 0 to π and Φ = 0 to 2π

So, Q =  ∫∫∫0.2r⁴sinθdθdrdΦ

Q =  ∫∫∫0.2r⁴[∫sinθdθ]drdΦ

Q =  ∫∫0.2r⁴[-cosθ]drdΦ

Q =  ∫∫0.2r⁴-[cosπ - cos0]drdΦ

Q =  ∫∫∫0.2r⁴-[-1 - 1]drdΦ

Q =  ∫∫0.2r⁴-[- 2]drdΦ

Q =  ∫∫0.2r⁴(2)drdΦ

Q =  ∫∫0.4r⁴drdΦ

Q =  ∫0.4r⁴dr∫dΦ

Q =  ∫0.4r⁴dr[Φ]

Q =  ∫0.4r⁴dr[2π - 0]

Q =  ∫0.4r⁴dr[2π]

Q =  ∫0.8πr⁴dr

Q =  0.8π∫r⁴dr

Q =  0.8π[r⁵/5]

Q = 0.8π[(0.5 m)⁵/5 - (0 m)⁵/5]

Q = 0.8π[0.125 m⁵/5 - 0 m⁵/5]

Q = 0.8π[0.025 m⁵ - 0 m⁵]

Q = 0.8π[0.025 m⁵]

Q = (0.02π mC/m⁵) m⁵

Q = 0.0628 mC

Q = 0.0628 × 10⁻³ C

Q = 62.8 × 10⁻³ × 10⁻³ C

Q = 62.8 × 10⁻⁶ C

Q = 62.8 μC

3 0
2 years ago
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