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photoshop1234 [79]
3 years ago
13

Henry mixed salt and water together in a cup until he observed a clear solution. He measured the mass of the solution. Then he p

laced the cup outside for several sunny days during the summer. After a week, he observed that only solid salt remained in the cup and the mass had decreased. Henry concluded that a physical and chemical change occurred in this investigation.
Which statements correctly defend or dispute his conclusion?

He is correct. Dissolving salt in water is a physical change, but evaporating the water is a chemical change. Formation of a solid is evidence that a chemical change occurred.
He is correct. Evaporation is a physical change, but dissolving salt in water is a chemical change. The change in mass is evidence that a chemical change occurred.
He is incorrect. Dissolving salt in water and evaporation of the water are both physical changes. The reappearance of salt is evidence that the change was reversible by a physical change, so it could not be a chemical change.
He is incorrect. Dissolving salt in water and evaporation of the water are both chemical
Physics
1 answer:
lapo4ka [179]3 years ago
6 0

Answer:

He is incorrect. Dissolving salt in water and evaporation of the water are both physical changes. The reappearance of salt is evidence that the change was reversible by a physical change, so it could not be a chemical change.

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Coulomb measured the deflection of sphere A when spheres A and B had equal charges and were a distance d apart. He then made the
sdas [7]

Answer:

The new distance is     d = 0.447 d₀

Explanation:

The electric out is given by Coulomb's Law

         F = k q₁ q₂ / r²

This electric force is in balance with tension.

We reduce the charge of sphere B to 1/5 of its initial value (q_{B}=q₂ = q₂ / 5) than new distance (d = n d₀)

dat

     q₁ = q_{A}

     q₂ = q_{B}

     r = d₀

In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points

         F = k q₁ q₂ / d₀²

         F = k q₁ (q₂ / 5) / (n d₀)²

         .k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)

          5 n² = 1

          n = √ 1/5

          n = 0.447

The new distance is

         d = 0.447 d₀

6 0
3 years ago
HELLO.
aivan3 [116]

Answer:

Hello! Thanks! I hope you are too! hope this helps!

Explanation:

The primary reason is to create more water pressure. ... If a municipal water tank is elevated to more than a hundred feet above the ground, there is an increase of . 43 psi per foot. A tank that has been elevated to an appropriate height can create water pressure similar to that created by a large pump

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3 years ago
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Maurinko [17]
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3 years ago
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What equation directly defines hookes law?
love history [14]

B. F<em>spring = k(triangle)</em> x

7 0
3 years ago
A particle's position is given by z(t) = −(6.50 m/s2)t2k for t ≥ 0. (Express your answer in vector form.) a. Find the particle's
blondinia [14]

Answer:

a) z'(t) =v(t) = -13t

Now we can replace the velocity for t=1.75 s

v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}

For t = 3.0 s we have:

v(3.0s) = -13*3.0 =-39 \frac{m}{s}

b) v_{avg}= \frac{z_f - z_i}{t_f -t_i}

And we can find the positions for the two times required like this:

z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m

z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m

And now we can replace and we got:

V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}

Explanation:

The particle position is given by:

z(t) = -(6.5 \frac{m}{s^2}) t^2, t\geq 0

Part a

In order to find the velocity we need to take the first derivate for the position function like this:

z'(t) =v(t) = -13t

Now we can replace the velocity for t=1.75 s

v(1.75s) = -13*1.75 =-22.75 \frac{m}{s}

For t = 3.0 s we have:

v(3.0s) = -13*3.0 =-39 \frac{m}{s}

Part b

For this case we can find the average velocity with the following formula:

v_{avg}= \frac{z_f - z_i}{t_f -t_i}

And we can find the positions for the two times required like this:

z_f = z(3.0s) = -(6.5 \frac{m}{s^2}) (3.0s)^2=-58.5m

z_i = z(1.75s) = -(6.5 \frac{m}{s^2}) (1.75s)^2=-19.906m

And now we can replace and we got:

V_{avg}= \frac{-58.5 -(-19.906) m}{3-1.75 s}= -30.875 \frac{m}{s}

8 0
3 years ago
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