I think you forgot to give the options along with the question. I am answering the question based on my knowledge and research. The criteria responsible for deciding whether a heterogeneous mixture is a colloid or a suspension is whether the <span>particles remain suspended for an extended period of time. I hope it helps you.</span>
<span> The boiling point of water at sea level is 100 °C. At higher altitudes, the boiling point of water will be.....
a) higher, because the altitude is greater.
b) lower, because temperatures are lower.
c) the same, because water always boils at 100 °C.
d) higher, because there are fewer water molecules in the air.
==> e) lower, because the atmospheric pressure is lower.
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Water boils at a lower temperature on top of a mountain because there is less air pressure on the molecules.
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I hope this is helpful. </span>
The solution would be like
this for this specific problem:
<span>
The force on m is:</span>
<span>
GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] ->
1
The force on 2m is:</span>
<span>
GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2]
-> 2
From (1), you’ll get M = 2mx^2 / L^2 and from
(2) you get M = m(L - x)^2 / L^2
Since the Ms are the same, then
2mx^2 / L^2 = m(L - x)^2 / L^2
2x^2 = (L - x)^2
xsqrt2 = L - x
x(1 + sqrt2) = L
x = L / (sqrt2 + 1) From here, we rationalize.
x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1)
x = L(sqrt2 - 1) / (2 - 1)
x = L(sqrt2 - 1) </span>
= 0.414L
<span>Therefore, the third particle should be located the 0.414L x
axis so that the magnitude of the gravitational force on both particle 1 and
particle 2 doubles.</span>
Answer:
The fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C
Explanation:
Given;
initial temperature of the liquid, t₁ = 76.3 +/- 0.4⁰C
final temperature of the liquid, t₂ = 67.7 +/- 0.3⁰C
The change in temperature of the liquid is calculated as;
Δt = t₂ - t₁
Δt = (67.7 - 76.3) +/- (0.3 - 0.4)
Δt = (-8.6) +/- (-0.1)
Δt = 8.6 +/- 0.1 ⁰C
Therefore, the fall in temperature of the liquid is 8.6 +/- 0.1 ⁰C
