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aivan3 [116]
3 years ago
13

Fitness results come from hard work, not an instant fix Question 1 options: True False

Physics
1 answer:
zhannawk [14.2K]3 years ago
5 0
<span>This is a true statement. Fitness is typically a long-term solution, not a short-term repair. When looking for fitness results, it will usually take weeks to months to see any tangible results, even though there could be some short-term benefits such as increased energy. This is why fitness is sometimes seen as a lifetime endeavor and not one that occurs overnight.</span>
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A swimming pool, 10.0 m by 4.0 m, is filled with water to a depth of 3.0 m at a temperature of 20.2°c. if the energy needed to r
Tomtit [17]

You need to find the mass of water in the pool.
Find the volume (10 x 4 x 3) = 120 m3

Water has a density of 1000g/m3,so 120 m3 = 120 x 1000 = 120 000 kg

[delta]H = 4.187 x 120 000 x 3.4 (and the units will be kJ)

You then use the heat of combustion knowing that each mole of methane releases 891 kJ of heat so if you divide 891 into the previous answer, you will get the number of moles of CH4


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A student stands at the edge of a cliff and throws a stone horizontally over the edge with a speed of v0 = 15.0 m/s. The cliff i
ankoles [38]

Answer:

(a) x0 = 0m and y0 = 49.0m

(b) Vox = 15.0m/s Voy = 0m/s

(c) Vx = Vo = 15.0m/s and Vy = -gt

(d) X = 15.0t and y = 49.0 - 4.9t²

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Explanation:

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3 years ago
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What are the units, if any, of the particle in a box wavefunction. What does this mean?
SIZIF [17.4K]

Answer:

  • [\psi]= [Length^{-3/2}]
  • This means that the integral of the square modulus over the space is dimensionless.

Explanation:

We know that the square modulus of the wavefunction integrated over a volume gives us the probability of finding the particle in that volume. So the result of the integral

\int\limits^{x_f}_{x_0} \int\limits^{yf}_{y_0} \int\limits^{z_f}_{z_0} |\psi|^2 \, dz \,  dy \,  dx

must be dimensionless, as represents a probability.

As the differentials has units of length

[dx]=[dy]=[dz]=[Length]

for the integral to be dimensionless, the units of the square modulus of the wavefunction has to be:

[\psi]^2 = [Length^{-3}]

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