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Harman [31]
3 years ago
14

How many balls are there in a ski ball machine?

Physics
1 answer:
gtnhenbr [62]3 years ago
8 0

Answer:

i think 2 per each sac

Explanation:

i dont know for sure...

Hope this helps

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Freezing Point Depression: Can someone explain this formula to me? ΔTf = Kfcm
Leya [2.2K]
If the solution is treated as an ideal solution, the extent of freezing point depression depends only on the solute concentration that can be estimated by a simple linear relationship with the cryoscopic constant: ΔTF = KF · m · i ΔTF, the freezing point depression, is defined as TF (pure solvent) - TF (solution). KF, the cryoscopic constant, which is dependent on the properties of the solvent, not the solute. Note: When conducting experiments, a higher KF value makes it easier to observe larger drops in the freezing point. For water, KF = 1.853 K·kg/mol.[1] m is the molality (mol solute per kg of solvent) i is the van 't Hoff factor (number of solute particles per mol, e.g. i = 2 for NaCl).
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3 years ago
Who was the second man to walk on the moon?
klemol [59]

Answer:

Buzz Aldrin

Explanation:

6 0
3 years ago
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A milliliter of very hot water is added to a liter of very cold water. Which of these events will occur? Assume the surrounding
prisoha [69]

Answer:b

Explanation:the water will not be hot nomore because of the cold water

5 0
3 years ago
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A mass is oscillating with amplitude A at the end of a spring.
Dmitry_Shevchenko [17]

A) x=\pm \frac{A}{2\sqrt{2}}

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

E=\frac{1}{2}kA^2 (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2 (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

U=\frac{1}{3}K

Using (2) we can rewrite this as

U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}

And using (1), we find

U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2

Substituting U=\frac{1}{2}kx^2 into the last equation, we find the value of x:

\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}

B) x=\pm \frac{3}{\sqrt{10}}A

In this case, the kinetic energy is 1/10 of the total energy:

K=\frac{1}{10}E

Since we have

K=E-U

we can write

E-U=\frac{1}{10}E\\U=\frac{9}{10}E

And so we find:

\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A

3 0
3 years ago
How did the magnet’s density measurement using the Archimedes’ Principle compare to the density measurement using the calculated
guapka [62]

Answer:

The two methods will yield different results as one is subject to experimental errors that us the Archimedes method of measurement, the the density measurement method will be more accurate

Explanation:

This is because the density method using the calculated volume will huve room for less errors that's occur in practical method i.e Archimedes method due to human error

5 0
3 years ago
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