When we convert 24 gallon/min to cubic centimeters per second (cm³/s), the result obtained is 1514.164 cm³/s
<h3>How to convert 24 gallon/min to cm³/min</h3>
We'll begin by converting 24 gallon/min to cm³/min. This can be obtained as follow:
1 gallon/min = 3785.41 cm³/min
Therefore,
24 gallon/min = (24 gallon/min × 3785.41 cm³/min) / 1 gallon/min
24 gallon/min = 90849.84 cm³/min
<h3>How to convert 90849.84 cm³/min to cm³/s</h3>
We can convert 90849.84 cm³/min to cm³/s as follow:
60 cm³/min = 1 cm³/s
Therefore,
90849.84 cm³/min = (90849.84 cm³/min × 1 cm³/s) / 60 cm³/min
90849.84 cm³/min = 1514.164 cm³/s
Thus,
24 gallon/min = 1514.164 cm³/s
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Answer:
The answer to your question is given below
Explanation:
From the question given above, we can see that the wave with a higher frequency has a shorter wavelength while that with a lower frequency has a longer wavelength. This is so because the frequency and wavelength of a wave has inverse relationship. This can further be explained by using the following formula:
Velocity = wavelength x frequency
Divide both side by wavelength
Frequency = Velocity /wavelength
Keeping the velocity constant, we have:
Frequency ∝ 1 / wavelength
From the above illustration, we can see clearly that the frequency and wavelength are in inverse relationship. This implies that the higher the frequency, the shorter the wavelength and the shorter the frequency, the higher the wavelength.
Answer:
200 nm is the thinnest film that produces a strong reflection for green light with a wavelength of 500 nm
Explanation:
If two reflected waves interfere constructively ,strong reflection is produced. Two reflected waves will experience a phase change
For constructive interference

for thinnest film m=1
refractive index should be taken for film n=1.25
thickness of the thinnest film is

Answer:
W = 1884J
Explanation:
This question is incomplete. The original question was:
<em>Consider a motor that exerts a constant torque of 25.0 N.m to a horizontal platform whose moment of inertia is 50.0kg.m^2 . Assume that the platform is initially at rest and the torque is applied for 12.0rotations . Neglect friction.
</em>
<em>
How much work W does the motor do on the platform during this process? Enter your answer in joules to four significant figures.</em>
The amount of work done by the motor is given by:


Where I = 50kg.m^2 and ωo = rad/s. We need to calculate ωf.
By using kinematics:

But we don't have the acceleration yet. So, we have to calculate it by making a sum of torque:

=> 
Now we can calculate the final velocity:

Finally, we calculate the total work:

Since the question asked to "<em>Enter your answer in joules to four significant figures.</em>":
W = 1884J