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Leviafan [203]
2 years ago
6

Is 1 milliliter/second a small or large flow? Why?

Physics
2 answers:
ozzi2 years ago
7 0

Answer:

1 ml/second is a small flow

Explanation:

NeTakaya2 years ago
7 0

It dependependsdependsdepends


Are you a blue whale or a brine shrimp ?

You might be interested in
Why is pseudoscience bad?
USPshnik [31]

Answer:

It is quite difficult to picture a pseudoscientist—really picture him or her over the course of a day, a year, or a whole career. What kind or research does he or she actually do, what differentiates him or her from a carpenter, or a historian, or a working scientist? In short, what do such people think they are up to?

… it is a significant point for reflection that all individuals who have been called “pseudoscientists” have considered themselves to be “scientists”, with no prefix.

The answer might surprise you. When they find time after the obligation of supporting themselves, they read papers in specific areas, propose theories, gather data, write articles, and, maybe, publish them. What they imagine they are doing is, in a word, “science”. They might be wrong about that—many of us hold incorrect judgments about the true nature of our activities—but surely it is a significant point for reflection that all individuals who have been called “pseudoscientists” have considered themselves to be “scientists”, with no prefix.

What is pseudoscience?

“Pseudoscience” is a bad category for analysis. It exists entirely as a negative attribution that scientists and non‐scientists hurl at others but never apply to themselves. Not only do they apply the term exclusively as a discrediting slur, they do so inconsistently. Over the past two‐and‐a‐quarter centuries since the term popped into the Western European languages, a great number of disparate doctrines have been categorized as sharing a core quality—pseudoscientificity, if you will—when in fact they do not. It is based on this diversity that I refer to such beliefs and theories as “fringe” rather than as “pseudo”: Their defining characteristic is the distance from the center of the mainstream scientific consensus in whichever direction, not some essential property they share.

Scholars have by and large tended to ignore fringe science as regrettable sideshows to the main narrative of the history of science, but there is a good deal to be learned by applying the same tools of analysis that have been used to understand mainstream science. This is not, I stress, to imply that there is no difference between hollow‐Earth theories and geophysics; on the contrary, the differences are the point of the analysis. Focusing on the historical and conceptual relationship between the fringe and the core of the various sciences as that blurry border has fluctuated over the centuries provides powerful analytical leverage for understanding where contemporary anti‐science movements come from and how mainstream scientists might address them.

As soon as professionalization blossomed, tagging competing theories as pseudoscientific became an important tool for scientists to define what they understood science to be

The central claim of this essay is that the concept of “pseudoscience” was called into being as the shadow of professional science. Before science became a profession—with formalized training, credentialing, publishing venues, careers—the category of pseudoscience did not exist. As soon as professionalization blossomed, tagging competing theories as pseudoscientific became an important tool for scientists to define what they understood science to be. In fact, despite many decades of strenuous effort by philosophers and historians, a precise definition of “science” remains elusive. It should be noted however that the absence of such definitional clarity has not seriously inhibited the ability of scientists to deepen our understanding of nature tremendously.

Explanation:

8 0
3 years ago
A solid sphere, a solid disk, and a thin hoop are all released from rest at the top of the incline (h0 = 20.0 cm).
Ede4ka [16]

Answer:

a. The object with the smallest rotational inertia, the thin hoop

b. The object with the smallest rotational inertia, the thin hoop

c.  The rotational speed of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

Explanation:

a. Without doing any calculations, decide which object would be spinning the fastest when it gets to the bottom. Explain.

Since the thin has the smallest rotational inertia. This is because, since kinetic energy of a rotating object K = 1/2Iω² where I = rotational inertia and ω = angular speed.

ω = √2K/I

ω ∝ 1/√I

since their kinetic energy is the same, so, the thin hoop which has the smallest rotational inertia spins fastest at the bottom.

b. Again, without doing any calculations, decide which object would get to the bottom first.

Since the acceleration of a rolling object a = gsinФ/(1 + I/MR²), and all three objects have the same kinetic energy, the object with the smallest rotational inertia has the largest acceleration.

This is because a ∝ 1/(1 + I/MR²) and the object with the smallest rotational inertia  has the smallest ratio for I/MR² and conversely small 1 + I/MR² and thus largest acceleration.

So, the object with the smallest rotational inertia gets to the bottom first.

c. Assuming all objects are rolling without slipping, have a mass of 2.00 kg and a radius of 3.00 cm, find the rotational and translational speed at the bottom of the incline of any one of these three objects.

We know the kinetic energy of a rolling object K = 1/2Iω²  + 1/2mv² where I = rotational inertia and ω = angular speed, m = mass and v = velocity of center of mass = rω where r = radius of object

The kinetic energy K = potential energy lost = mgh where h = 20.0 cm = 0.20 m and g = acceleration due to gravity = 9.8 m/s²

So, mgh =  1/2Iω²  + 1/2mv² =  1/2Iω²  + 1/2mr²ω²

Let I = moment of inertia of sphere = 2mr²/5 where r = radius of sphere = 3.00 cm = 0.03 m and m = mass of sphere = 2.00 kg

So, mgh = 1/2Iω²  + 1/2mr²ω²

mgh = 1/2(2mr²/5 )ω²  + 1/2mr²ω²

mgh = mr²ω²/5  + 1/2mr²ω²

mgh = 7mr²ω²/10

gh = 7r²ω²/10

ω² = 10gh/7r²

ω = √(10gh/7) ÷ r

substituting the values of the variables, we have

ω = √(10 × 9.8 m/s² × 0.20 m/7) ÷ 0.03 m

= 1.673 m/s ÷ 0.03 m

= 55.77 rad/s

≅ 55.8 rad/s

So, its rotational speed is 55.8 rad/s

Its translational speed v = rω

= 0.03 m × 55.8 rad/s

= 1.67 m/s

So, its rotational speed is of the sphere is 55.8 rad/s and Its translational speed is 1.67 m/s

6 0
2 years ago
What happens to the magnetic domains in a material when the material is placed in a strong magnetic field?
Allisa [31]
If it is diamagnetic then it magnetise opposite to magnetic field
if paramagnetic it weekly magnetise in direction of magnetic field
if ferromagnetic it strongly magnetise in direction of magnetic field
6 0
3 years ago
If a sound travels 343m/s through air and has a frequency of 800Hz, what would the wavelength be of the same sound traveling thr
VARVARA [1.3K]

Answer:

0.625m

Explanation:

Now Velocity of a wave ,V = frequency × wavelength

Wavelength =velocity /frequency

=500/800 =0.625m

7 0
3 years ago
A solid concrete block weighs 150. N and is resting on the ground. Its dimensions are 0.400 m ✕ 0.200 m ✕ 0.100 m. A number of i
N76 [4]

Answer:

27 blocks

Explanation:

First, the expression to use here is the following:

P = F/A

Where:

P: pressure

F: Force exerted

A: Area of the block.

Now , we need to know the number of blocks needed to exert a pressure that equals at least 2 atm. To know this, we should rewrite the equation. We know that certain number of blocks, with the same weight and dimensions are putting one after one over the first block, so we can say that:

P = W/A

P = n * W1 / A

n would be the number of blocks, and W1 the weight of the block.We have all the data, and we need to calculate the area of the block which is:

A = 0.2 * 0.1 = 0.02 m²

Solving now for n:

n = P * A / W1

The pressure has to be expressed in N/m²

P = 2 atm * 1.01x10^5 N/m² atm = 2.02x10^5 N/m²

Finally, replacing all data we have:

n = 2.02x10^5 * 0.02 / 150

n = 26.93

We can round this result to 27. So the minimum number of blocks is 27.

5 0
3 years ago
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