air at 1 atmosphere and 20∘ c flows in a 3 centimeter diameter pipe. the maximum velocity of air to keep the flow laminar is

The following measurement means "no limitation". This measurement indicates the range of motion for wrist flexion, which is the movement of the wrist towards the palm of the hand.

What is wrist flexion?

Wrist flexion is the movement of the wrist towards the palm of the hand. It is the opposite movement of wrist extension, which is the movement of the wrist away from the palm of the hand. Wrist flexion is an important motion for many daily activities such as typing, writing, and holding objects.

What does 15-85 degrees of wrist flexion mean?

When measuring wrist flexion, the range of motion is measured in degrees. In this case, the measurement is 15-85 degrees. This means that the normal range of motion for wrist flexion is between 15 and 85 degrees. If the measurement falls within this range, then there is no limitation in wrist flexion. However, if the measurement falls outside of this range, then there may be a limitation in wrist flexion.A limitation in extension refers to a decreased range of motion when moving the wrist away from the palm of the hand. A limitation in flexion refers to a decreased range of motion when moving the wrist towards the palm of the hand. A limitation in both flexion and extension refers to a decreased range of motion in both movements.

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Answer:

F = 20.61 N

Explanation:

Given that,

Mass of the object, m = 2 kg

It is moving with a speed of 10 m/s

It is acted upon by a 5 N of friction.

The force of gravity acts in downward direction which is given by :

W = mg

W = 2 × 10 = 20 N

5 N of force is acting in horizonal surface.

The net force is given by the resultant of two vectors as follows :

\(F=\sqrt{20^2+5^2} \\\\=20.61\ N\)

So, the net force acting on the object is 20.61 N.

The net force acting on the mass is equal to 13.6 Newton.

Given the following data:

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 \(m/s^2\).

To calculate the net force:

First of all, we would determine the normal force acting on the mass.

Mathematically, the normal force is given by the formula:

\(F_N = mg\)

Where:

Substituting the given parameters into the formula, we have;

\(F_N = 2 \times 9.8\)

Normal force = 19.6 Newton

To find net force:

\(Net\;force = 19.6 - 5\)

Net force = 13.6 Newton.

Note: We subtracted because friction opposes the motion of an object and as such acts in opposite direction.

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Answer:

V = 50 volts

Explanation:

Given that,

Resistance, R = 10 ohms

Current, I = 5 A

We need to find the potential difference across the circuit. We know that,

V = IR

Put all the values,

V = 5 × 10

V = 50 volts

Hence, the potential difference is equal to 50 volts.

Answer:3

Explanation:

Answer:

I had this awhile ago if I'm not mistaking its A

Answer:

Given: V = 220V, Pmin = 360W, Pmax = 840W

For minimum heating case:

We know that

Pmin = VI

360 = 220 X I

I = 1.63 amp

R = V/I

R = 220/1.63

R = 134.96ohms

For maximum heating case:

We know that

Pmax = VI

840 = 220 X I

I = 3.81 amp

R = V/I

R = 220/3.81

R = 57.74 ohms

Answer: It would take 5 seconds for an object traveling at 12 m/s to go 60 m. Rounded to the nearest whole number, the answer is 5 seconds.

Explanation:

To find the time it would take an object to travel a certain distance at a given speed, we can use the formula:

time = distance / speed

Plugging in the given values, we get:

time = 60 m / 12 m/s

time = 5 seconds

Advantages of using Fourier transform spectrometers (FTIR) instead of dispersive spectrometers for IR spectroscopy include higher resolution, faster acquisition time, and higher sensitivity.

The three main advantages of using Fourier transform spectrometers (FTIR) over dispersive spectrometers for IR spectroscopy are:

Speed: FTIR spectrometers can acquire a complete spectrum in a few seconds, whereas dispersive spectrometers require several minutes or even hours.

Resolution: FTIR spectrometers can achieve high resolution, allowing for the detection of small differences in spectra that are not possible with dispersive spectrometers.

Sensitivity: FTIR spectrometers are more sensitive than dispersive spectrometers, which is particularly important when analyzing small sample sizes or trace amounts of a sample. FTIR can also detect weak IR signals that are beyond the detection limit of dispersive spectrometers.

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Answer: yes

Explanation: Having a higher metabolism can help you lose weight and keep it off, while also giving you more energy.

Answer:

organisms that lived at a given time.

Explanation:

Answer:

organisms that lived at a given time.

Explanation:

c. Objects below this mass are not hot enough to fuse normal hydrogen.

The lower mass limit of 0.08 solar masses is the minimum mass required for a star to sustain nuclear fusion reactions in its core and enter the main sequence phase of its life cycle. Objects below this mass are not hot enough to fuse normal hydrogen and hence cannot become main sequence stars. They instead become brown dwarfs or sub-brown dwarfs, which do not have enough mass and pressure in their cores to sustain nuclear fusion reactions.

Option B is wrong because objects below the 0.08 solar mass limit can still form in molecular clouds that contain molecular hydrogen (H2) rather than atomic hydrogen (HI). These clouds can collapse and fragment to become brown dwarfs or sub-stellar objects.

Option D is also not the primary reason for the lower mass limit. The formation process of stars and brown dwarfs is influenced by various factors, including the presence of nearby hot stars. However, it is not the main factor determining the lower mass limit of main sequence stars.

The lower mass limit is well understood and explained by the limitations of hydrogen fusion in objects below 0.08 solar masses, so Option A is incorrect.

Option E does not apply since the lower mass limit is determined by physical processes occurring inside stars rather than telescope capabilities.

In summary, the lower mass limit of 0.08 solar masses for main sequence stars exists because objects below this mass are not hot enough to initiate and sustain normal hydrogen fusion in their cores.

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The force exerted on the block by the spring is 89.7 N and it will experience an acceleration of 35.8 m/s².

Mass of the block = 2.5 kg

Spring constant of the spring = 115 N/m

Change in the length of the spring = 78.5 cm = 0.78 m

Friction = 0

The force exerted on the block by the spring is,

\(Spring \: constant = \frac{Force }{ Change \: in \: length \: of \: the \: spring}\)

\(k = \frac{F}{x}\)

\(F = x \times k\)

\(F = 0.78 \times 115\)

\( = 89.7 \: N\)

The force exerted on the block by the spring is 89.7 N.

It will experience an acceleration of,

\(Force= mass \times acceleration\)

\(F = m \times a\)

\(a = \frac{F}{m} \)

\(a = \frac{89.7}{2.5} \)

\( = 35.8 \: m/s ^{2} \)

Thus, it will experience an acceleration of 35.8 m/s².

Therefore, the force exerted on the block by the spring is 89.7 N and it will experience an acceleration of 35.8

m/s².

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Answer:

Equal

Explanation:

The law of conservation of energy states that the total momentum before and after a collision are equal

Answer:

Equal to the total momentum of the two objects after the collision.

The time required to move from the equilibrium position directly to a point a will be 0.879 sec.

The period is the amount of time it takes for a particle to perform one full oscillation. T is the symbol for it. Taking the reciprocal of the frequency yields the frequency of the oscillation.

The standard equation for the SHM is found as;

\(\rm x = asin \omega t \\\\ \rm 12.3 \times 10^{-2} = 18.0 \times 10^{-2}sin \omega t \\\\ 0.6833=sin \omega t \\\\ \omega t =sin^{-1}0.6833 \\\\ \omega t =0.7522 \\\\ t = \frac{0.7522}{0.855 } \\\\ t=0.879\ sec\)

Hence the time required to move from the equilibrium position directly to a point a will be 0.879 sec.

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The sky on Earth is blue because blue light is scattered through the atmosphere as the other colors of light pass through undisturbed. Dark nebulae  have something similar going on.

Nebulae are interstellar dust and gas clouds that get backlighting from stars both inside and behind them. The clouds are illuminated by the photons from these stars that are dispersed throughout them. A nebula does not generate most of the light that appears to come from it; rather, it reflects and refracted light.

A dark nebula, also known as an absorption nebula, is a form of interstellar cloud, particularly a molecular cloud, that is so dense that it absorbs all of the light from objects behind it, including emission or reflection nebulae and background stars, at visible wavelengths.

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Unmanned aerial vehicles (UAVs) is the best aerial reconnaissance method for dealing with extremely large scale sites.

Drones, also known as unmanned aerial vehicles (UAVs), are the most effective approach for aerial reconnaissance of big scale sites.

They have been used effectively for commercial applications such as inspections, surveys, and mapping.

They are also utilized in military reconnaissance and surveillance to gather intelligence information and track hostile locations, structures, or activities.

UAVs have a number of advantages over other conventional aerial reconnaissance techniques.

They are less expensive, safer, and faster to deploy than manned aircraft.

They can be deployed over difficult terrain and areas that are difficult to reach.

They can be remotely controlled and can fly at low altitude to provide high-resolution images and video footage that can be used to analyze the site in detail.

UAVs are equipped with advanced sensors and cameras that enable them to capture high-quality images and videos that can be analyzed in real-time or later to produce detailed maps and 3D models of the site.

This makes them the best approach for aerial reconnaissance of big scale sites.

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The maximum current that can safely be applied to a 1/4 watt 3 ohm resistor is approximately 0.54 amperes.


To obtain this expression, we first need to use Ohm's law, which states that the current flowing through a resistor is directly proportional to the voltage applied across it and inversely proportional to its resistance. Thus, we can write:
I = V/R
where I is the current, V is the voltage, and R is the resistance.
Next, we can use the power law, which states that the power dissipated by a resistor is equal to the product of the voltage and current flowing through it. Thus, we can write:
P = VI
where P is the power.
For a 1/4 watt resistor, the maximum power it can safely dissipate is 1/4 watt. Therefore, we can substitute this into the equation for power and solve for the maximum current as follows:
1/4 = V x I
I = 1/4V
Since the resistance is 3 ohms, we can also substitute this into the equation for current:
I = V/3
Now, we can combine these two equations to obtain an expression for the maximum current:
1/4V = V/3
V = sqrt(3/4)
I = V/3 = sqrt(3/4)/3 ≈ 0.54 A
Therefore, the maximum current that can safely be applied to a 1/4 watt 3 ohm resistor is approximately 0.54 amperes.

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Answer:

  96 m

Explanation:

The average velocity over those 4 seconds is the velocity after 2 seconds:

  30 m/s -(2 s)(3.0 m/s^2) = 24 m/s

The displacement is the product of this average velocity and the duration of the acceleration:

  d = (24 m/s)(4 s) = 96 m

The boat covered 96 m while being accelerated.

Answer:

The gravity helps the planets stay together near the Sun, without it Earth would be floating away with the planet eventually becoming frozen up, thus the role gravity have in the motion of planets around the sun is by keeping them together.

Explanation:

Hope this helps!

The weather that is least likely to occur when a cold air mass moves in and overtakes a warm front in the presence of a low-pressure system is clear skies ( D ).

When a cold air mass moves in and overtakes a warm front in the presence of a low-pressure system, it is called as occluded front. During this the colder air is pushed beneath the warmer air. Depending on the direction of occlusion, it can be of two types of of occluded fronts such as

During occluded fronts the possible weather conditions are

Therefore, the weather that is least likely to occur with these conditions is clear skies ( D ).

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Answer:

10 m/s

Explanation:

a minimum length of 133.33 meters of runway is needed for takeoff if a speed of 40 m/s is required.

using the formula:

v² = u² + 2as

where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.

To find the minimum length of runway, we need to calculate the distance traveled, which can be obtained by rearranging the above formula as:

s = (v² - u²) / (2a)

Here, u = 0 m/s (since the plane starts from rest), v = 40 m/s (since a speed of 40 m/s is required for takeoff), and we can assume a constant acceleration of 3 m/s² (which is a typical value for an airplane's acceleration during takeoff). Therefore, the minimum length of runway required is:

s = (40² - 0²) / (2 × 3)

= 800 / 6= 133.33 meters (rounded to two decimal places)

Therefore, a minimum length of 133.33 meters of runway is needed for takeoff if a speed of 40 m/s is required.

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Therefore, the radius of the loop is approximately 1458.7 meters.

To find the radius of the loop, we can start by considering the forces acting on the pilot at the top of the loop. At this point, the pilot feels three times their normal weight, which means the net force acting on them is greater than just their weight.
The force at the top of the loop can be calculated using the formula:
Net force = mass × acceleration.
Since the pilot is at the top of the loop, the net force is the difference between the weight (mg) and the centripetal force (mv² / r). Here, "m" is the mass of the pilot, "g" is the acceleration due to gravity, "v" is the speed of the plane, and "r" is the radius of the loop.
The weight of the pilot is given by the formula:
Weight = mass × gravity.
Now, since the pilot feels three times their normal weight, we can write:
Net force = 3 × (mass × gravity) - (mass × velocity² / radius).
Substituting the given values, we have:
3 × (mass × gravity) - (mass × velocity² / radius) = mass × acceleration.
Simplifying the equation, we get:
3 × gravity - (velocity² / radius) = acceleration.
To find the acceleration, we can use the centripetal acceleration formula:
Acceleration = velocity² / radius.
Substituting this into the previous equation, we have:
3 × gravity - (velocity² / radius) = velocity² / radius.
Rearranging the equation, we get:
2 × gravity = (velocity²2 / radius).
Finally, solving for the radius, we have:
radius = velocity² / (2 × gravity).
Plugging in the given values, we get:
radius = (170 m/s)² / (2 × 9.8 m/s²).
Calculating this, we find:
radius = 1458.7 m.
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The height of the radio antenna is approximately 36.61 meters.

To solve this problem, we can use trigonometry and the properties of right triangles. Let's draw a diagram to visualize the situation.
We have a right triangle with the radio antenna as the hypotenuse, the guy wire as one leg, and the ground as the other leg. The angle between the guy wire and the ground is given as 14 degrees.
Using trigonometric functions, we can find the length of the guy wire and then subtract it from the total height of the tower to get the height of the radio antenna.
First, let's find the length of the guy wire. We know that the opposite side (the guy wire) is the side opposite the given angle, and the adjacent side (the ground) is the side adjacent to the given angle. Therefore, we can use the tangent function:
tan(14) = opposite/adjacent
tan(14) = guy wire/48
guy wire = 48 tan(14) ≈ 12.51 m
Next, let's find the height of the radio antenna. We know that the hypotenuse (the tower) is the longest side of the right triangle, so we can use the Pythagorean theorem:
tower^2 = guy wire^2 + ground^2
tower^2 = (12.51)^2 + (48)^2
tower ≈ 49.12 m
Finally, we can subtract the length of the guy wire from the height of the tower to get the height of the radio antenna:
radio antenna = tower - guy wire
radio antenna ≈ 36.61 m
Therefore, the height of the radio antenna is approximately 36.61 meters.

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The integral of 1/(1 + x²) is (1/2)ln|1 + x²| + C where C is the constant of integration.

Integration is a mathematical process of finding the antiderivative of a function. To integrate the given expression 1/(1 + x²), we will use the substitution method.

Let u = 1 + x², du/dx = 2x dx, then dx = du/2x and the integral becomes:

∫1/(1 + x²) dx = ∫1/u * (1/2x) du= (1/2)∫1/u du

The antiderivative of 1/u is ln|u| + C, where C is the constant of integration.

Therefore, the final solution of the integral is (1/2)ln|1 + x²| + C.

Let us work through the steps:

Step 1:Let u = 1 + x² and then differentiate both sides with respect to x to obtain du/dx. du/dx = 2x

Substitute 2x dx = du into the integral ∫1/(1 + x²) dx to get the integral in terms of u:∫1/u * (1/2x) du = (1/2) ∫1/u du

Step 2:Calculate the antiderivative of 1/u, which is ln|u|. Thus, the final solution is (1/2)ln|1 + x²| + C, where C is the constant of integration. The constant C will vary depending on the initial conditions of the problem.

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Answer:

Roundworms:

The size of roundworms ranges from microscopic to as long as a meter long. Most roundworms have a hard, sharp lance on their head, and some have bristles on the head. Most of the roundworms are protected by an adaptable yet solid, furrowed body covering.  

Roundworms are very primitive living beings, having no respiratory or circulatory system of a host. They have an empty body with two openings at the closures. Be that as it may, the head isn't particular or recognizable to the unaided eye.  

Roundworms reproduce sexually and can finish their life cycle in immature puppies, however as the little pupps immunity system develops (for the most part by a half-year-old enough), the larval phases of the roundworm will become captured and will encyst (become encased in a sore) in the puppy's muscles. They can stay encysted in the pooch's tissues for a considerable length of time or years.

Answer:

The Positions are, Pitcher, Catcher, First base, second base, third base, shortstop, left field, Center field, and right field. Some pitches that they throw are, 4-seam fastball, Curveball, 2-seam fastball, Slider, Knuckleball, 12-6 Curve, Knucklecurve, Slurve, Changeup, Splitter, Cutter, Split-finger fastball, Sinker, Palm ball, Eephus pitch, Screwball, Cut Fastball, Circle Changup, and Forkball. I think thats all the piches i know.

Answer:

67.9 kg*m/s

Explanation:

Pi = 38 kgm/s

F = 88.3N and ∆t = 0.338s

Final momentum Pf = Pi + F∆t = 38 + (88.3)(0.338) = 38 + 29.8454

=) Pf = 67.8454 kgm/s = 67.85kg*m/s

Your answer is 67.9kg*m/s with three significant figures

hope this helps your troubles!

Answer:

Obviously, the 5lb ball when thrown will go farther . Newton second law of motion F = mass x acceleration .

So the more the mass , the more force will be used to throw the ball.

Explanation:

Answer:

Speaking hypothitically, since I dont have a lot of knowledge around this subject, but yes i think it's possible, and probably not as rare as one might think.

Explanation: