1. A fluid with viscosity 1.2 Pa s moves through a 200-m-long pipe with a 20 cm diameter. What difference in pressure between the ends of the pipe is needed for a flow of 0.1 m^3/s?

1. 611 kPa

2. 380 kPa

3. 38 mPa

4. 0.61 Pa

2. A torpedo with characteristic length L=1.2 m is traveling through water (density 1000 kg/m^3, viscosity 1.0 mPa s) at speed 10 m/s. What is Reynold's number for this torpedo?

1. 1.2x10^4

2. 1.2

3. 1.2x10^7

4. 1.2x10^-3

3. A torpedo with characteristic length L=1.2 m is traveling through water (density 1000 kg/m^3, viscosity 1.0 mPa s) at speed 10 m/s. Is the flow of water around the torpedo laminar?

1. It is unstable, so it could be either laminar or turbulent.

2. It is neither because water isn't real.

3. No, it is turbulent.

4. Yes, it is laminar.

Answer:

True im pretty sure

Explanation:

Answer:

False

Explanation:

The answer is false, I tested it on IXL

The true statement about the electric potential for the equipotential surface is  \(V_A = V_B = V_C\)

A surface with an equipotential potential is one where all points on the surface have the same electric potential. .

That is an equipotential surface is that surface at every point of which, the electric potential is the same.

The formula for the potential across every point on the surface is given as;

V = F/Q x R

V = ER

where;

Since the surface is equipotential with uniform electric across the surface, the electric potential at any point across the surface will be the same.

So \(V_A = V_B = V_C\)

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Answer:

- They need oxygen to burn fuel

- Aerodynamics

- Extreme temperatures

- Radiation

- Pressure issues

Explanation:

A airplane is a heavier-than-air aircraft kept aloft by the upward thrust exerted by the passing air on its fixed wings and driven by propellers, jet propulsion, etc.

Aeroplanes cannot travel in space for several reasons:

They need oxygen to burn fuel - Aeroplane engines rely on the oxygen in the atmosphere to burn fuel and generate thrust. In space, there is no atmosphere so there is no oxygen for the engines to work.

Aerodynamics - Aeroplane wings generate lift by interacting with the air. In space, there is no air so wings would be unable to generate any lift. Aeroplanes rely on aerodynamics to fly which does not work in space.

Extreme temperatures - In space, temperatures can range from -150 degrees Celsius to 150 degrees Celsius. Aeroplanes are designed to operate within a much narrower temperature range. The extreme cold and heat of space could damage aeroplane components.

Radiation - In space, there are high levels of radiation from the Sun and cosmic rays. Aeroplane bodies are not designed to shield against this type of radiation and it could damage electronics and affect aeroplane systems.

Pressure issues - Aeroplanes are designed to withstand air pressures at altitudes up to around 12 kilometers. In low-Earth orbit and beyond, the air pressure is essentially zero. This extreme change in pressure could cause structural damage to the aeroplane.

In summary, while aeroplanes are designed to fly through the Earth's atmosphere, they lack the key features needed to operate in the extreme environment of outer space like spaceships. Aeroplanes require things like oxygen, aerodynamics and being able to withstand changes in pressure - all of which do not exist or work the same way in space.

Explanation:

The wing is pushed up by the air under it. Large planes can only fly as high as about 7.5 miles. The air is too thin above that height. It would not hold the plane up.

If one ball is dropped at rest from a height of h = 65 m.  The acceleration of the two balls is: -9.8 m/s^2 for both balls.

Let's start by finding the time it takes for the two balls to meet. We know that the ball thrown upward starts from rest, so its initial velocity is 0 m/s. We can use the following kinematic equation:

y = v_i*t + (1/2)at^2

where y is the displacement (in this case, it is the distance between the two balls), v_i is the initial velocity, a is the acceleration, and t is the time. We can set y equal to the initial height of the dropped ball, which is h = 65 m. For the ball thrown upward, the initial position is y = 0.

For the dropped ball:

y = h = 65 m

v_i = 0 m/s

For the ball thrown upward:

y = 0

v_i = 0 m/s

Using the given information, we can solve for t:

h = (1/2)at^2

65 m = (1/2)*(-9.8 m/s^2)t^2

t = sqrt(65 m / (1/2(-9.8 m/s^2))) ≈ 3.64 s

So it takes about 3.64 seconds for the two balls to meet.

Now, we can find the acceleration of the two balls. For the dropped ball, the acceleration is simply the acceleration due to gravity, which is -9.8 m/s^2. For the ball thrown upward, the acceleration is also the acceleration due to gravity, but with a negative sign since it is moving in the opposite direction of gravity. Therefore, the acceleration of the two balls is:

a = -9.8 m/s^2 for both balls

This means that both balls experience the same acceleration due to gravity, regardless of their initial velocities.

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Answer:

The thickness is  \(t = 0.5615 \ cm\)    

Explanation:

From the question we are told that

    The wavelength of the of the rader waves is  \(\lambda = 2.92 \ cm\)

     The index of refraction of the polymer is  \(n = 1.30\)

The thickness is mathematically represented as

             \(t = \frac{\lambda }{4 n }\)

Substituting values

            \(t = \frac{2.92}{4 * 1.30 }\)

          \(t = 0.5615 \ cm\)

Answer:

108 N

Explanation:

Divide the force by uk to get the normal force

(25.7)/(0.238) = 107.98 N = 108 N

Answer:

107

Explanation:

The equations for rotary motion is based on motion in a circular path

a) Time taken by clothes to come up to speed during washing is approximately 1.325 s

b) The tub turns approximately 24.016 revolutions in the 20.0 s interval

Reasons:

Known parameter are;

Operating speed of an automatic washing machine = 5.5 rad·s⁻¹

Average angular acceleration of the washing machine, α = 4.15 rad·s⁻²

Angular speed in the spin-dry mode, ω = 6.0 revolution per second

Time it takes to reach angular speed in spin-dry mode = 7.0 s

a) The relationship between angular acceleration and time are;

\(\alpha = \dfrac{\Delta \omega}{\Delta t}\), \(\Delta t = \dfrac{\Delta \omega}{\alpha}\)

Where;

Δω = ω₂ - ω₁ = 5.5 rad·s⁻¹

Δt = The time it takes to accelerate

Therefore;

\(\Delta t = \dfrac{5.5 \ rad\cdot s^{-1}}{4.15 \ rad \cdot s^{-2}} \approx 1.325 \ s\)

The time it takes for the clothes to come up to speed during the washing mode, Δt ≈ 1.325 s

b) The angular acceleration of the spin-dry mode is given as follows;

6 revolutions per second = 2·π ×6  rad per second = 12·π rad/s

\(\alpha = \dfrac{\Delta \omega}{\Delta t}\)

The acceleration during spin up

\(\therefore \alpha = \dfrac{12 \cdot \pi }{7.0} \ rad \cdot s^{-2} \approx 5.39 \ rad \cdot s^{-2}\)

The acceleration during slowing down;

\(\therefore \alpha = \dfrac{12 \cdot \pi }{13.0} \ rad \cdot s^{-2} \approx 2.9 \ rad \cdot s^{-2}\)

The angle turned in the 20.0 second interval is therefore;

\(\dfrac{1}{2} \times 5.39 \times 7^2 + 12 \cdot \pi \times 7-\dfrac{1}{2} \times 2.9 \times 13^2 \approx 150.9\)

The angle turned in the 20.0 s is approximately 150.9 radians

\(Number \ of \ revolutions \ =\dfrac{\theta}{2 \cdot \pi}\)

Therefore;

\(Number \ of \ revolutions \ in \ 20.0 \ s =\dfrac{150.9 }{2 \cdot \pi} \approx 24.016\)

The number of revolutions of the tub in the 20.0 s interval is approximately 24.016 revolutions

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Answer:

All molecules of one substance are identical (exactly the same)(2)

(a) The amplitude of the wave is 0.2 m.

(b) The period of the wave is  4 s.

(c) The wavelength of the wave is 100 m.

(a) The amplitude of the wave is the maximum displacement of the wave.

amplitude of the wave = 0.2 m

(b) The period of the wave is the time taken for the wave to make one complete cycle.

period of the wave = 5.5 s - 1.5 s = 4 s

(c) The wavelength of the wave is calculated as follows;

λ = v / f

where;

f = 1/t = 1 / 4s = 0.25 Hz

λ = ( 25 m/s ) / 0.25 Hz

λ = 100 m

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A 120 ft. long, 2 in. diameter steel rod is expected to fail at 80,000 psi. If a safety factor of 2 is required then the largest allowable axial loading of the rod is 1.92 inches.

First, let's determine the rod's cross-sectional area.

Given

D = 2 inches

Area =\(\pi (\frac{D}{2}) ^{2}\)  = \(\pi (\frac{2}{2}) ^{2}\)= 3.14 sq inches.

We know that tensile stress

\(\sigma =\frac{P}{A}\)

P =\(\sigma \times A\)

Now, stress at failure is = 80000 psi

and Load at failure P = 80000 × 3.14 = 251200lbs

Now,

FOS = 2

\(\sigma _{allow } =\frac{ \sigma _{y}}{FOS }\)= \(\frac{80000}{2}\)= 40000 psi

equivalent axial load

P =\(\sigma \times A\)

and stress at failure = 40000 psi

Load at failure P = 40000 × 3.14 = 125600lbs

Elongation

\(\delta =\frac{\rho \times L}{A \times E}\)

Given length = 120 feet = 1440 inches

\(\delta =\frac{125600 \times 1440}{3.14 \times 30 \times 10^6}\)

\(\delta\) = 1.92 inches at the axial loading

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Answer:

a.26

Explanation:

39 divide by 1.5=26

Answer:

iron nail rusting

Explanation:

Answer:

Option C

Explanation:

Revolution: When an object moves around another object it is called revolution.

Rotation: When an object spins around its axis it's called rotation

Answer:C

Explanation:

Good luck!

The principal stresses and corresponding principal directions for the stresses in a problem can be found using various methods, one of which is Mohr's circle.

What is Mohr's circle?

Mohr's circle is a graphical representation of a two-dimensional stress state. Given the stress tensor components, the Mohr's circle can be used to visualize and determine the maximum and minimum normal stresses (i.e., the principal stresses) and the orientation of the planes on which they act (i.e., the principal directions).

It is important to note that finding the principal stresses and directions requires a thorough understanding of stress analysis, including stress transformation and the use of Mohr's circle. If you have access to the full problem statement, I would be happy to help you work through the solution.

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Answer:

0.749 m

Explanation:

1 m = 39.37 inches

Answer:

1st speed 2.velocity 3.speed4.velocity5.velocity6.speed

The average acceleration of the motorcycle over the given distance is approximately 9.39 m/s².

To calculate the average acceleration of the motorcycle, we can use the formula:

Average acceleration = (final velocity - initial velocity) / time

First, let's convert the final velocity from km/h to m/s since the distance is given in meters. We know that 1 km/h is equal to 0.2778 m/s.

Converting the final velocity:

Final velocity = 78 km/h * 0.2778 m/s = 21.67 m/s

Since the motorcycle starts from rest (initial velocity is zero), the formula becomes:

Average acceleration = (21.67 m/s - 0 m/s) / time

To find the time taken to reach this velocity, we need to use the formula for average speed:

Average speed = total distance/time

Rearranging the formula:

time = total distance / average speed

Plugging in the values:

time = 50 m / 21.67 m/s ≈ 2.31 seconds

Now we can calculate the average acceleration:

Average acceleration = (21.67 m/s - 0 m/s) / 2.31 s ≈ 9.39 m/s²

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Answer:

\(a=1.46\ m/s^2\)

Explanation:

Given that,

Mass of a rocket, m = 5100 kg

Thrust force acting on the rocket, F = 7450 N

We need to find the acceleration of the rocket. The net force acting on an object is given by the formula as follows :

F = ma

a is acceleration of the rocket

\(a=\dfrac{F}{m}\\\\a=\dfrac{7450}{5100}\\\\a=1.46\ m/s^2\)

So, the acceleration is \(1.46\ m/s^2\).

Answer:

Force = -50000 Newton

Explanation:

Given the following data;

Initial velocity, u = 30 m/s

Final velocity, v = 20 m/s

Time = 0.2secs

Mass = 1000 kg

First of all, we have to solve for its acceleration.

Acceleration = (v - u)/t

Substituting into the equation, we have

Acceleration = 20 - 30)/0.2

Acceleration = -10/0.2

Acceleration = -50m/s²

To find the force;

Force = mass * acceleration

Force = 1000 * (-50)

Force = -50000 Newton. The value of the force is negative because it stopped the motion of the car.

Therefore, the barrier applied a force of -5,0000 Newtons to stop the car.

Answer: -50000

Explanation:

Answer:

2.405 m/s

Explanation:

Given that,

Mass of a women, m₁ = 60 kg

Mass of a ball, m₂ = 3.9 kg

Velocity of the ball, v₂ = 37 m/s

We need to find the velocity of the woman. It is a concept based on the conservation of linear momentum. Let v₁ is the velocity of the woman. So,

\(m_1v_1=m_2v_2\\\\v_1=\dfrac{m_2v_2}{m_1}\\\\v_1=\dfrac{3.9\times 37}{60}\\\\v_1=2.405\ m/s\)

So, the velocity of the woman is 2.405 m/s.

Answer:

Trawler B should set off on a course of 120°.

Trawler A is traveling at 20km/h on a bearing of 60°. This means that Trawler A is traveling north-east. Trawler B is traveling at 25km/h. In order to intercept Trawler A, Trawler B must travel on a bearing of 120°. This means that Trawler B must travel south-east.

The two trawlers will eventually meet at a point that is 40km east of Trawler A's starting position and 20km north of Trawler B's starting position.

Answer:

Analog to digital converters

Explanation:

An analog-to-digital converter (ADC)  is a device that converts analog signals such as sound into digital signals. Analog information is transmitted by modulating the signal and then amplifying the signal's strength. The conversion from analog to digital involves quantization of the input thereby reducing error or noise. The ADC produces the output data as a series of digital values (0's and 1's) with fixed precision.

Answer:

Analog to digital converter.

Explanation:

Three important type that covert analog signal to digital

1 flash ADC

2 Digital Ramp ADC

3 successive Approximation

Answer:

A. New space missions show that Pluto is much larger than originally thought.

Explanation:

The new definition of a planet that was adopted in 2006, defined planet as  an object that orbits the sun, with sufficient mass to be round, not a satellite of another object, and has removed debris and small objects from the area around its orbit.

This new definition of a planet that was adopted in 2006, classified Pluto as "dwarf planet", because Pluto meets planetary criteria except that it has not cleared debris from its orbital neighborhood.

However, new Horizons spacecraft flew by Pluto in 2015, revealed that Pluto is much larger than originally thought

Therefore, the correct option is "A"

A. New space missions show that Pluto is much larger than originally thought.

Answer: it means that we now technically have over 100 planets

Explanation:

it’s not New space missions show that Pluto is much larger than originally thought!!!

Answer:

As shown in the figure below, a uniform solid sphere rotates about a vertical support on a frictionless bearing. A light cord passes around the equator of the sphere, over a uniform solid disk, and is attached to a hanging mass. The sphere, disk and hanging mass all have equal mass M. The sphere and disk have the same radius R. Find the acceleration of the hanging mass if the string does not slip on the sphere or the disk. Express your answer in terms of M, R and g as needed...

Explanation:

As shown in the figure below, a uniform solid sphere rotates about a vertical support on a frictionless bearing. A light cord passes around the equator of the sphere, over a uniform solid disk, and is attached to a hanging mass. The sphere, disk and hanging mass all have equal mass M. The sphere and disk have the same radius R. Find the acceleration of the hanging mass if the string does not slip on the sphere or the disk. Express your answer in terms of M, R and g as needed...

\( \boxed{ \sf \: better \: luck \: at \: your \: class}\)

A uniform solid sphere revolves around a vertical support on a frictionless bearing.

A solid disk is covered with a light string that circles the sphere's equator, over which is suspended a mass. The mass M of the hanging mass, disk, and sphere is the same.

The radius R of the sphere and disk is the same. If the string does not slip on the sphere or disk, determine the acceleration of the hanging mass. If necessary, explain your response in terms of M, R, and g.

A uniform solid sphere revolves around a vertical support on a frictionless bearing, as seen in the illustration below. Around the sphere's equator, a thin cable travels over a solid surface.

Therefore, A uniform solid sphere revolves around a vertical support on a frictionless bearing.

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We can use the formula for angular deceleration to find the time interval needed for the wheel to come to a complete stop:

α = (ωf - ωo) / Δt

where ωf is the final angular velocity, ωo is the initial angular velocity, and Δt is the time interval. Rearranging the formula, we get:

Δt = (ωf - ωo) / α

Since the wheel comes to a complete stop, the final angular velocity is zero:

ωf = 0

Substituting the given values, we get:

Δt = (0 - 34.28) / (-1.77) ≈ 19.37 s

The formula for angular displacement is:

θ = ωo t + (1/2) α t^2

where θ is the angular displacement, ωo is the initial angular velocity, α is the angular acceleration, and t is the time interval. When the wheel comes to a complete stop, the final angular velocity is zero, so the formula simplifies to:

θ = ωo t + (1/2) α t^2

Substituting the given values, we get:

θ = (34.28 rad/s)(19.37 s) + (1/2)(-1.77 rad/s^2)(19.37 s)^2 ≈ -2003.9 rad

The negative sign indicates that the wheel has rotated in the opposite direction of its initial motion.

To get the total angular distance traveled by the wheel during this time interval, we take the absolute value of θ:

|θ| = |-2003.9 rad| = 2003.9 rad

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Answer:

  B)  100 J

Explanation:

Assuming the distance given is measured along the incline, the vertical change in height is (5 m)(sin 30°) = 2.5 m. Then the change in potential energy is ...

  ∆PE = mg(∆h) = (4 kg)(10 m/s^2)(2.5 m) = 100 J

The force is defined as the product of the mass and the acceleration. The energy stored in an object which is converted into kinetic energy is called potential energy.

The formula for the potential energy is mgh.

The correct answer is B that is 100J.

As the surface is inclined the vertical change in height is\((5 m)(sin 30^o) = 2.5\)m.

Therefore the potential energy after solving it is as follows:-

\(PE = mg(h) \\= (4 kg)(10 m/s^2)(2.5 m)\\\\= 100 J\)

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Answer:

you need to provide a figure that came with the problem.

Answer:

B. Northern Canada

Explanation:

A continental polar air mass can form over the land during the winter months. In the Northern Hemisphere, it originates in northern Canada or Alaska. As it moves southward, it brings dry weather conditions to the United States. Temperature and humidity levels are both low. Hope this helps :)

The missing species of the nuclear reaction obtained is ³⁰₁₅P

The missing species of the equation can be obtain as follow:

Now, we can obtain the value of x, y and Z as follow:

²⁷₁₃Al + ⁴₂He -> ʸₓZ + ¹₀n

For x

13 + 2 = x + 0

15 = x

x = 15

For y

27 + 4 = y + 1

31 = y + 1

Collect like terms

y = 31 - 1

y = 30

For Z

ʸₓZ => ³⁰₁₅Z

From the period table, the element with atomic number of 15 is phosphorus, P. Thus, we have

ʸₓZ => ³⁰₁₅Z => ³⁰₁₅P

Therefore, we can write the complete equation as:

²⁷₁₃Al + ⁴₂He -> ³⁰₁₅P + ¹₀n

Thus, the missing species is ³⁰₁₅P

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1. The force between the two charges is 1.78 × 10⁻⁵ pounds, with opposite signs indicating attraction between the charges.

2. The resistance of 10 gauge aluminum wire over a 40 ft distance would be 0.506 ohms.

3. The cost of electricity from the automobile battery is 38.6 cents per kWh.

4. The current that would be carried through the body is 0.8 mA if dry.

1. The force between two point charges can be calculated using Coulomb's law, which states that the force is proportional to the product of the charges and inversely proportional to the square of the distance between them.

Using the values given, the force can be calculated as F = (k * q1 * q2) / r², where k is Coulomb's constant, q1 and q2 are the charges, and r is the distance between them. Plugging in the values, the force can be calculated as 1.78 × 10⁻⁵ pounds, with opposite signs indicating attraction between the charges.

2. The resistance of a wire is determined by its length, cross-sectional area, and resistivity. The resistivity of aluminum is higher than that of copper, so a larger cross-sectional area is required to achieve the same resistance. Using the gauge size conversion chart, 10 gauge aluminum wire has a cross-sectional area of 5.26 mm², which is approximately 83% of the cross-sectional area of 12 gauge copper wire.

Thus, the resistance of 10 gauge aluminum wire over a 40 ft distance can be calculated as R = (rho * L) / A, where rho is the resistivity of aluminum, L is the length, and A is the cross-sectional area. Plugging in the values, the resistance can be calculated as 0.506 ohms.

3. To calculate the cost of electricity per kWh, the total cost and the total amount of energy supplied must be known. Since the battery supplies 12 V and 51 A for one hour, the total energy supplied can be calculated as E = V * I * t, where V is the voltage, I is the current, and t is the time.

Plugging in the values, the total energy supplied can be calculated as 612 watt-hours (Wh). Since one kWh is equal to 1000 Wh, the total energy supplied can be converted to 0.612 kWh. Dividing the total cost by the total energy supplied gives the cost per kWh, which is 38.6 cents.

4. The current through the body can be calculated using Ohm's law, which states that current is equal to voltage divided by resistance. Using the values given, the resistance can be either 100,000 ohms or 300 ohms depending on whether the skin is dry or wet.

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