For a very rough pipe wall the friction factor is constant at high Reynolds numbers. For a length L1 the pressure drop over the length is p1. If the length of the pipe is then doubled, what is the relation of the new pressure drop p2 to the original pressure drop p1 at the original mass flow rate?

Answers

Answer 1

Answer:

Answer: ∆p2 = 2* ∆p1

Explanation:

Given that all other factors remain constant. The pressure drop across the pipeline is directly proportional to the length.

i.e ∆p ~ L

Therefore,

∆p2/L2 = ∆p1/L1

Since L2 = 2 * L1

∆p2/2*L1 = ∆p1/L1

Eliminating L1 we have,

∆p2/2 = ∆p1

Multiplying both sides by 2

∆p2 = 2 * ∆p1

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Give some reasons for our knowledge of the solar system has increased considerably in the past few years. Support your response with at least 3 reasons with details regarding concepts from the units learned in this course.

Answers

Answer:

Improvement in observational, and exploratory technology

Rapid increase in knowledge

International collaboration

Explanation:

Our knowledge of the solar system has increased greatly in the past few years due to to some factors which are listed below.

Improvement in observational, and exploratory technology: In recent years, developments in technology has led to the invention of advanced observational instruments and probes, that are used to study the solar system. Also more exploratory units are now developed to go out into the solar system and gather useful data which is then further processed to yield more results about our solar system.

Rapid increase in knowledge: The past few years has seen an increased number of theories proposed to explain phenomena in the solar system. Some of these theories have been seen to be accurate under experimentation, leading to newer and fresher insights into our solar system. Also, new experiments and research are carried out, all these leading to an exponential growth in our knowledge of the solar system.

International Collaboration: The sharing of knowledge by scientists all over has led to a better, quick understanding of the solar system. Also, scientists from different countries, working together on different experiment and data sharing regarding our solar system now allows our knowledge of the solar system to deepen faster.

A uniform, 4.5 kg, square, solid wooden gate 2.0 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.2 kg raven flying horizontally at 4.5 m/s flies into this door at its center and bounces back at 1.5 m/s in the opposite direction. What is the angular speed of the gate just after it is struck by the unfortunate raven?

Answers

Answer:

Explanation:

Mass of the gate, $m_1 = 4.5 kg$

Mass of the raven, $m_2 = 1.2 kg$

Initial speed of raven, $v_1 = 4.5 m/s$

Final speed of raven, $v_2 = - 1.5 m/s$

Moment of Inertia of the gate about the axis passing through one end:

$I = (1)/(3) m_1 a^2\nI = (1)/(3) *4.5 * 2^2\nI = 6 kg m^2$

Angular momentum of the gate, $L = I \omega$

$L = 5.33 \omega$

Using the law of conservation of angular momentum:

$m_2 v_f (a/2) + I\omega = m_2v_i (a/2)\nI\omega = m_2 (a/2)(v_i - v_f)\n$

The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the object's charge q? ( k = 1/4πε 0 = 8.99 × 10 9 N ∙ m 2/C 2)

Answers

Answer:

Charge, $Q=1.56* 10^(-8)\ C$

Explanation:

It is given that,

Electric field strength, E = 180000 N/C

Distance from a small object, r = 2.8 cm = 0.028 m

Electric field at a point is given by :

$E=(kQ)/(r^2)$

Q is the charge on an object

$Q=(Er^2)/(k)$

$Q=(180000\ N/C* (0.028\ m)^2)/(9* 10^9\ Nm^2/C^2)$

$Q=1.56* 10^(-8)\ C$

So, the charge on the object is $1.56* 10^(-8)\ C$ . Hence, this is the required solution.

"For the lowest harmonic of pipe with two open ends, how much of a wavelength fits into the pipe’s length?"

Answers

Answer:

0.5 lambda(wavelength)

Explanation:

We know that

The first harmonic for both side open ended pipe is

L= 1/2lambda

So L = 0.5*wavelength

Two point charges are placed on the x axis.The firstcharge, q1= 8.00 nC, is placed a distance 16.0 mfromthe origin along the positive x axis; the second charge,q2= 6.00 nC, is placed a distance 9.00 mfrom the originalong the negative x axis.[Give the x and y components of the electric fieldas an ordered pair. Express your answer innewtons per coulomb to three significant figures.Keep in mind that an x component that points tothe right is positive and a y component thatpoints upward is positive.]

Answers

Answer:

E = (0, 0.299) N

Explanation:

Given,

Charge $q_1\ =\ 8.0\ nC$
Charge $q_2\ =\ 6.0\ nC$
Distance of the first charge from the origin = (16m, 0)
Distance of the second charge from the origin = (-9, 0)
Point where the electric field required = (0, 12m)

Let $\theta_1\ and\ theta_2$ be the angle of the electric fields by first and second charge at the point A.

$\therefore sin\theta_1\ =\ (12)/(20)\n\Rightarrow \theta_1\ =\ sin^(-1)\left ((12)/(20)\ \right )\n\Rightarrow \theta_1\ =\ 36.87^o\n\n\therefore sin\theta_1\ =\ (12)/(9)\n\Rightarrow \theta_1\ =\ sin^(-1)\left ((12)/(9)\ \right )\n\Rightarrow \theta_1\ =\ 53.13^o\n$

Electric field by charge $q_1$ at point A,

$F_1\ =\ (kq_1)/(r_1^2)\n\Rightarrow F_1\ =\ (9* 10^9* 8* 10^(-9))/(20^2)\n\Rightarrow F_1\ =\ 0.18\ N/C$

Electric field by the charge $q_2$ at point A,

$F_1\ =\ (kq_1)/(r_1^2)\n\Rightarrow F_1\ =\ (9* 10^9* 6.0* 10^(-9))/(16^2)\n\Rightarrow F_1\ =\ 0.24\ N/C$

Now,

Net electric field in horizontal direction at point A $F_x\ =\ F_(1x)\ +\ F_(2x)\n\Rightarrow F_x\ =\ F_1cos\theta_1\ +\ F_2cos\theta_2\n\Rightarrow F_x\ =\ 0.18*( -cos36.87^o)\ +\ 0.24* cos53.13^o\n\Rightarrow F_x\ =\ -0.144\ +\ 0.144\ N/C\n\Rightarrow F_x\ =\ 0\ N/C$

Net electric field in vertical direction at point A.

$F_y\ =\ F_(1y)\ +\ F_(2y)\n\Rightarrow F_y\ =\ F_1sin\theta_1\ +\ F_2sin\theta_2\n\Rightarrow F_y\ =\ 0.18* sin36.87^o\ +\ 0.24* sin53.13^o\n\Rightarrow F_y\ =\ 0.180\ +\ 0.192\n\Rightarrow F_y\ =\ 0.299\ N/C$