In order to keep a leaking ship from sinking, it is necessary to pump 12.0 lb of water each second from below deck up a height of 2.00 m and over the side. What is the minimum horse-
power motor that can be used to save the ship?

Answers

Answer 1

Answer:

P = 0.14 hp

Explanation:

The power required by the ship is given as:

[tex]P = \frac{Work}{Time} = \frac{Potential\ Eenrgy}{t}\\\\P = \frac{mgh}{t}[/tex]

where,

P = Power = ?

m = mass to pump = (12 lb)(1 kg/2.20 lb) = 5.44 kg

g = acceleration due to gravity = 9.81 m/s²

h = height = 2 m

t = time = 1 s

Therefore,

[tex]P = \frac{(5.44\ kg)(9.81\ m/s^2)(2\ m)}{1\ s}\\\\P = 106.8\ W[/tex]

Converting to horsepower (hp):

[tex]P = (106.8\ W)(\frac{1\ hp}{746\ W})[/tex]

P = 0.14 hp


Related Questions

Q 26.12: Assume current flows in a cylindrical conductor in such a way that the current density increases linearly with radius, from zero at the center to 1.0 A/m2 at the surface of the conductor. If the conductor has a cross sectional area of 1.0 m2, what can you say about the current in this conductor

Answers

Answer:

The current is 0.67 A.

Explanation:

Density, J = 1 A/m^2

Area, A = 1 m^2

Let the radius is r. And outer is R.

Use the formula of current density

[tex]I = \int J dA = \int J 2\pi r dr\\\\I = \int_{0}^{R}\frac{2\pi r^2}{R} dr\\\\I = \frac{2 \pi R^2}{3}.... (1)Now A = \pi R^2\\\\1 =\pi R^2\\\\R^2 = \frac{1}{\pi}\\\\So, \\\\I = \frac{2\pi}{3}\times \frac{1}{\pi}\\\\I = 0.67 A[/tex]

Soap bubble coloring example:
(reflection, interference, refraction, diffraction)​

Answers

Explanation:

Interference is the example of soap bubble colouring

EXTRA INFO:

(LOOK AT THE IMAGE)

An incoming light ray is partly reflected by the top surface of the soap film and partly reflected by the bottom surface. The wave reflected from the bottom surface has traveled further (an extra distance equal to twice the thickness of the film) so emerges out of step with the top wave. When the two waves meet, they add together, and some colors are removed by destructive interference. Where the film is thickest, the bubble appears more blueish; where it's thinner, it will look more violet or magenta.

[tex]\huge\bold\color{black}{ANSWER}[/tex]

Soap bubble coloring example: INTERFERENCE

A 2.0 kg wood block is launched up a wooden ramp that is inclined at a 30˚ angle. The block’s initial speed is 10 m/s. What vertical height does the block reach above its starting point? Use the coefficients μk=0.20 andμs=0.50.

Answers

Answer:

The wood block reaches a height of 4.249 meters above its starting point.

Explanation:

The block represents a non-conservative system, since friction between wood block and the ramp is dissipating energy. The final height that block can reach is determined by Principle of Energy Conservation and Work-Energy Theorem. Let suppose that initial height has a value of zero and please notice that maximum height reached by the block is when its speed is zero.

[tex]\frac{1}{2}\cdot m \cdot v^{2} = m \cdot g\cdot h + \mu_{k}\cdot m\cdot g\cdot s \cdot \sin \theta[/tex]

[tex]\frac{1}{2}\cdot v^{2} = g\cdot h + \mu_{k}\cdot g\cdot \left(\frac{h}{\sin \theta} \right)\cdot \sin \theta[/tex]

[tex]\frac{1}{2}\cdot v^{2} = g\cdot h +\mu_{k}\cdot g\cdot h[/tex]

[tex]\frac{1}{2}\cdot v^{2} = (1 +\mu_{k})\cdot g\cdot h[/tex]

[tex]h = \frac{v^{2}}{2\cdot (1 + \mu_{k})\cdot g}[/tex] (1)

Where:

[tex]h[/tex] - Maximum height of the wood block, in meters.

[tex]v[/tex] - Initial speed of the block, in meters per second.

[tex]\mu_{k}[/tex] - Kinetic coefficient of friction, no unit.

[tex]g[/tex] - Gravitational acceleration, in meters per square second.

[tex]m[/tex] - Mass, in kilograms.

[tex]s[/tex] - Distance travelled by the wood block along the wooden ramp, in meters.

[tex]\theta[/tex] - Inclination of the wooden ramp, in sexagesimal degrees.

If we know that [tex]v = 10\,\frac{m}{s}[/tex], [tex]\mu_{k} = 0.20[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the height reached by the block above its starting point is:

[tex]h = \frac{\left(10\,\frac{m}{s} \right)^{2}}{2\cdot (1+0.20)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}[/tex]

[tex]h = 4.249\,m[/tex]

The wood block reaches a height of 4.249 meters above its starting point.

~~~~NEED HELP ASAP~~~~

Block A slides into block B along a frictionless surface. They are moving in the direction from left o the right.

Block A= 3kg

Block B= 4kg

Block A velocity before collision =30m/s.

Block B velocity before collision = 15 m/s

The velocity of block B after the collision is 20m/s.


a.) What is the velocity of block A after collision?

b.) Is the collision elastic? Show work to explain answer why or why not.

Answers

Answer:

Block A velocity is 23.33 m/s and the collission is not elastic.

Explanation:

a) m1v1 + m2v2 = m1v1' + m2v2'

Plug in givens

90+60=3v1'+80

solve for v1'= 23.33m/s

b) Find the initial and final kinetic energy of Block B

Ki= 1/2(4)(15)^2 + 1/2(3)(30)^2 = 1800 J

Kf= 1/2(4)(20)^2 + 1/2(3)*(23.33)^2= 1616.433J

Since Ki does not equal Kf the collision is not elastic

А pressure gauge with a measurement range of 0-10 bar has a quoted inaccuracy of £1.0% f.s. (+1% of full-scale reading). (a) What is the maximum measurement error expected for this instrument? (b) What is the likely measurement error expressed as a percentage of the or reading if this pressure gauge is measuring a pressure of 1 bar?​

Answers

Answer:

I am not able to answer this question please don't mind...

Explanation:

please marks me as brainliests...

While an object near the earths surface is in free fall, its
A) velocity increases
B) acceleration increases

Answers

Answer:

a

Explanation:

The rate of change of an object's location with relation to a reference point is its velocity, which is dependent on time. when an object is dropped from space at rest (t = 0) under the influence of gravity, the velocity of the object changes and increases with time while the acceleration decreases.

A gymnast falls from a height onto a trampoline. For a moment, both the gymnast’s kinetic energy and gravitational potential energy are zero. How is the gymnast’s mechanical energy stored for that moment? Question 12 options: rest energy chemical energy elastic energy thermal energy

Answers

Answer:

elastic energy

Explanation:

When a gymnast falls on a trampoline from a height, after coming in contact with the trampoline, both the gymnast and the trampoline start to move down due to the elastic property of the trampoline.

During this stretching of the trampoline there comes a maximum point up to which the trampoline is stretched. At this point, both the kinetic energy and the gravitational potential energy of the gymnast are zero due to zero speed and zero height, respectively.

The only energy stored in the gymnast's body at this point is the elastic potential energy due to stretching of the trampoline. Hence,the correct option is:

elastic energy

A room has dimensions of 15 ft by 15 ft by 20 ft contains air with a density of 0.0724 pounds-mass per cubic feet. The weight of air in the room in pounds-force is

Answers

Answer:

the weight of the air in pound-force (lb-f) is 325.8 lbf

Explanation:

Given;

dimension of the room, = 15 ft by 15 ft by 20 ft

density of air in the room, ρ = 0.0724 lbm/ft³

The volume of air in the room is calculated as;

Volume = 15 ft x 15 ft x 20 ft = 4,500 ft³

The mass of the air is calculated as;

mass = density x volume

mass = 0.0724 lbm/ft³  x  4,500 ft³

mass = 325.8 lb-m

The weight of the air is calculated as;

Weight = mass x gravity

Weight = 325.8 lb-m x 32.174 ft/s²

Weight = 10482.29 lbm.ft/s²

The weight of the air in pound-force (lb-f) is calculated as;

1 lbf = 32.174 lbm.ft/s²

[tex]Weight =10,482.29\ lbm.ft/s^2\times \frac{1 \ lbf}{32.174 \ lbm.ft/s^2} \\\\Weight = 325.8 \ lbf[/tex]

Therefore, the weight of the air in pound-force (lb-f) is 325.8 lbf

A 1,200kg roller coaster car starts rolling up a slope at a speed of 15m/s. What is the highest point it could reach

Answers

Answer: 11.36 m

Explanation:

Given

Mass of roller coaster is m=1200 kg

Initial speed of roller coaster is v=15 m/s

Energy at bottom and at the top is same i.e.

[tex]\Rightarrow \dfrac{1}{2}mv^2=mgh\\\\\Rightarrow \dfrac{1}{2}\times 1200\times 15^2=1200\times 9.8\times h\\\\\Rightarrow h=\dfrac{15^2}{2\times 9.8}\\\\\Rightarrow h=11.36\ m[/tex]

Thus, the highest point reach by the roller coaster is 11.36 m

Answer:

11.36m

Explanation:

Mary and her younger brother Alex decide to ride the carousel at the State Fair. Mary sits on one of the horses in the outer section at a distance of 2.0 m from the center. Alex decides to play it safe and chooses to sit in the inner section at a distance of 1.1 m from the center. The carousel takes 5.8 s to make each complete revolution.

Required:
a. What is Mary's angular speed %u03C9M and tangential speed vM?
b. What is Alex's angular speed %u03C9A and tangential speed vA?

Answers

Answer:

you can measure by scale beacause we dont no sorry i cant help u but u can ask me some other Q

A ball is launched from the ground with a horizontal speed of 30 m/s and a vertical speed of 30 m/s. How long will it take to get to the top of its trajectory? A. 3 seconds B. 4 seconds C. 2 seconds D. 6 seconds

Answers

Answer:

A (3 seconds)

Explanation:

Well here we have a type of motion called projectile motion and it is pretty similar to an upside down parabola. The top of the trajectory is the vertex of the parabola and is also when v=0.

Lets identify our givens.

Givens:

Horizontal speed= 30m/s

Vertical Speed= 30 m/s

Since the ball is in freefall after being launched ay=-g(take up to be positive) and ax=0

The ball is launched from the ground so y0=0

Final vertical velocity= 0

This problem is now relatively easy because we only need to find the vertical distance so we can ignore horizontal speed and use

vy=vy0+ayt

Plug in our givens

0=30-10t

solve for t

t=3 seconds

Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth to study the earliest stages of star formation, before a star begins to emit visible light. Suppose an infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz. Calculate the wavelength of the infrared radiation.

Answers

Answer:

[tex]\lambda=6.83\times 10^{-5}\ m[/tex]

Explanation:

Given that,

An infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz.

We know that,

1 THz = 10¹² Hz

So,

f = 4.39 × 10¹² Hz

We need to find the wavelength of the infrared radiation.

We know that,

[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{4.39\times 10^{12}}\\\\=6.83\times 10^{-5}\ m[/tex]

So, the wavelength of the infrared radiation is [tex]6.83\times 10^{-5}\ m[/tex].

190 students sit in an auditorium listening to a physics lecture. Because they are thinking hard, each is using 125 W of metabolic power, slightly more than they would use at rest. An air conditioner with a COP of 5.0 is being used to keep the room at a constant temperature. What minimum electric power must be used to operate the air conditioner?

Answers

Answer:

W = 4.75 KW

Explanation:

First, we will calculate the heat to be removed:

Q = (No. of students)(Metabolic Power of Each Student)

Q = (190)(125 W)

Q = 23750 W = 23.75 KW

Now the formula of COP is:

[tex]COP = \frac{Q}{W}\\\\W = \frac{Q}{COP}\\\\W = \frac{23.75\ KW}{5}\\\\[/tex]

W = 4.75 KW

A resident of a lunar colony needs to have her blood pressure checked in one of her legs. Assume that we express the systemic blood pressure as we do on earth and that the density of blood does not change. Suppose also that normal blood pressure on the moon is still 120/80 (which may not actually be true).

Required:
If a lunar colonizer has her blood pressure taken at a point on her ankle that is 1.5 m below her heart, what will be her systemic blood-pressure reading, expressed in the standard way, if she has normal blood pressure? The acceleration due to gravity on the moon is 1.67 m/s^2

Answers

Answer:

The pressure is 2505 Pa.  

Explanation:

Height, h = 1.5 m

density of blood, d = 1000 kg/cubic meter

Gravity, g = 1.67 m/s^2

let the pressure is P.  

The pressure due to the fluid is given by

P = h d g

P = 1.5 x 1000 x 1.67

P = 2505 Pa

A rope, under a tension of 221 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement of the rope is given by y = (0.10 m)(sin πx/2) sin 12πt, where x = 0 at one end of the rope, x is in meters, and t is in seconds.

What are:
a. the length of the rope.
b. the speed of the waves on the rope
c. the mass of the rope
d. If the rope oscillates in a third-harmonic standing wave pattern, what will be the period of oscillation.

Answers

Answer:

sup qwertyasdfghjk

Explanation:

For a research project, a student needs a solenoid that produces an interior magnetic field of 0.0100 T. She decides to use a current of 1.00 A and a wire 0.500 mm in diameter. She winds the solenoid in layers on an insulating form 1.00 cm in diameter and 20.0 cm long.
Determine the number of layers of wire needed. (Round your answer up to the nearest integer.)
Determine the total length of the wire. (Use the integer number of layers and the average layer diameter.)

Answers

Answer:

[tex]n=3.8[/tex]

Explanation:

From the question we are told that:

Magnetic Field [tex]B=0.01T[/tex]

Current [tex]I=1.00[/tex]

Wire Diameter [tex]d_w=0.5*10^3m[/tex]

Layers Diameter [tex]d_l=1*10^2m[/tex]

Length [tex]l=0.2m[/tex]

Generally the equation for number of layers is mathematically given by

[tex]n=\frac{Bd_w}{\mu_o I}[/tex]

Where

[tex]Vacuum\ permeability=\mu_0[/tex]

[tex]n= \frac{0.01*0.5*10^3m}{4 \pi *10^{-7}*1 }[/tex]

[tex]n=3.8[/tex]

A system gains 1500J of heat and 2200J of work is done by the system on its surroundings. Determine the change in internal energy of the system

Answers

Answer:

-700

formula is heat gained - work done

The change in internal energy if A system gains 1500J of heat and 2200J of work is done by the system on its surroundings, is 700 joules.

What is Energy?

Energy is the ability to perform work in physics. It could exist in several different forms, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc.

Additionally, there is heat and work, which is energy being transferred from one body to another. Energy is always assigned based on its nature once it has been transmitted. Thus, heat transmitted may manifest as thermal energy while work performed may result in mechanical energy.

Given:

A system gains 1500J of heat and 2200J of work is done by the system on its surroundings,

Calculate the change in internal energy as shown below,

The change in internal energy = heat gained - work done

The change in internal energy = 1500 - 2200

The change in internal energy = -700 J

Thus, the change in internal energy is 700 joules.

To know more about Energy:

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PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. What is the decay constant for this decay?

Answers

Explanation:

hope this will help u

mark me as brinalist or vote me

Thankyou

A bullet with mass 5.35 g is fired horizontally into a 2.174-kg block attached to a horizontal spring. The spring has a constant 6.17 102 N/m and reaches a maximum compression of 6.34 cm.
(a) Find the initial speed of the bullet-block system.
(b) Find the speed of the bullet.

Answers

Answer:

a)[tex]V=1.067\: m/s[/tex]

b)[tex]v=434.65\: m/s [/tex]  

Explanation:

a)

Using the conservation of energy between the moment when the bullet hit the block and the maximum compression of the spring.

[tex]\frac{1}{2}MV^{2}=\frac{1}{2}k\Delta x^{2}[/tex]

Where:

M is the bullet-block mass (0.00535 kg + 2.174 kg = 2.17935 kg)V is the speed of the systemk is the spring constant (6.17*10² N/m)Δx is the compression of the spring (0.0634 m)

Then, let's find the initial speed of the bullet-block system.

[tex]V^{2}=\frac{k\Delta x^{2}}{M}[/tex]

[tex]V=\sqrt{\frac{6.17*10^{2}*0.0634^{2}}{2.17935}}[/tex]

[tex]V=1.067\: m/s[/tex]

b)

Using the conservation of momentum we can find the velocity of the bullet.

[tex]mv=MV[/tex]

[tex]v=\frac{MV}{m}[/tex]

[tex]v=\frac{2.17935*1.067}{0.00535}[/tex]

[tex]v=434.65\: m/s [/tex]  

I hope it helps you!

             

 

Your cell phone typically consumes about 300 mW of power when you text a friend. If the phone is operated using a lithium-ion battery with a voltage of 3.5 V, what is the current (in A) flowing through the cell-phone circuitry under these circumstances

Answers

Answer:

I = 0.0857 A

Explanation:

Given that,

Power consumed by the cellphone, P = 300 mW

The voltage of the battery, V = 3.5 V

Let I is the current flowing through the cell-phone. We know that,

P = VI

Where

I is the current

So,

[tex]I=\dfrac{P}{V}\\\\I=\dfrac{300\times 10^{-3}}{3.5}\\\\I=0.0857\ A[/tex]

So, the current flowing the cell-phone is 0.0857 A.

Part B
What is the approximate amount of thrust you need to apply to the lander to keep its velocity roughly constant? Explain why, using Newton's first
law of motion.

Answers

Answer:

Force is zero.

Explanation:

According to the Newton's second law, when an object is moving with an acceleration the force acting on the object is directly proportional to the rate of change of momentum of the object.

F = m a

if the object is moving with uniform velocity, the acceleration is zero, and thus, the force is also zero.  

Answer: Near the moon’s surface, a thrust over 11,250 N but under 13,500 N would make it travel at a constant vertical velocity.

Explanation: .Newton’s first law of motion states that an object in motion continues to move in a straight line at a constant velocity unless acted upon by an unbalanced force. In accordance with this law, the lunar lander moves in a downward direction toward the surface of the moon under the influence of force due to gravity. A thrust somewhere between 11,250 and 13,500 balances this gravitational force out.

as the ball rises the vertical component of it's velocity_____. explain​

Answers

Answer:

Decreases

Explanation:

because its moving against gravitational attraction and at maximum height its velocity will be and it will decrease until it reaches maximum height and the start to increase again

How are Newton’s 1 and 2 law related?

Answers


Newton's first law states that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force. This postulate is known as the law of inertia.


Newton's Second Law of Motion. According to NASA, this law states, "Force is equal to the change in momentum per change in time. For a constant mass, force equals mass times acceleration."

At an airport, luggage is unloaded from a plane into the three cars of a luggage carrier, as the drawing shows. The acceleration of the carrier is 0.12 mls2, and friction is negligible. The coupling bars have negligible mass. By how much would the tension in each of the coupling bars A, B, and C change if 39 kg of luggage were removed from car 2 and placed in (a) car I and (b) car 3

Answers

Answer:

a) ΔT₁ = -4.68 N,   ΔT₂ = 4.68 N, b) ΔT₂ = 4.68 N, ΔT₁ = 4.68 N

Explanation:

In this exercise we will use Newton's second law.

         ∑F = m a

Let's start with the set of three cars

         F_total = M a

         F_total = M 0.12

where the total mass is the sum of the mass of each charge

          M = m₁ + m₂ + m₃

 

This is the force with which the three cars are pulled.

Now let's write this law for each vehicle

car 1

         F_total - T₁ = m₁ a

         T₁ = F_total - m₁ a

car 2

         T₁ - T₂ = m₂ a

         T₂ = T₁ - m₂ a

car 3

         T₂ = m₃ a

         

note that tensions are forces of action and reaction

a) They tell us that 39 kg is removed from car 2 and placed on car 1

         m₂’= m₂ - 39

         m₁'= m₁ + 39

         m₃ ’= m₃

they ask how much each tension varies, let's rewrite Newton's equations

         

The total force does not change since the mass of the set is the same F_total ’= F_total

car 1

           F_total ’- T₁ ’= m₁’ a

           T₁ ’= F_total - m₁’ a

           T₁ ’= (F_total - m₁ a) - 39 a

           T₁ '= T₁ - 39 0.12

           ΔT₁ = -4.68 N

car 2

           T₁’- T₂ ’= m₂’ a

           T₂ ’= T₁’- m₂’ a

           T₂ '= (T₁'- m₂ a) + 39 a

           T₂ '= T₂ + 39 0.12

           ΔT₂ = 4.68 N

b) in this case the masses remain

            m₁ '= m₁

           m₂ ’= m₂ - 39

           m₃ ’= m₃ + 39

we write Newton's equations

car 3

          T₂ '= m₃' a

          T₂ ’= (m₃ + 39) a

          T₂ '= m₃ a + 39 a

          T₂ '= T₂ + 39 0.12

          ΔT₂ = 4.68 N

car 1

            F_total - T₁ ’= m₁’ a

            T₁ ’= F_total - m₁ a

car 2

            T₁' -T₂ '= m₂' a

            T₁ ’= T₂’- m₂’ a

            T₁ '= (T₂'- m₂ a) + 39 a

            T₁ '= T₁ + 39 0.12

            ΔT₁ = 4.68 N

The tension in each of the coupling bars A, B, and C of the luggage carrier changes as,

When luggage were removed from car 2 and placed in car 1, the tension is A and C does not change and the tension in B is decreased by 4.68 N.When luggage were removed from car 2 and placed in car 3, the tension is A and B does not change and the tension in C is increased by 4.68 N.

What is tension force?

Tension is the pulling force carried by the flexible mediums like ropes, cables and string.

Tension in a body due to the weight of the hanging body is the net force acting on the body.

At an airport, luggage is unloaded from a plane into the three cars of a luggage carrier, as the drawing shows. The acceleration of the carrier is 0.12 m/s², and friction is negligible.

The acceleration is the same, Tension due to the horizontal component of the forces for car 1, 2 and 3 can be given as,

[tex]\sum F_{1h}=T_A-T_B=m_1a\\\sum F_{2h}=T_B-T_C=m_2a\\\sum F_{3h}=T_C=m_3a[/tex]

On solving the above 3 equation, we get the values of tension in each bar as,

[tex]T_A=(m_1+m_2+m_3)a\\T_B=(m_3+m_2)a\\T_C=m_3a[/tex]

Case 1- When 39 kg of luggage were removed from car 2 and placed in car I

The tension is A and C does not change for this case. The acceleration of the carrier is 0.12 m/s². Thus, the change in tension is B is,

[tex]\Delta T_B=39\times0.12\\\Delta T_B=4.68\rm \;N[/tex]

Case 2- When 39 kg of luggage were removed from car 2 and placed in car III

The tension is A and B does not change for this case. The acceleration of the carrier is 0.12 m/s². Thus, the change in tension is B is,

[tex]\Delta T_C=39\times0.12\\\Delta T_C=4.68\rm \;N[/tex]

Hence, the tension in each of the coupling bars A, B, and C of the luggage carrier changes as,

When luggage were removed from car 2 and placed in car 1, the tension is A and C does not change and the tension in B is decreased by 4.68 N.When luggage were removed from car 2 and placed in car 3, the tension is A and B does not change and the tension in C is increased by 4.68 N.

 

Learn more about the tension here;

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Find a parametric representation for the surface. The plane through the origin that contains the vectors i - j and j - k

Answers

Answer:

parametric representation: x = u, y = v - u ,  z = - v

Explanation:

Given vectors :

i - j ,  j - k

represent the vector equation of the plane as:

r ( u, v ) = r₀ + ua + vb

where:  r₀ = position vector

            u and v = real numbers

             a and b = nonparallel vectors

expressing the nonparallel vectors as :

a = i -j , b = j - k , r = ( x,y,z ) and r₀ = ( x₀, y₀, z₀ )

hence we can express vector equation of the plane as

r(u,v) = ( x₀ + u, y₀ - u + v,  z₀ - v )

Finally the parametric representation of the surface through (0,0,0) i.e. origin = 0

( x, y , z ) = ( x₀ + u,  y₀ - u + v,   z₀ - v )

x = 0 + u ,

y = 0 - u + v

z = 0 - v

parametric representation: x = u, y = v - u ,  z = - v

A transverse sine wave with an amplitude of 2.50 mm and a wavelength of 1.80 m travels, from left to right along a long, horizontal stretched string with a speed of 36.0 m s. I Take the origin at the left end of the undisturbed string. At time t = 0 the left end of the string has its maximum upward displacement,
(a) What is the frequency of the wave?
(b) What is the angular frequency of the wave?
(c) What is the wave number of the wave?
(d) What is the function y(x,t) that describes the wave?
(e) What is y(t) for a particle at the left end of the string?
(f) What is y(t) for a particle 1.35 m to the right of the origin?
(g) What is the maximum magnitude of transverse velocity of any particle of the string?
(h) Find the transverse displacement of a particle 1.35 m to the right of the origin at time t = 0.0625 s.
(i) Find the transverse velocity of a particle 1.35 m to the right of the origin at time t = 0.0625 s.

Answers

Explanation:

Given that,

Amplitude, A = 2.5 nm

Wavelength,[tex]\lambda=1.8\ m[/tex]

The speed of the wave, v = 36 m/s

At time t = 0 the left end of the string has its maximum upward displacement.

(a) Let f is the frequency. So,

[tex]f=\dfrac{v}{\lambda}\\\\f=\dfrac{36}{1.8}\\\\f=20\ Hz[/tex]

(b) Angular frequency of the wave,

[tex]\omega=2\pi f\\\\=2\pi \times 20\\\\=125.7\ rad/s[/tex]

(c) The wave number of the wave[tex]=\dfrac{1}{\lambda}[/tex]

[tex]=\dfrac{1}{1.8}\\\\=0.56\ m^{-1}[/tex]

Container A and container B hold samples of the same ideal gas. The volume and the pressure of container A is equal to the volume and pressure of container B, respectively. If Container A has half as many molecules of the ideal gas in it as Container B does, then which of the following mathematical statements is correct regarding the absolute temperatures TA and TB in Container A and Container B. respectively?
A. TA = TB/2.
B. TA = 4TB.
C. TA = TB/4.
D. TA = 2TB.
E. TA = TB

Answers

Answer:

A. TA = TB/2.

Explanation:

Since container A has half as many molecules of the ideal gas in it as container B. Therefore, container A will have half the volume of gas as in container B:

[tex]V_A = \frac{1}{2}V_B[/tex]

Now, from Charle's Law:

[tex]\frac{V_A}{T_A}=\frac{V_B}{T_B}\\\\\frac{1}{2}\frac{V_B}{T_A}=\frac{V_B}{T_B}\\\\T_A = \frac{T_B}{2}[/tex]

Hence, the correct option is:

A. TA = TB/2.

A uniformly dense solid disk with a mass of 4 kg and a radius of 4 m is free to rotate around an axis that passes through the center of the disk and perpendicular to the plane of the disk. The rotational kinetic energy of the disk is increasing at 21 J/s. If the disk starts from rest through what angular displacement (in rad) will it have rotated after 3.3 s?

Answers

Answer:

3.44 rad

Explanation:

The rotational kinetic energy change of the disk is given by ΔK = 1/2I(ω² - ω₀²) where I = rotational inertia of solid sphere = MR²/2 where m = mass of solid disk = 4 kg and R = radius of solid disk = 4 m, ω₀ = initial angular speed of disk = 0 rad/s (since it starts from rest) and ω = final angular speed of disk

Since the kinetic energy is increasing at a rate of 21 J/s, the increase in kinetic energy in 3.3 s is  ΔK = 21 J/s × 3.3 s = 69.3 J

So, ΔK = 1/2I(ω² - ω₀²)

Since ω₀ = 0 rad/s

ΔK = 1/2I(ω² - 0)

ΔK = 1/2Iω²

ΔK = 1/2(MR²/2)ω²

ΔK = MR²ω²/4

ω² = (4ΔK/MR²)

ω = √(4ΔK/MR²)

ω = 2√(ΔK/MR²)

Substituting the values of the variables into the equation, we have

ω = 2√(ΔK/MR²)

ω = 2√(69.3 J/( 4 kg × (4 m)²))

ω = 2√(69.3 J/[ 4 kg × 16 m²])

ω = 2√(69.3 J/64 kgm²)

ω = 2√(1.083 J/kgm²)

ω = 2 × 1.041 rad/s

ω = 2.082 rad/s

The angular displacement θ is gotten from

θ = ω₀t + 1/2αt² where ω₀ = initial angular speed = 0 rad/s (since it starts from rest), t = time of rotation = 3.3 s and α = angular acceleration = (ω - ω₀)/t = (2.082 rad/s - 0 rad/s)/3.3 s = 2.082 rad/s ÷ 3.3 s = 0.631 rad/s²

Substituting the values of the variables into the equation, we have

θ = ω₀t + 1/2αt²

θ = 0 rad/s × 3.3 s + 1/2 × 0.631 rad/s² (3.3 s)²

θ = 0 rad + 1/2 × 0.631 rad/s² × 10.89 s²

θ = 1/2 × 6.87159 rad

θ = 3.436 rad

θ ≅ 3.44 rad

What is the total surface charge qint on the interior surface of the conductor (i.e., on the wall of the cavity)

Answers

Answer: hello your question is incomplete below is the missing part

A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q.

answer:

- q

Explanation:

Since the spherical cavity was carved out of a neutral conducting sphere hence the electric field inside this conductor = zero

given that there is a point charge +q at the center of the spherical cavity hence for the electric field inside the conductor to be = zero the total surface charge qint on the wall of the cavity will be -q

Distillation is the separation of multiple Choose... components based on their different Choose... . As the mixture is heated and the first component Choose... , its Choose... form travels through the distillation set-up and Choose... into a different container.

Answers

Answer:

Explanation:

Distillation is the separation of multiple LIQUID components based on their different BOILING POINT. As the mixture is heated and the first component SEPARATES, its PURE form travels through the distillation set-up and GOES into a different container

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