A string that is under 50.0N of tension has linear density 5.0g/m. A sinusoidal wave with amplitude 3.0cm and wavelength 2.0m travels along the string. What is the maximum speed of a particle on the string

Answers

Answer 1

Answer:

9.42 m/s

Explanation:

Applying,

V' = Aω.............. Equation 1

Where V' = maximum speed of the string, A = Amplitude of the wave, ω = angular velocity.

But,

ω = 2πf................. Equation 2

Where f = frequency, π = pie

And,

f = v/λ................ Equation 3

Where, λ = wave length, v = velocity

Also,

v = √(T/μ)................. Equation 4

Where T = Tension, μ = linear density.

From the question,

Given: T = 50.0 N, μ = 5.0 g/m = 0.005 kg/m

Substitute into equation 4

v = √(50/0.005)

v = √(10000)

v = 100 m/s

Also Given: λ = 2.0 m

Substitute into equation 3

f = 100/2

f = 50 Hz.

Substitute the value of f into equation 2

Where π = constant = 3.14

ω = 2(3.14)(50)

ω = 314 rad/s

Finally,

Given: A = 3.0 cm = 0.03 m

Substitute into equation 1

V' = 0.03(314)

V' = 9.42 m/s


Related Questions

help asap PLEASE I will give u max everything all that

steps if possible

Answers

Explanation:

2. [tex]R_T = R_1 + R_2 + R_3 = 625\:Ω + 330\:Ω + 1500\:Ω[/tex]

[tex]\:\:\:\:\:\:\:= 2455\:Ω = 2.455\:kΩ[/tex]

3. Resistors in series only need to be added together so

[tex]R_T = 8(140\:Ω) = 1120\:Ω = 1.12\:kΩ[/tex]

what is entrapersonal environment​

Answers

Answer:

The interpersonal environment is considered to be a subset of the organizational environment – defined as the employee’s perception of the practices, policies, and processes of an organization

Explanation:

why material selection is important to design and manufacturing?​

Answers

Answer:

. You want your product to be as strong and as long lasting as possible. There are also the safety implications to consider. You see, dangerous failures arising from poor material selection are still an all too common occurrence in many industries. yep that the answer have a Great day

Explanation:

(◕ᴗ◕✿)

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and her takeoff point is 1.80 m above the pool. We assume the upward direction to be positive, and the downward direction to be negative.
(a) How long are her feet in the air?(b) What is her highest point above the board?(c) What is her velocity when her feet hit the water?

Answers

Answer:

(a) t = 1.14 s

(b) h = 0.82 m

(c) vf = 7.17 m/s

Explanation:

(b)

Considering the upward motion, we apply the third equation of motion:

[tex]2gh = v_f^2 - v_i^2[/tex]

where,

g = - 9.8 m/s² (-ve sign for upward motion)

h = max height reached = ?

vf = final speed = 0 m/s

vi = initial speed = 4 m/s

Therefore,

[tex](2)(9.8\ m/s^2)h = (0\ m/s)^2-(4\ m/s)^2\\[/tex]

h = 0.82 m

Now, for the time in air during upward motion we use first equation of motion:

[tex]v_f = v_i + gt_1\\0\ m/s = 4\ m/s + (-9.8\ m/s^2)t_1\\t_1 = 0.41\ s[/tex]

(c)

Now we will consider the downward motion and use the third equation of motion:

[tex]2gh = v_f^2-v_i^2[/tex]

where,

h = total height = 0.82 m + 1.8 m = 2.62 m

vi = initial speed = 0 m/s

g = 9.8 m/s²

vf = final speed = ?

Therefore,

[tex]2(9.8\ m/s^2)(2.62\ m) = v_f^2 - (0\ m/s)^2\\[/tex]

vf = 7.17 m/s

Now, for the time in air during downward motion we use the first equation of motion:

[tex]v_f = v_i + gt_1\\7.17\ m/s = 0\ m/s + (9.8\ m/s^2)t_2\\t_2 = 0.73\ s[/tex]

(a)

Total Time of Flight = t = t₁ + t₂

t = 0.41 s + 0.73 s

t = 1.14 s

g you hang an object of mass m on a spring with spring constant k and find that it has a period of T. If you change the spring to one that has a spring constant of 2 k, the new period is

Answers

Answer:

a)   T = 2π [tex]\sqrt{\frac{m}{k} }[/tex],  b)  T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]

Explanation:

a) A system formed by a mass and a spring has a simple harmonic motion with angular velocity

          w² = k / m

angular velocity and period are related

          w = 2π /T

     

we substitute

          4π²/ T² = k / m

           T = 2π [tex]\sqrt{\frac{m}{k} }[/tex]

b) We change the spring for another with k ’= 2 k, let's find the period

           T ’= 2π [tex]\sqrt{\frac{m}{k'} }[/tex]

           T ’= 2π [tex]\sqrt{ \frac{m}{2k} }[/tex]

           T ’= [tex]\frac{1}{\sqrt{2} } T[/tex]

what is the frequency of a wave related to​

Answers

Answer:

Frequency is the number of complete oscillations or cycles or revolutions made in one second.

Increasing the surfactant concentration above the critical micellar concentration
will result in: Select one:
1.An increase in surface tension
2. A decrease in surface tension
3. No change in surface tension
4.None of the above​

Answers

Answer:

Explanation:no change in surface tension

An increase in the surfactant concentration above the critical micellar concentration will result in no change in surface tension.

In water-gas interface, surfactant reduces the surface tension of water by adsorbing at the liquid–gas interface.

Also, in oil-water interface, surfactant reduces the interfacial tension between oil and water by adsorbing at the oil-water interface.

The concentration of the surfactant can increase to a level called critical micellar concentration, which is an important characteristic of a surfactant.

As the concentration of the surfactant increases before critical micellar concentration, the surface tension changes strongly with an increase in the concentration of the surfactant. After reaching the critical micellar concentration, any further increase in the concentration will result in no change of the surface tension, that is the surface tension will be constant.

Thus, increasing the surfactant concentration above the critical micellar concentration will result in no change in surface tension.

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A camera lens with focal length f = 50 mm and maximum aperture f>2
forms an image of an object 9.0 m away. (a) If the resolution is limited
by diffraction, what is the minimum distance between two points on the
object that are barely resolved? What is the corresponding distance
between image points? (b) How does the situation change if the lens is
“stopped down” to f>16? Use λ= 500 nm in both cases

Answers

Answer:

The minimum distance between two points on the  object that are barely resolved is 0.26 mm

The corresponding distance between the  image points = 0.0015 m

Explanation:

Given  

focal length f = 50 mm and maximum aperture f>2

s =  9.0 m

aperture = 25 mm = 25 *10^-3 m

Sin a = 1.22 *wavelength /D  

Substituting the given values, we get –  

Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m

Sin a = 2.93 * 10 ^-5 rad

Now  

Y/9.0 m = 2.93 * 10 ^-5

Y = 2.64 *10^-4 m = 0.26 mm

Y’/50 *10^-3 = 2.93 * 10 ^-5  

Y’ = 0.0015 m

Give the missing ammeter reading a and b. suggest why more current flow through some bulbs than through others Grade 10 question and Answer

Answers

Answer:

becaude of electricity

E=kq/r^2 chứng minh điện thế V=kq/r từ mối liên hệ giữa điện trường E và điện thế V

Answers

Answer:

hindi ko maintindihan teh

why are you teachers regarded as professionals​

Answers

Answer:

coz teaching is their profession.

help me with this question ​

Answers

Explanation:

Let's set the x-axis to be parallel to the and positive up the plane. Likewise, the y-axis will be positive upwards and perpendicular to the plane. As the problem stated, we are going to assume that m1 will move downwards so its acceleration is negative while m2 moves up so its acceleration is positive. There are two weight components pointing down the plane, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex] and two others pointing up the plane, the two tensions T along the strings. There is a normal force N pointing up from the plane and two pointing down, [tex]m_1g \sin \theta[/tex] and [tex]m_2g \sin \theta[/tex]. Now let's apply Newton's 2nd law to this problem:

x-axis:

[tex]m1:\:\:\:\displaystyle \sum_i F_i = T - m_1g \sin \theta = - m_1a\:\:\:\:(1)[/tex]

[tex]m2:\:\:\:\displaystyle \sum_i F_i = T - m_2g \sin \theta = m_2a\:\:\:\:(2)[/tex]

y-axis:

[tex]\:\:\:\displaystyle \sum_i F_i = N - m_1g \cos \theta - m_2g \cos \theta = 0[/tex]

Use Eqn 1 to solve for T,

[tex]T = m_1(g \sin \theta - a)[/tex]

Substitute this expression for T into Eqn 2,

[tex]m_1g \sin \theta - m_1a - m_2g \sin \theta = m_2a[/tex]

Collecting all similar terms, we get

[tex](m_1 + m_2)a = (m_1 - m_2)g \sin \theta[/tex]

or

[tex]a = \left(\dfrac{m_1 - m_2}{m_1 + m_2} \right)g \sin \theta[/tex]

A police car in hot pursuit goes speeding past you. While the siren is approaching, the frequency of the sound you hear is 5500 Hz. When the siren is receding away from you, the frequency of the sound is 4500 Hz. Use the Doppler formula to determine the velocity of the police car. Use vsound=330 m/s.
What is the velocity v of the police car ?

Answers

Vs = 34m/s
I don’t have an explanation my apologies.

When a police car in hot pursuit goes speeding past you, the velocity v of the police car is 33 m/s.

What is the Doppler formula?

The formula is used when there exists a Doppler shift. The Doppler shift is due to the relative motion of sound waves between the source and observer.

The frequency increase by the Doppler effect is represented by the formula

f' = [tex]\dfrac{v-v_{o} }{v-v_{s} }[/tex]× f

Given the frequency of source f' is 5500 Hz . Velocity of the observer v₀  is 0.

Substituting the value into the equation will give us the velocity of the police car.

[tex]5500 = \dfrac{330}{330-v} \times f[/tex]...........(1)

When the car is receding, the frequency of the receiving signal f = 4500 Hz.

[tex]4500 = \dfrac{330}{330+v} \times f[/tex]..........(2)

Solving both equation, we get the velocity of a police car.

v = 33 m/s

Therefore, the velocity v of the police car is 33 m/s.

Learn more about Doppler equation.

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who is the biggest man in the world​

Answers

Answer:

Sultan Kösen

here is a pic

During which radioactive decay process does a neutron change into a proton?
A. Alpha decay
B. Gamma decay
C. Beta decay (positron)
D. Beta decay (electron)

Answers

A neutron changes into a proton during a radioactive decay process called beta decay (positron), which is option D.

What is beta decay?

A beta decay in physics is a nuclear reaction in which a beta particle (electron or positron) is emitted.

A positron is an electron with a positive charge.

During a beta decay, a neutron in the nucleus of the radioactive material suddenly changes into a proton, causing an increase in the atomic number of an element.

Learn more about beta decay at: https://brainly.com/question/25455333

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If the loading is 0.4, the coinsurance rate is 0.2, the number of units of medical care is 100, and the number of units of medical care is 1. What is the premium of this insurance?

Answers

Answer:

72  is the premimum of the insurance.

Explanation:

Below is the given values:

The loading = 0.4

Coinsurance rate = 0.2

Number of units = 100

Total number of units = 100 * 0.4 = 40

Remaining units = 60 * 0.2 = 12

Add the 60 and 12 values = 60 + 12 = 72

Thus, 72  is the premimum of the insurance.

nariz (am
miria amy
0 = 0 +260 + (0)
U= 29 mb
6= ut +1 (04)
Car I was sitting at rest when it nous hit from
the rear by car 2 of identical mass. Both cant had
their heaks on and they stidled together Guy
in the original directioned of motion. If the stopping
force is notx (Combined weight of the cars), die
u=0 to find the approximate speed of car a just
before the collision took place on​

Answers

Answer:

33 mph

Explanation:

My best guess

What does it mean when work is positive?
O Velocity is greater than kinetic energy.
O Kinetic energy is greater than velocity.
O The environment did work on an object.
O An object did work on the environment.

Answers

Answer:

O The environment did work on an object

Explanation:

A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from zero to 9.41 m/s in 4.24 s. What is the magnitude of the linear impulse experienced by a 67.0 kg passenger in the car during this time

Answers

Answer:

the impulse experienced by the passenger is 630.47 kg

Explanation:

Given;

initial velocity of the car, u = 0

final velocity of the car, v = 9.41 m/s

time of motion of the car, t = 4.24 s

mass of the passenger in the car, m = 67 kg

The impulse experienced by the passenger is calculated as;

J = ΔP = mv - mu = m(v - u)

           = 67(9.41 - 0)

           = 67 x 9.41

           = 630.47 kg

Therefore, the impulse experienced by the passenger is 630.47 kg

Pete is investigating the solubility of salt (NaCl) in water. He begins to add 50 grams of salt to 100 grams of
room temperature tap water in a beaker. After adding all of the salt and stirring for several minutes, Pete
notices a solid substance in the bottom of the beaker. Which statement best explains why there is a solid
substance in the bottom of the beaker?
A. The salt he is using is not soluble in water.
B. The salt is changing into a new substance that is not soluble in water,
C. The dissolving salt is causing impurities in the water to precipitate to the bottom
D. The water is saturated and the remaining salt precipitates to the bottom

Answers

Answer:

would the answer be c

Explanation: that what i think in my opian

Answer:

A

Explanation:

what change occurs to the mass of an object when a unbalanced

Answers

Answer:

The mass decreases

Explanation:

Just smart

An object with mass m = 0.56 kg is attached to a string of length r = 0.72 m and is rotating with an angular velocity ω = 1.155 rad/s. What is the centripetal force acting in the object?

Answers

Answer:

The centripetal force is 0.54 N.

Explanation:

mass, m = 0.56 kg

radius, r = 0.72 m

angular speed, w = 1.155 rad/s

The centripetal force is given by

[tex]F = m r w^2\\\\F =0.56\times 0.72\times 1.155\times 1.155\\\\F = 0.54 N[/tex]

A lens with a focal length of 15 cm is placed 45 cm in front of a lens with a focal length of 5.0 cm .

Required:
How far from the second lens is the final image of an object infinitely far from the first lens?

Answers

Answer:

the required distance is 6 cm

Explanation:

Given the data in the question;

f₁ = 15 cm

f₂ = 5.0 cm

d = 45 cm

Now, for first lens object distance s = ∝

1/f = 1/s + 1/s' ⇒ 1/5 = 1/∝ + 1/s'

Now, image distance of first lens s' = 15cm  

object distance of second lens s₂ will be;

s₂ = 45 - 15 = 30 cm

so

1/f₂ = 1/s₂ + 1/s'₂

1/5 = 1/30 + 1/s'₂

1/s'₂ = 1/5 - 1/30  

1/s'₂ = 1 / 6

s'₂ = 6 cm

Hence, the required distance is 6 cm

 

The distance of the final image from the first lens will be is 6 cm.

What is mirror equation?

The mirror equation expresses the quantitative connection between object distance (do), image distance (di), and focal length (fl).

The given data in the problem is;

f₁ is the focal length of lens 1= 15 cm

f₂ s the focal length of lens 2= 5.0 cm

d is the distance between the lenses = 45 cm

From the mirror equation;

[tex]\frac{1}{f} = \frac{1}{s} +\frac{1}{s'} \\\\ \frac{1}{5} = \frac{1}{\alpha} +\frac{1}{s'} \\\\[/tex]

If f₁ is the focal length of lens 1 is 15 cm then;

[tex]s'=15 cm[/tex]

f₂ s the focal length of lens 2= 5.0 cm

s₂ = 45 - 15 = 30 cm

From the mirror equation;

[tex]\frac{1}{f_2} = \frac{1}{s_1} +\frac{1}{s_2'} \\\\ \frac{1}{5} = \frac{1}{30} +\frac{1}{s_2'} \\\\ \frac{1}{s_2'}= \frac{1}{5} -\frac{1}{30} \\\\ \frac{1}{s_2'}= \frac{1}{6} \\\\ \rm s_2'= 6 cm[/tex]

Hence the distance of the final image from the first lens will be is 6 cm.

To learn more about the mirror equation refer to the link;

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Two children stretch a jump rope between them and send wave pulses back and forth on it. The rope is 3.3 m long, its mass is 0.52 kg, and the force exerted on it by the children is 47 N. (a) What is the linear mass density of the rope (in kg/m)

Answers

Answer:

The linear mass density of rope is 0.16 kg/m.

Explanation:

mass, m = 0.52 kg

force, F = 47 N

length, L = 3.3 m

(a) The linear mass density of the rope is defined as the mass of the rope per unit length.

Linear mass density = m/L = 0.52/3.3 = 0.16 kg/m

Your car breaks down in the middle of nowhere. A tow truck weighing 4000 lbs. comes along and agrees to tow your car, which weighs 2000 lbs., to the nearest town. The driver of the truck attaches his cable to your car at an angle of 20 degrees to horizontal. He tells you that his cable has a strength of 500 lbs. He plans to take 10 secs to tow your car at a constant acceleration from rest in a straight line along a flat road until he reaches the maximum speed of 45 m.p.h. Can the driver carry out the plan

Answers

Answer:

F = 1010 Lb

the tension on the cable is greater than its resistance, which is why the plan is not viable

Explanation:

For this exercise we can use the kinematic relations to find the acceleration and with Newton's second law find the force to which the cable is subjected.

          v = v₀ + a t

how the car comes out of rest v₀ = 0

          a = v / t

let's reduce to the english system

          v = 45 mph (5280 ft / 1 mile) (1h / 3600) = 66 ft / s

let's calculate

          a = 66/10

          a = 6.6 ft / s²

now let's write Newton's second law

X axis

         Fₓ = ma

with trigonometry

         cos 20 = Fₓ / F

         Fₓ = F cos 20

we substitute

          F cos 20 = m a

          F = m a / cos20

          W = mg

          F = [tex]\frac{W}{g} \ \frac{a}{cos 20}[/tex]

let's calculate

          F = [tex]\frac{2000}{32} \ \frac{6.6 }{cos20}[/tex](2000/32) 6.6 / cos 20

          F = 1010 Lb

Under these conditions, the tension on the cable is greater than its resistance, which is why the plan is not viable.

which of the following is a correct statement. a. In dc steady state conditions, the voltages across the capacitors are constant and the currents through the capacitance are zero. The current through the inductors are constant and the voltage across the inductances are constant. b. In dc steady state conditions, the voltages across the capacitors are zero and the currents through the capacitance are constant. The current through the inductors are constant and the voltage across the inductances are zero. c. In dc steady state conditions, the voltages across the capacitors are constant and the currents through the capacitance are zero. The current through the inductors are zero and the voltage across the inductances are constant. d. WIn dc steady state conditions, the voltages across the capacitors are constant and the currents through the capacitance are zero. The current through the inductors are constant and the voltage across the inductances are zero.

Answers

Answer:

d. In dc steady state conditions, the voltages across the capacitors are constant and the currents through the capacitance are zero. The current through the inductors are constant and the voltage across the inductances are zero.

Explanation:

The current through a capacitor is given by i = CdV/dt where C = capacitance of capacitor and V = voltage across capacitor. At steady state dV/dt = 0 and V = constant. So, i = CdV/dt = C × 0 = 0.

So, in dc steady state, the voltage across a capacitor is constant and the current zero.

The voltage across an inductor is given by V = Ldi/dt where L = inductance of inductor and i = current through inductor. At steady state di/dt = 0 and V = constant. So, V = Ldi/dt = L × 0 = 0.

So, in dc steady state, the voltage across an inductor is zero and the current constant.

So, In dc steady state conditions, the voltages across the capacitors are constant and the currents through the capacitance are zero. The current through the inductors are constant and the voltage across the inductances are zero.

The answer is d.

Kulsum’s TV uses 45 W. How much does it cost her to watch TV for one month (30 days). She watches TV for 4 hours/day during mid-peak time (10.4 cents/kWh).

Answers

Answer:

Total cost = 56.16 cents

Explanation:

Given the following data;

Power = 45 Watts

Time = 4 hours

Number of days = 30 days

Cost = 10.4 cents

To find how much does it cost her to watch TV for one month;

First of all, we would determine the energy consumption of the TV;

Energy = power * time

Energy = 45 * 4

Energy = 180 Watt-hour = 180/1000 = 0.18 Kwh (1 Kilowatts is equal to 1000 watts).

Energy consumption = 0.18 Kwh

Next, we find the total cost;

Total cost = energy * number of days * cost

Total cost = 0.18 * 30 * 10.4

Total cost = 56.16 cents

A 5.0-kg mass is placed at (3.0, 4.0) m, and a 6.0-kg mass is placed at (3.0, -4.0) m. What is the moment of inertia of this system of masses about the y-axis?

Answers

Answer:

the moment of inertia of this system of masses about the y-axis is 99 kgm²

Explanation:

Given the data in the question;

mass m₁ = 5.0 kg at point ( 3.0, 4.0 )

mass m₂ = 6.0 kg at point ( 3.0, -4.0 )

Now, Moment of inertia [tex]I[/tex] of this system of masses about the y-axis will be;

Moment of inertia [tex]I[/tex]ₓ = mixi²

Moment of inertia [tex]I[/tex] = m₁x₁² + m₂x₂²

we substitute

Moment of inertia [tex]I[/tex] = [ 5.0 × ( 3 )² ]  + [ 6.0 × ( 3 )² ]

Moment of inertia [tex]I[/tex] = [ 5.0 × 9 ]  + [ 6.0 × 9 ]

Moment of inertia [tex]I[/tex] = 45 + 54

Moment of inertia [tex]I[/tex] = 99 kgm²

Therefore, the moment of inertia of this system of masses about the y-axis is 99 kgm²

plz answer the question

Answers

Answer:

Ray A = Incidence ray

Ray B = Reflected ray

Explanation:

From the law of reflection,

Normal: This is the line that makes an angle of 90° with the reflecting surface.

Ray A is the incidence ray: This is the ray that srikes the surface of a reflecting surface. The angle formed between the normal and the incidence ray is called the incidence angle

Ray B is the reflected ray: This is the ray leaves the surface of a reflecting surface. The angle formed between the reflected ray and the normal is called reflected angle

Which of the following elements has the largest atomic radius?
Silicon
Aluminum
Sulfur
Phosphorous

Answers

aluminum has the largest atomic radius

Answer:

francium

Atomic radii vary in a predictable way across the periodic table. As can be seen in the figures below, the atomic radius increases from top to bottom in a group, and decreases from left to right across a period. Thus, helium is the smallest element, and francium is the largest.

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