The mass per unit length of the rope is 0.0500 kg/m. Find the tension. Express your answer in newtons.

Answers

Answer 1

Complete question:

A transverse wave on a rope is given by [tex]y \ (x, \ t) = (0.75 \ cm) \ cos \ \pi[(0.400 \ cm^{-1}) x + (250 \ s^{-1})t][/tex]. The mass per unit length of the rope is 0.0500 kg/m. Find the tension. Express your answer in newtons.

Answer:

The tension on the rope is 1.95 N

Explanation:

The general equation of a progressive wave is given as;

[tex]y \ (x,t) = A \ cos(kx \ + \omega t)[/tex]

Compare the given equation with the general equation of wave, the following parameters will be deduced.

A = 0.75 cm

k = 0.400π cm⁻¹

ω = 250π s⁻¹

The frequency of the wave is calculated as;

ω = 2πf

2πf = 250π

2f = 250

f = 250/2

f = 125 Hz

The wavelength of the wave is calculated as;

[tex]\lambda = \frac{2\pi}{k} \\\\\lambda = \frac{2\pi }{0.4 \pi} = 5 \ cm = 0.05 \ m[/tex]

The velocity of the wave is calculated as;

v = fλ

v = 125 x 0.05

v = 6.25 m/s

The tension on the rope is calculated as;

[tex]v = \sqrt{\frac{T}{\mu}} \\\\where;\\\\T \ is \ the \ tension \ of \ the \ rope\\\\\mu \ is \ the \ mass \ per \ unit \ length = 0.05 \ kg/m\\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (6.25)^2\times (0.05)\\\\T = 1.95 \ N[/tex]

Therefore, the tension on the rope is 1.95 N


Related Questions

Fast please!!. Serious answers only.

Considering that Susan is 38 years old, what is the probability that she lives to the age of 85?

Answers

Answer:

The probability that Susan turns 85 years old is 45,436/97,825= 0.4644.

Explanation:

Edmentum

Answer:

45,436/97,825= 0.4644.

A photon has an energy of 2.09×10^-18 kJ .What is its wavelength?​

Answers

Answer:

wavelength= 1.05 × 10^ -46 m

Explanation:

the formula : λ= hc/E

where; "h" = Planck's constant [6.626 × 10^ -34]

c= speed of light [3.0 × 10^ 8]

you first have to convert the energy of the photon to Joules by dividing the constant by 1000

2.09 × 10^ -18 / 1000 = 2.09 × 10^ -21

then you replace you data into the equation

λ= 6.626 × 10^ -34 × 3.0 × 10^ 8 / 2.09 × 10 ^ -21

first multiply the Planck's constant and the speed of light then divide it by the energy which is in "Joules"

:. λ = 1.05 × 10^ -46

hope this helps

an alternating voltage of 100V, 50HZ Is Applied across an impedance of (20-j30) calculate the resistance, the capacitance, current, the phase angle between current and voltage

Answers

The resistance R = 20 Ω

The capacitance C = 106.1 μF

The current, I is 2.773 A at 56.31°.

The phase angle of the between the current and the voltage is 56.31° leading.

Since the impedance Z = 20 - j30 Ω, the resistance, R is the real part of the impedance. So R = ReZ = 20 Ω

So, the resistance R = 20 Ω

To find the capacitance, we need first to find the reactance of the capacitor X. Since the impedance Z = 20 - j30, the reactance of the capacitor X. is the imaginary part of the impedance. So X = ImZ = 30 Ω.

Now the reactance of the capacitor X = 1/ωC where ω = angular frequency of the circuit = 2πf where f = frequency of the circuit = 50 Hz and C = capacitance  

So, C = 1/ωX = 1/2πfX

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

C = 1/2πfX

C = 1/(2π × 50 Hz × 30 Ω)

C = 1/3000π

C = 1/9424.778

C = 1.061 × 10⁻⁴ F

C = 106.1 × 10⁻⁶ F

C = 106.1 μF

So, the capacitance is 106.1 μF

The current I = V/Z where V = voltage = 100 V at 0° and Z = impedance.

The magnitude of Z = √(20² + (-30)²)

= √(400 + 900)

= √1300

= 36.06 Ω

and its angle Φ = tan⁻¹(ImZ/ReZ)

= tan⁻¹(-30/20)

= tan⁻¹(-1.5) = -56.31°

So, V = 100 ∠ 0° and Z = 36.06 ∠ -56.31°

So, the current, I = V/Z =  (100 ∠ 0°)/36.06 ∠ -56.31°

= 100/36.06 ∠(0° - (-56.31° ))

= 2.773 ∠ 56.31° A

So, the current is 2.773 A at 56.31°.

Since the current is 2.773 A at 56.31°, the phase angle of the between the current and the voltage is 56.31° leading.

So, the phase angle of the between the current and the voltage is 56.31° leading.

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Light takes 1.2 sec to get from the moon to the Earth. Assume you are looking at the moon with noticeable earth shine. If the Sun burned out, you would eventually see the crescent of the moon disappear. The earth shine part of the moon would disappear Answer 2.4 s after the crescent disappeared.

Answers

Answer:

1.2 seconds

Explanation:

Answer to the following question is 1.2 seconds

Because light from the moon takes 1.2 seconds to reach Earth, the light released from the crescent immediately before it vanishes will also take 1.2 seconds to reach Earth. As a result, the earth-shine portion of the moon will vanish 1.2 seconds after the crescent has vanished.

why do atom absorb photon since it makes it unstable??​

Answers

[tex]\textsf{When an electron is hit by a }[/tex] [tex]\textsf{photon of light, it absorbs the quanta}[/tex] [tex]\textsf{of energy the photon was carrying}[/tex] [tex]\textsf{and moves to a higher energya}[/tex] [tex]\textsf{ state. Electrons therefore have to }[/tex] [tex]\textsf{jump around within the atom as }[/tex] [tex]\textsf{they either gain or lose energy. }[/tex]

When an electron is hit by a photon of light, it absorbs the quanta of energy the photon was carrying and moves to a higher energy state. Electrons therefore have to jump around within the atom as they either gain or lose energy.

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A/An is a type of blood cell that's also called a red blood cell. a) Jeukocyte O b) thrombocyte c) plasma d) erythrocyte

Answers

Answer:

red blood cell, also called erythrocyte

Explanation:

Hope it helps

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A car is traveling at 118 km/h when the driver sees an accident 85 m ahead and slams on the brakes. What minimum constant deceleration is required to stop the car in time to avoid a pileup

Answers

Answer:

The constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²

Explanation:

From the question, the car is traveling at 118 km/h, that is the initial velocity, u = 118km/h

The distance between the car and the accident at the moment when the driver sees the accident is 85 m, that is s = 85 ,

Since the driver slams on the brakes and the car will come to a stop, then the final velocity, v = 0 km/h = 0 m/s

First, convert 118 km/h to m/s

118 km/h = (118 × 1000) /3600 = 32.7778 m/s

∴ u = 32.7778 m/s

Now, to determine the deceleration, a, required to stop,

From one of the equations of motion for linear motion,

v² = u² + 2as

Then

0² = (32.7778)² + 2×a×85

0 = 1074.3841 + 170a

∴ 170a = - 1074.3841

a = - 1074.3841 / 170

a = - 6.3199

a ≅ - 6.32 m/s²

Hence, the constant minimum deceleration required to stop the car in time to avoid pileup is 6.32 m/s²

help me I m stuck on this question

Answers

Answer:

Answer is in the picture.

Explanation:

Answer is in the picture.

Meaning of power in physics

Answers

Answer:

The rate of doing work is called power.

Answer:

The amount of energy transported or transformed per unit time is referred to as power in physics. The watt, which is equal to one joule per second in the International System of Units, is the unit of power.

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A 1-cm long wire carrying 15 A is inside a solenoid 4 cm in radius with 800 turns/m carrying a current of 40 mA. The wire segment is oriented perpendicularly to the axis of the solenoid. What is the magnitude of the magnetic force on this wire segment in ???? N?

Answers

Answer:

the magnitude of the magnetic force on the wire segment is 6.03 x 10⁻⁶ N

Explanation:

Given;

length of the conductor, L = 1 cm = 0.01 m

current carried by the solenoid, I₁ = 15 A

radius of the solenoid, r = 4 cm

number of turns per length of the solenoid, n = 800 turns/m

current carried by the solenoid, I₂ = 40 mA = 0.04 A

The magnetic field of the solenoid is calculated as;

B = μnI₂

where;

μ is the permeability of free space = 4π x 10⁻⁷ Tm/A

B = ( 4π x 10⁻⁷) x (800) x (0.04)

B = 4.022 x 10⁻⁵ T

The magnitude of the magnetic force on the wire segment is calculated as;

F = BI₁L sinθ

where

θ is the angle made by the wire segment against the solenoid = 90⁰

F = (4.022 x 10⁻⁵) x (15) x (0.01) x sin(90)

F = 6.03 x 10⁻⁶ N

Therefore, the magnitude of the magnetic force on the wire segment is 6.03 x 10⁻⁶ N

Your little sister (mass 25 kg) is sitting in her little red wagon (mass
8.5 kg) at rest. You begin pulling her forward, accelerating her with a
constant force for 2.35 s to a speed of 1.8 m/s. Calculate the impulse
you imparted to the wagon and its passenger.

Answers

Answer:

p = 60.6N*s

Explanation:

v_f = v_0+a*t

a = (v_f-v_0)/t

a = (1.8m/s)/2.35s

a = 0.77m/s²

F = m*a

F = (25kg+8.5kg)*0.77m/s²

F = 25.8N

^p = F*t

p = 25.8N*2.35s

p = 60.6N*s

what effect does the force of gravity have on a stone thrown vertically upwards​

Answers

Answer:

rock go down

Explanation:

what comes up must come down.

2. What is the average speed of an athlete who runs 1500 m in 4 minutes?

Answers

Answer:

375 is the answer.

Explanation:

Speed : Distance / Time taken

S: m/ s

s: 1500/4

375 m / s answer

Answer:

375m per minute

Explanation:

if you are looking for a diffrent unit just multiply your answer by however many minutes are in that time frame

An organ pipe of length 3.0 m has one end closed. The longest and next-longest possible wavelengths for standing waves inside the pipe are

Answers

Answer:

The longest wavelength for closed at one end and open at the other is

y / 4      where y is the wavelength - that is node - antinode

The next possible wavelength is 3 y / 4 -    node - antinode - node -antinode

y / 4 = 3 m     y = 12 meters    the longest wavelength

3 y / 4 = 3 m      y = 4 meters   1 / 3 times as long

what is the major difference between the natural frequency and the damped frequency of oscillation.​

Answers

Answer:

This causes the amplitude of the oscillation to decay over time. The damped oscillation frequency does not equal the natural frequency. Damping causes the frequency of the damped oscillation to be slightly less than the natural frequency

Vặt nhỏ được ném lên từ điểm A trên mặt đất với vận tốc đầu 20m/s theo phương thẳng đứng. Xác định độ cao của điểm O mà vật đạt được. Bỏ qua ma sát

Answers

Explanation:

mặt đất với vận tốc ban đầu 20m/s. Bỏ qua mọi ma sát, lấy g = 10 m/s2. Độ cao cực đại mà vật đạt được là.

A block of mass 2 kg starts from rest at the top of a friction quarter of a circle of radius R. The block then slides over frictionless curved surface in the shape of a eventually comes to rest 8 m from the beginning s a horizontal rough surface where e of the horizontal surface. The coefficient kinetic friction between the rough surface and the block is 0.4 . determine the acceleration of the block over the rough surface length 8m​

Answers

The acceleration of the block over the rough surface is 1.22625 m/s²

The process through which the acceleration is obtained is presented as follows of approach to

The given parameters are;

Mass of block, m = 2 kg

Nature of the surface of the quarter circle = Frictionless

The length of the horizontal, d = 8 m

The coefficient of friction of the horizontal surface, μ = 0.4

The unknown parameter;

The acceleration of the block over the rough surface

Method;

Find the work done by friction to stop the block and divide the result by the mass of the block

The work done by friction, [tex]W_f[/tex] = (Force of friction) × (Distance the block moves on the rough surface before coming to rest)

[tex]\mathbf{W_f}[/tex] = [tex]\mathbf{F_f}[/tex] × d

[tex]F_f[/tex] = Normal reaction of surface on block, [tex]N_r[/tex] × μ

Normal reaction on block, [tex]\mathbf{N_r}[/tex] = Weight of block

[tex]\mathbf{N_r}[/tex] ≈ 2 kg × 9.81 m/s² = 19.62 N

Therefore;

The work done by friction [tex]\mathbf{W_f}[/tex] = [tex]\mathbf{F_f}[/tex] × d = [tex]\mathbf{N_r}[/tex] × μ × d

[tex]\mathbf{W_f}[/tex] = 19.62 N × 0.4 × 8 m = 62.784 J

The work done by the block, W = Force, F × d

Force, F = m × a

Where;

a = The acceleration of the block

According to the principle of conservation of energy, we have;

[tex]\mathbf{W_f}[/tex]  = W

∴ 19.62 J = 2 kg × a × 8 m

a = 19.62/(2 kg × 8 m) = 1.22625 m/s²

The acceleration of the block over the rough surface, a = 1.22625 m/s²

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A projectile is launched straight upwards at 75 m/s. Three seconds later, its velocity is...?

Answers

Answer:

V = V0 + a t

V = 75 - 9.8 * 3 = 45.6 m/s

The final velocity of the projectile after 3 seconds is equal to 45.6 m/s.

What is the equation of motion?

The equations of motion can be defined as the relation of the motion of a physical system as the function of time and set up the relationship between the displacement (s), acceleration, velocity (v & u), and time of a moving system.

Given, the initial velocity of the projectile, u = 75 m/s

The time taken by the projectile, t = 3 sec

The acceleration due to gravity upward, g = - 9.8 m/s²

From the first equation of motion we can calculate the final velocity of the projectile:

v = u + at

v = u - gt

v = 75 - 9.8 ×(3)

v = 75 - 29.4

v = 45.6 m/s

Therefore, the final velocity of the projectile after three seconds is 45.6 m/s.

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You have 150 W/m^2 hitting your roof each day. You can convert 13% of it into
usable energy, and you need 3.5 kW to run your house for a day. Show the MATH,
answer and units, to determine the size solar panel you will need to succeed.

Answers

Answer:

Energy = .13 W / m^2      energy of incident energy

N = 3500 Watts / day       power needed

N = 3500 Watts  (3600 * 24 sec) = .0405 Watts/sec

The problem must mean that one needs 3.5 Kw-days

3.5 Kw-days = 3500 watts * 86400 sec = 3.02E8 joules

150 J/sec-m^2 * .13 = 19.5 J / sec-m^2      usable energy

In one day 19.5 J/sec-m^2 = 1.68E6 J/m^2    usable energy received

Area = 3.028E8 J / 1.68E6 J/m2 = 180 m^2

One would need 180 m^2 of solar panels

That's quite a lot of energy

A 1100 watt microwave oven uses 1.1 kW while running so 3.5 kW for 24 hours seems to be quite a lot.

A sound wave made up of large number of unrelated frequencies superposted on each other is

Answers

Since the frequencies are unrelated, and there are a large number of them, I'll say this represents an example of noise.

You are outdoors when you hear the constant chirp of a still cricket. You start walking toward the cricket and at some point you are able to detect that the intensity of the chirp of the cricket has increased by a factor of 4. What of the following statements is true at your new position with respect to the cricket?

a. The power delivered by the sound wave you hear has doubled.
b. The speed of the sound wave emitted by the cricket has decreased by a factor of 4.
c. The distance between you and the cricket has decreased by a factor of 2

Answers

Answer:

C

Explanation:

intensity = Power delivered by the sound (Watt)/ Surounding Area (m²)

I = P/A

A = πr²

r = is the distance between you and the cricket.

so in other form we can get

I = P/πr²

let take I(1) as first intensitilynyou heard and I(2) as the increased intensity.

I(1) / I(2) = r(2)² / r(1)²

1/4 = r(2)²/r(1)²

1/2 = r(2) / r(1)

r(2) = ½ r(1)

or r(2) is decreaases by a factor of 2.

A projectile is fired into the air from the top of a 200-m cliff above a valley as shown below. Its initial velocity is 60 m/s at 60° above the horizontal. Calculate (a) the maximum height, (b) the time required to reach its highest point, (c) the total time of flight, (d) the components of its velocity just before striking the ground, and (e) the horizontal distance traveled from the base of the cliff.​

Answers

a) y(max)  = 337.76 m

b) t₁ = 5.30 s  the time for y maximum

c)t₂ =  13.60 s  time for y = 0 time when the fly finish

d) vₓ = 30 m/s        vy = - 81.32 m/s

e)x = 408 m

Equations for projectile motion:

v₀ₓ = v₀ * cosα          v₀ₓ = 60*(1/2)     v₀ₓ = 30 m/s   ( constant )

v₀y = v₀ * sinα           v₀y = 60*(√3/2)     v₀y = 30*√3  m/s

a) Maximum height:

The following equation describes the motion in y coordinates

y  =  y₀ + v₀y*t - (1/2)*g*t²      (1)

To find h(max), we need to calculate t₁ ( time for h maximum)

we take derivative on both sides of the equation

dy/dt  = v₀y  - g*t

dy/dt  = 0           v₀y  - g*t₁  = 0    t₁ = v₀y/g

v₀y = 60*sin60°  = 60*√3/2  = 30*√3

g = 9.8 m/s²

t₁ = 5.30 s  the time for y maximum

And y maximum is obtained from the substitution of t₁  in equation (1)

y (max) = 200 + 30*√3 * (5.30)  - (1/2)*9.8*(5.3)²

y (max) = 200 + 275.40 - 137.64

y(max)  = 337.76 m

Total time of flying (t₂)  is when coordinate y = 0

y = 0 = y₀  + v₀y*t₂ - (1/2)* g*t₂²

0 = 200 + 30*√3*t₂  - 4.9*t₂²            4.9 t₂² - 51.96*t₂ - 200 = 0

The above equation is a second-degree equation, solving for  t₂

t =  [51.96 ±√ (51.96)² + 4*4.9*200]/9.8

t =  [51.96 ±√2700 + 3920]/9.8

t =  [51.96 ± 81.36]/9.8

t = 51.96 - 81.36)/9.8         we dismiss this solution ( negative time)

t₂ =  13.60 s  time for y = 0 time when the fly finish

The components of the velocity just before striking the ground are:

vₓ = v₀ *cos60°       vₓ = 30 m/s  as we said before v₀ₓ is constant

vy = v₀y - g *t        vy = 30*√3  - 9.8 * (13.60)

vy = 51.96 - 133.28         vy = - 81.32 m/s

The sign minus means that vy  change direction

Finally the horizontal distance is:

x = vₓ * t

x = 30 * 13.60  m

x = 408 m

A wheel is rotating freely at angular speed 530 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initially at rest and with 9 times the rotational inertia of the first, is suddenly coupled to the same shaft. (a) What is the angular speed of the resultant combination of the shaft and two wheels

Answers

Answer: [tex]53\ rev/min[/tex]

Explanation:

Given

angular speed of wheel is [tex]\omega_1 =530\ rev/min[/tex]

Another wheel of 9 times the rotational inertia is coupled with initial wheel

Suppose the initial wheel has moment of inertia as I

Coupled disc has [tex]9I[/tex] as rotational inertia

Conserving angular momentum,

[tex]\Rightarrow I\omega_1=(I+9I)\omega_2\\\\\Rightarrow \omega_2=\dfrac{I}{10I}\times 530\\\\\Rightarrow \omega_2=53\ rev/min[/tex]

HELP ME PLZ FAST
There is more than 1 answer,
The picture is down

Answers

Answer:

test her prototype and collect data about its flight


The temperature of a body falls from 30°C to 20°C in 5 minutes. The air
temperature is 13°C. Find the temperature after a further 5 minutes.

Answers

Answer:

15.88

is the correct answer

A parallel plate air capacitor has a circular disc of diameter 0.1 m, 2 mm apart and potential difference of 300 V is connected between the plates Calculate: (i) Energy of the capacitor and (ii) Electric intensity between the plates. ​

Answers

Your Answer Is in this attachment

A diffraction grating has 6000 lines per centimeter ruled on it. What is the angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe when the grating is illuminated with a beam of light of wavelength 500 nm

Answers

Explanation:

Hope it Will help he hsuejwoamxgehanwpalasmbwfwfqoqlmdbehendalmZbgevzuxwllw. yeh we pabdvddxhspapalw. X

The angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is  27.29°.

What is diffraction?

Waves spreading outward around obstructions are known as diffraction. Sound, electromagnetic radiation like light, X-rays, and gamma rays, as well as very small moving particles like atoms, neutrons, and electrons that exhibit wavelike qualities all exhibit diffraction.

Given:

The number of lines = 6000 per cm,

The Wavelength, λ = 500 nm = 500 × 10 ⁻⁹ m

Calculate the diffraction grating,

[tex]d = 1 / no\ of\ lines[/tex]

d = 10⁻² / 6000 m,

Calculate the second-order maxima angle and third-order maxima angle by the formula given below,

[tex]dsin\theta_1 = n_1 \lambda[/tex]

[tex]sin\theta_1 = n_1\lambda / d[/tex]

[tex]\theta _1 = sin^{-1}[2\times 500\times 10 ^{-9}/10^{-2}\times 6000][/tex]

θ₁ = sin⁻¹(0.6)

θ₁ = 36.87°

Similarly, for θ₂,

θ₂ = sin⁻¹(3 × 500 × 10 ⁻⁹ / 10⁻² × 6000)

θ₂ = sin⁻¹(0.9)

θ₂ = 64.16°

Calculate the separation as follows,

θ₂ - θ₁ = 64.16° - 36.87°

θ₂ - θ₁ =  27.29°

Therefore, the angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is  27.29°.

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When a charged particle moves at an angle of 26.1 with respect to a magnetic field, it experiences a magnetic force of magnitude F. At what angle (less than 90o) with respect to this field will this particle, moving at the same speed?

Answers

Answer:

The angle is 153.9 degree.

Explanation:

Let the magnetic field is B and the charge is q. Angle = 26.1 degree

The force is F.

Let the angle is A'.

Now equate the magnetic forces

[tex]q v B sin 26.1 = q v B sin A'\\\\A' = 180 - 26.1 = 153.9[/tex]

a stone is thrown vertically upwards with a velocity of 20 m per second determine the total time of flight of stone in air​

Answers

Answer:

Explanation:

The best way to do this is to remember the rule about the halfway mark in a parabolic path. At a trajectory's half way point in its travels, it will be at its max height. To get the total time in the air, we take that time at half way and double it. Here's what we know that we are told:

initial velocity is 20 m/s

Here's what we know that we are NOT told:

a = -9.8 m/s/s and

final velocity is 0 at an object's max height in parabolic motion.

We will use the equation:

[tex]v=v_0+at[/tex] where v is final velocity and v0 is initial velocity. Filling in:

0 = 20 + (-9.8)t and

-20 = -9.8t so

t = 2 seconds. The stone reaches its max height 2 seconds after it is thrown; that means that after another 2 seconds it will be on the ground. Total air time is 4 seconds.

: A fan is placed on a horizontal track and given a slight push toward an end stop 1.80 meters away. Immediately after the push, the fan of the cart engages and slows the cart with an acceleration of -0.45 m/s2. What is the maximum possible velocity (magnitude) the cart can have after the push so that the cart turns around just before it hits the end-stop

Answers

Answer:

The initial velocity is 1.27 m/s.

Explanation:

distance, s = 1.8 m

acceleration, a = - 0.45 m/s^2

final velocity, v = 0

let the initial velocity is u.

Use third equation of motion

[tex]v^2 = u^2 + 2 a s \\\\0 = u^2 - 2 \times 0.45\times 1.8\\\\u = 1.27 m/s[/tex]

We have that the Initial velocity  is mathematically given as

u=1.27m/s

Maximum possible velocity

Question Parameters:

a slight push toward an end stop 1.80 meters away

he fan of the cart engages and slows the cart with an acceleration of -0.45 m/s2

Generally the equation for the  third equation of motion    is mathematically given as

Vf^2 = Vi^2 + 2ad

Therefore

0=u^2+0.45*1.8

u=1.27m/s

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