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2 years ago

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A small town has decided to forego the use of electrical power and send energy through town via mechanical waves on ropes. They

use rope with a mass per length of 1.50 kg/m under 6000 N tension. If they are limited to a wave amplitude of 0.500 m, what must be the frequency of waves necessary to transmit power at the average rate of 2.00 kW

1 answer:

mojhsa [17]2 years ago

3 0

Answer:

the required frequency of waves is 2.066 Hz

Explanation:

Given the data in the question;

μ = 1.50 kg/m

T = 6000 N

Amplitude A = 0.500 m

P = 2.00 kW = 2000 W

we know that, the average power transmit through the rope can be expressed as;

p = $\frac{1}{2}$ vμω²A²

p = $\frac{1}{2}$ √(T/μ)μω²A²

so we solve for ω

ω² = 2P / √(T/μ)μA²

we substitute

ω² = 2(2000) / √(6000/1.5)(1.5)(0.500)²

ω² = 4000 / 23.71708

ω² = 168.65

(2πf)² = ω²

so

(2πf)² = 168.65

4π²f² = 168.65

f² = 168.65 / 4π²

f² = 4.27195

f = √4.27195

f = 2.066 Hz

Therefore, the required frequency of waves is 2.066 Hz

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Block A, mass 250 g , sits on top of block B, mass 2.0 kg . The coefficients of static and kinetic friction between blocks A and

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Answer:

F = 69.3 N

Explanation:

For this exercise we use Newton's second law, remembering that the static friction force increases up to a maximum value given by

fr = μ N

We define a reference system parallel to the floor

block B ( lower)

Y axis

N - W₁-W₂ = 0

N = W₂ + W₂

N = (M + m) g

X axis

F -fr = M a

for block A (upper)

X axis

fr = m a (2)

so that the blocks do not slide, the acceleration in both must be the same.

Let's solve the system by adding the two equations

F = (M + m) a (3)

a = $\frac{F}{ M+m}$

the friction force has the formula

fr = μ N

fr = μ (M + m) g

let's calculate

fr = 0.34 (2.0 + 0.250) 9.8

fr = 7.7 N

we substitute in equation 2

fr = m a

a = fr / m

a = 7.7 / 0.250

a = 30.8 m / s²

we substitute in equation 3

F = (2.0 + 0.250) 30.8

F = 69.3 N

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2 years ago

Consider a father pushing a child on a playground merry-go-round. the system has a moment of inertia of 84.4 kg · m2. the father

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two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i

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A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 - 14.7t - 4.9t^2$

$4.9t^2 + 14.7t - 19.6 = 0$

$t^2 + 3t - 4 = 0$

$(t + 4)(t - 1) = 0$

$(t - 1) = 0$

$\boxed {t = 1 ~ second}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2$

$0 = 19.6 + 14.7t - 4.9t^2$

$4.9t^2 - 14.7t - 19.6 = 0$

$t^2 - 3t - 4 = 0$

$(t - 4)(t + 1) = 0$

$(t - 4) = 0$

$\boxed {t = 4 ~ seconds}$

The difference in the two ball's time in the air is:

$\Delta t = 4 ~ seconds - 1 ~ second$

$\large {\boxed {\Delta t = 3 ~ seconds} }$

<h2>Question B:</h2><h3>First Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (-14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

<h3>Second Ball</h3>

$v^2 = u^2 - 2gH$

$v^2 = (14.7)^2 + 2(-9.8)(-19.6)$

$v^2 = 600.25$

$v = \sqrt {600.25}$

$\boxed {v = 24.5 ~ m/s}$

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

$h = H - ut - \frac{1}{2}gt^2$

$h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 11.025 ~ m}$

<h3>Second Ball</h3>

$h = H + ut - \frac{1}{2}gt^2$

$h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2$

$\boxed {h = 25.725 ~ m}$

The difference in the two ball's height after 0.500 s is:

$\Delta h = 25.725 ~ m - 11.025 ~ m$

$\large {\boxed {\Delta h = 14.7 ~ m} }$

<h3>Learn more</h3>

Velocity of Runner : brainly.com/question/3813437
Kinetic Energy : brainly.com/question/692781
Acceleration : brainly.com/question/2283922
The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

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2 years ago

A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V. What are (a) the total energy stored i

Debora [2.8K]

Answer:

(A) Total energy will be equal to $0.044\times 10^{-5}J$

(b) Energy density will be equal to $0.0175J/m^3$

Explanation:

We have given diameter of the plate d = 2 cm = 0.02 m

So area of the plate $A=\pi r^2=3.14\times 0.02^2=0.001256m^2$

Distance between the plates d = 0.50 mm = $0.50\times 10^{-3}m$

Permitivity of free space $\epsilon _0=8.85\times 10^{-12}F/m$

Potential difference V =200 volt

Capacitance between the plate is equal to $C=\frac{\epsilon _0A}{d}=\frac{8.85\times 10^{-12}\times 0.001256}{0.50\times 10^{-3}}=0.022\times 10^{-9}F$

(a) Total energy stored in the capacitor is equal to

$E=\frac{1}{2}CV^2$

$E=\frac{1}{2}\times 0.022\times 10^{-9}\times 200^2=0.044\times 10^{-5}J$

(b) Volume will be equal to $V=Ad$ , here A is area and d is distance between plates

$V=0.001256\times 0.02=2.512\times 10^{-5}m^3$

So energy density $=\frac{Energy}{volume}=\frac{0.044\times 10^{-5}}{2.512\times 10^{-5}}=0.0175J/m^3$

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2 years ago

A yo-yo is made from two uniform disks, each with mass m and radius R, connected by a light axle of radius b. A light, thin stri

schepotkina [342]

Answer:

linear acceleration

$a = \frac{2g}{2 + \frac{R}{b}}$

angular acceleration

$\alpha = \frac{2g}{R(2 + \frac{R}{b})}$

Explanation:

As we know that the force due to tension force is upwards while weight of the disc is downwards

so we will have

$2mg - T = 2ma$

also we have

$Tb = (\frac{1}{2}mR^2 + \frac{1}{2}mR^2)\alpha$

now we have

$Tb = mR^2(\frac{a}{R})$

$T = \frac{mRa}{b}$

now we have

$2mg = (2ma + \frac{mRa}{b})$

$a(2 + \frac{R}{b}) = 2g$

so we have

linear acceleration

$a = \frac{2g}{2 + \frac{R}{b}}$

angular acceleration

$\alpha = \frac{2g}{R(2 + \frac{R}{b})}$

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2 years ago