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2 years ago

8

Water waves in a small tank are .06 m long. They pass a given point at a rate of 14.8 waves every three seconds. What is the spe

ed of the water waves? What is the period of these water waves?

1 answer:

snow_lady [41]2 years ago

4 0

Answer:

Speed = 0.296m/2

Period = 0.203 s

Explanation:

If by 'long' you mean the wavelength of the waves, then the wavelength $\lambda=0.06m$ .

The frequency $f$ of the waves is 14.8 waves every 3 seconds or

$f=14.8/3 =4.33Hz$ .

Now the relationship between wavelength $\lambda$ , frequency $f$ and speed $v$ of the waves is:

$v=\lambda f$

We put in the values $\lambda=0.06m$ and $f=4.933Hz$ and get:

$\boxed{v=0.06*4.922=0.296m/s}$

Now the period $T$ is just the inverse of the frequency, or

$T=\frac{1}{f}$

$\boxed{T=\frac{1}{4.933}=0.203\:seconds }$

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Two objects (45.0 and 21.0 kg) are connected by a massless string that passes over a massless, frictionless pulley. The pulley h

Oksanka [162]

a) The acceleration of the objects is $3.56 m/s^2$

b) The tension in the string is 280.8 N

Explanation:

a)

We start by writing the equations of motion for the two masses attached to the pulley.

For the heavier mass, we have:

$m_1 g - T = m_1 a$ (1)

where

$m_1 = 45.0 kg$ is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

T is the tension in the string

a is the acceleration of the system (here we assumed that the heavier mass accelerates downward)

For the lighter mass, we have

$T-m_2 g = m_2 a$ (2)

where

T is the tension in the string

$m_2 = 21.0 kg$ is the mass

$g=9.8 m/s^2$ is the acceleration of gravity

a is the acceleration of the system (here we assumed that the lighter mass accelerates upward)

From (1) we get

$T=m_1g - m_1 a$

And substituting into (2),

$(m_1 g - m_1 a)-m_2 g = m_2 a\\(m_1 -m_2)g = (m_1+m_2)a\\a=\frac{m_1 - m_2}{m_1+m_2}g=\frac{45-21}{45+21}(9.8)=3.56 m/s^2$

b)

From the previous part of the problem we got an expression for the tension in the string:

$T=m_1g - m_1 a$

Where we have

$m_1 = 45.0 kg$

$g=9.8 m/s^2$

$a=3.56 m/s^2$ is the acceleration, found in part a)

Susbtituting, we find

$T=(45.0)(9.8)-(45.0)(3.56)=280.8 N$

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2 years ago

Which number can each term of the equation be multiplied by to eliminate the fractions before solving? 6 – x + = 6 minus StartFr

zysi [14]

Answer:

We need to multiply 12 to each term to eliminate fractions.

Explanation:

Given expression:

$6-\frac{3}{4}x+\frac{1}{3}=\frac{1}{2}x+5$

To eliminate the fraction we need to multiply each term by least common multiple of the denominators of the fraction.

The denominators in the above expressions are:

4, 3 and 2

The multiples of each can be listed below.

2⇒ 2,4,6,8,10,<u>12</u>,14,16.....

3⇒ 3,6,9,<u>12</u>,15,18

4⇒ 4,8,<u>12</u>.......

From the list of the multiples stated, we can see the least common multiple is 12.

So we will multiply each term by 12.

Multiplying 12 to both sides.

$12(6-\frac{3}{4}x+\frac{1}{3})=12(\frac{1}{2}x+5)$

Using distribution,

$72-9x+4=6x+60$

Thus we successfully eliminated the fractions.

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2 years ago

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Table 2.4 shows how the dispacement of a runner changed during a sprint race. Draw a dispacement-time graph to show this data, a

GalinKa [24]

4. Table 2.4 shows how the displacement of a runner changed
during a sprint race. Draw a displacement–time graph to show
this data, and use it to deduce the runner’s speed in the middle
of the race.
Table 2.4 Data for a sprinter during a race
Displacement
(m)
0 4 10 20 50 80 105
Time (s) 1 2 3 6 9 12

8 0

1 year ago

Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci

Rudiy27

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

$\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W$

$\dot m( h1+\frac{v1^2}{2}) = \dot m( h2+\frac{v2^2}{2})$

$h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}$

$2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]$

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

2671.85 = 467.13 +x2*2226.0

x2 = 0.99

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2 years ago

A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp

Ahat [919]

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

at time t = 0:

amplitude A = 0.02m

unknown: velocity v at position y = 0.01 m

$y = A cos(\omega t + \phi)\\v = -\omega A sin(\omega t + \phi)\\ \omega = \sqrt{\frac{k}{m}}$

1. Finding Ф from the initial conditions:

$-0.02 = 0.02cos(0 + \phi) => \phi = \pi$

2. Finding time t at position y = 1 cm:

$0.01 =0.02cos(\omega t + \pi)\\ \frac{1}{2}=cos(\omega t + \pi)\\t=(acos(\frac{1}{2})-\pi)\frac{1}{\omega}$

3. Find velocity v at time t from equation 2:

$v =-0.02\sqrt{\frac{k}{m}}sin(acos(\frac{1}{2}))$

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1 year ago

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