Ask question

Ask question

All categories

2 years ago

11

A pipe that is 20.0 m long and 10.0 cm in diameter contains olive oil. the density of the olive oil is 890 kg/m3 and the bulk mo

dulus is 1.3 × 109 pa. a 3.4-hz longitudinal wave is transmitted in the oil. how many milliseconds does it take for the wave to travel the length of the pipe in the oil?

1 answer:

xenn [34]2 years ago

8 0

Velocity of wave through the oil medium, C = Sqrt (M_b/Sigma)

Where, M_b = Bulk modulus of oil; Sigma = Density of oil
Therefore,
C Sqrt [(1.3*10^9)/(890)] = 1208.58 m/s

Now,
C = d/t, where d = distance = length of pipe; t = time taken.

Therefore,
t = d/C = 20/1208.58 = 0.017 seconds = 0.017*1000 = 17 milliseconds.

You might be interested in

Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d

Step2247 [10]

Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that $F = k \frac{Q_1Q_2}{d^2}$ so if we double the charge on $Q_1$ and double the distance to $(2d)$ we plug these into the equation to find

<span> $F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}$ </span>

So we see the new force is exactly 1/2 of the old force so your answer should be $\frac{1}{2}F$ if I can remember my physics correctly.

9 0

2 years ago

Read 2 more answers

You want to move a heavy box with mass 30.0 kg across a carpeted floor. You pull hard on one of the edges of the box at an angle

charle [14.2K]

Answer:

a= $5.54m/s^{2}$

Explanation:

The net force, $F_{net}$ of the box is expressed as a product of acceleration and mass hence

$F_{net}=ma$ where m is mass and a is acceleration

Making a the subject, a= $\frac {F_{net}}{m}$

From the attached sketch,

∑ $F_{net}=Fcos\theta-F_{f}$ where $F_{f}$ is frictional force and $\theta$ is horizontal angle

Substituting ∑ $F_{net}$ as $F_{net}$ in the equation where we made a the subject

a= $\frac {Fcos\theta-F_{f}}{m}$

Since we’re given the value of F as 240N, $F_{f}$ as 41.5N, $\theta$ as $30^{o}$ and mass m as 30kg

a= $\frac {240cos30-41.5}{30.0}=\frac {166.346}{30.0}=5.54m/s^{2}$

6 0

2 years ago

The planets how and block are near each other in the Dorgon system. the Dorgons have very advanced technology, and a Dorgon scie

BlackZzzverrR [31]

Answer:

Decreasing the distance between Hox and Blox, increasing the mass of Hox, or increasing the mass of Hox and Blox.

Explanation:

The gravity force is directly proportional to the mass of the bodies and inversely proportional to the square of the distance that separates them.

Or

If we decrease the distance between both planets (Hox and Blox), the gravitational pull between them will increase.

On the other hand, if we keep the distance between Hox and Blox, but we increase the mass of one of them, or increase the mass of both, the gravitational pull between them will also increase.

4 0

2 years ago

A 65 kg students is walking on a slackline, a length of webbing stretched between two trees. the line stretches and sags so that

polet [3.4K]

Answer : Tension in the line = 936.7 N

Explanation :

It is given that,

Mass of student, m = 65 kg

The angle between slackline and horizontal, $\theta=20^0$

The two forces that acts are :

(i) Tension

(ii) Weight

So, from the figure it is clear that :

$mg=2T\ sin\ \theta$

$T=\dfrac{mg}{2\ sin\theta}$

$T=\dfrac{65\ kg\times 9.8\ m/s^2}{2(sin\ 20)}$

$T=936.7\ N$

Hence, this is the required solution.

5 0

2 years ago

Read 2 more answers

An all female guitar septet is getting ready to go on stage. The lead guitarist, Kira,who is always in tune, plucks her low E st

matrenka [14]

Answer:

Buffy > Aiko > Chandra > Freja > Evita

Explanation:

Beats occur when two waves of nearby frequencies overlap. The number of beats per second is equal to the difference in the frequency.

Let

$f=$ Frequency of Kira

$f_1=$ Frequency of Aiko

The beat frequency of Kira and Aiko is therefore,

$f_{beat}=|f_2-f_1|$

Substituting $f$ for $f_2$ and $3Hz$ for $f_{beat}$ , we get

$3Hz=|f-f_1|\\f_1=f+3Hz$

From the above equation, we see that beats increase

when

$f_2 =$ Frequency of Chandra

The beat frequency of Kira and Chandra is therefore,

$1Hz=|f-f_2|\\f_2=f-1Hz$

From the above equation, we see that beats decrease

When

$f_3=$ Frequency of Evita

The beat frequency of Kira and Evita is therefore,

$5Hz=|f-f_3|\\f_3=f-5Hz$

From the above equation, we see that beats decrease

When

$f_4=$ Frequency of Freja

The beat frequency of Kira and Freja is therefore,

$3Hz=|f-f_4|\\f_4=f-3Hz$

From the equation above, we see that beats decrease

When

$f_5=$ Frequency of Buffy

The beat frequency of Kira and Buffy becomes,

$4Hz=|f-f_5|\\f_5=f+4Hz$

From the equation above, we see that beats increase

Hence, the rank of members based on initial frequencies from largest to smallest is Buffy > Aiko > Chandra > Freja > Evita

6 0

2 years ago