Ask question

Ask question

All categories

2 years ago

13

The speed v of a sound wave traveling in a medium that has bulk modulus b and mass density ρ (mass divided by the volume) is v=b

ρ−−√ . what are the units of the bulk modulus in si base units?

1 answer:

PilotLPTM [1.2K]2 years ago

3 0

As it is given that Bulk modulus and density related to velocity of sound

$v = \sqrt{\frac{B}{\rho}}$

by rearranging the equation we can say

$B = \rho * v^2$

now we need to find the SI unit of Bulk modulus here

we can find it by plug in the units of density and speed here

$B = \frac{kg}{m^3} * (\frac{m}{s})^2$

so SI unit will be

$B = \frac{kg}{m* s^2}$

SO above is the SI unit of bulk Modulus

You might be interested in

Two friends, barbara and neil, are out rollerblading. with respect to the ground, barbara is skating due south at a speed of 5.9

Semmy [17]

<span>As seen by Barbara, Neil is traveling at a velocity of 6.1 m/s at and angle of 76.7 degrees north from due west. Let's assume that both Barbara and Neil start out at coordinate (0,0) and skate for exactly 1 second. Where do they end up? Barbara is going due south at 5.9 m/s, so she's at (0,-5.9) Neil is going due west at 1.4 m/s, so he's at (-1.4,0) Now to see Neil's relative motion to Barbara, compute a translation that will place Barbara back at (0,0) and apply that same translation to Neil. Adding (0,5.9) to their coordinates will do this. So the translated coordinates for Neil is now (-1.4, 5.9) and Barbara is at (0,0). The magnitude of Neil's velocity as seen by Barbara is sqrt((-1.4)^2 + 5.9^2) = sqrt(1.96 + 34.81) = sqrt(36.77) = 6.1 m/s The angle of his vector relative to due west will be atan(5.9/1.4) = atan(4.214285714) = 76.7 degrees So as seen by Barbara, Neil is traveling at a velocity of 6.1 m/s at and angle of 76.7 degrees north from due west.</span>

5 0

2 years ago

Read 2 more answers

Air has a density of 1.3 kg/m³. Calculate the mass of 36 m³ of air in kilograms. Give your answer to 1 decimal place.

vredina [299]

Answer:

46.8 kg

Explanation:

Mass = (density)(volume)

= (1.3)(36)

<u>M</u><u>a</u><u>s</u><u>s</u><u> </u><u>=</u><u> </u><u>4</u><u>6</u><u>.</u><u>8</u><u> </u><u>k</u><u>g</u>

3 0

2 years ago

Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two 1.8

Tema [17]

Answer:

The value of developed electric force is $3.516\times 10^{- 7} N$

Solution:

As per the question:

Mass of the droplet = 1.8 mg = $1.8\times 10^{- 6} kg$

Charge on droplet, Q = $25 pC = 25\times 10^{- 12} C$

Distance between the 2 droplets, D = 0.40 cm = 0.004 m

Now, the Electrostatic force given by Coulomb:

$F_{E} = \frac{1}{4\pi epsilon_{o}}.\frac{Q^{2}}{D^{2}}$

$\frac{1}{4\pi epsilon_{o}} = 9\times 10^{9} m/F$

$F_{E} = (9\times 10^{9}).\frac{(25\times 10^{- 12})^{2}}{0.004^{2}}$

$F_{E} = 3.516\times 10^{- 7} N$

The magnitude of force is too low to be noticed.

8 0

2 years ago

Read 2 more answers

Three different planet-star systems, which are far apart from one another, are shown above. The masses of the planets are much l

alex41 [277]

a) 4F0

b) Speed of planet B is the same as speed of planet A

Speed of planet C is twice the speed of planet A

Explanation:

a)

The magnitude of the gravitational force between two objects is given by the formula

$F=G\frac{m_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the masses of the 2 objects

r is the separation between the objects

For the system planet A - Star A, we have:

$m_1=M_p\\m_2 = M_s\\r=R$

So the force is

$F_A=G\frac{M_p M_s}{R^2}=F_0$

For the system planet B - Star B, we have:

$m_1 = 4 M_p\\m_2 = M_s\\r=R$

So the force is

$F=G\frac{4M_p M_s}{R^2}=4F_0$

So, the magnitude of the gravitational force exerted on planet B by star B is 4F0.

For the system planet C - Star C, we have:

$m_1 = M_p\\m_2 = 4M_s\\r=R$

So the force is

$F=G\frac{M_p (4M_s)}{R^2}=4F_0$

So, the magnitude of the gravitational force exerted on planet C by star C is 4F0.

b)

The gravitational force on the planet orbiting around the star is equal to the centripetal force, therefore we can write:

$G\frac{mM}{r^2}=m\frac{v^2}{r}$

where

m is the mass of the planet

M is the mass of the star

v is the tangential speed

We can re-arrange the equation solving for v, and we find an expression for the speed:

$v=\sqrt{\frac{GM}{r}}$

For System A,

$M=M_s\\r=R$

So the tangential speed is

$v_A=\sqrt{\frac{GM_s}{R}}$

For system B,

$M=M_s\\r=R$

So the tangential speed is

$v_B=\sqrt{\frac{GM_s}{R}}=v_A$

So, the speed of planet B is the same as planet A.

For system C,

$M=4M_s\\r=R$

So the tangential speed is

$v_C=\sqrt{\frac{G(4M_s)}{R}}=2(\sqrt{\frac{GM_s}{R}})=2v_A$

So, the speed of planet C is twice the speed of planet A.

3 0

2 years ago

Jenny and Alyssa are members of the cross-country team. On a training run, Jenny starts off and runs at a constant 3.8 m/s. Alys

guapka [62]

Answer:

285 seconds

Explanation:

Jenny speed is 3.8 m/s

Alyssa speed in 4.0 m/s

Alyssa starts after 15 seconds

Find the distance covered by Jenny, when Alyssa starts

Distance=Speed*time

Distance covered by Jenny in 15 seconds= 3.8×15=57m

Relative speed of the two members heading same direction will be;

4.0m/s-3.8m/s=0.2m/s

To find the time Alyssa catch up with Jenny you divide the distance to be covered by Alyssa by the relative speed of the two

Distance=57m, relative speed=0.2m/s t=57/0.2 =285 seconds

=4.75 minutes

5 0

2 years ago

Read 2 more answers