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

Robin Hood wishes to split an arrow already in the bull's-eye of a target 40 m away.

tamaranim1 [39]

Answer:

5.843 m

Explanation:

suppose that the arrow leave the bow with a horizontal speed , towards he bull's eye.

lets consider that horizontal motion

distance = speed * time

time = 40/ 37 = 1.081 s

arrow doesnot have a initial vertical velocity component. but it has a vertical motion due to gravity , which may cause a miss of the target.

applying motion equation

(assume g = 10 m/s²)

$s=ut+\frac{1}{2}*gt^{2} \\= 0+\frac{1}{2}*10*1.081^{2}\\= 5.843 m$

Arrow misses the target by 5.843m ig the arrow us split horizontally

4 0

2 years ago

Which of these discoveries contradict components of Dalton’s atomic theory? Check all of the boxes that apply. Atoms contain sma

EastWind [94]

Explanation:

According to Dalton's atomic theory, all the atoms are individual, all the atoms of the same element are identical in properties and mass, the compound is formed from two or more kinds of the atoms, all the matter is made up of small atoms and the chemical reaction is a rearrangement of the atoms.

The discoveries which contradicts the components of Dalton's atomic theory from the given discoveries are:

Nuclear reactions can change an atom of one element into an atom of another element.

Atoms of a given element can have different numbers of neutrons.

Atoms contain smaller particles: protons, neutrons, and electrons.

4 0

2 years ago

Read 2 more answers

Consider a string of length 1.0 meter, fixed at both ends, with mass 100 grams and tension 100 newtons. part a give the number o

Bond [772]

To answer the problem we would be using this formula which isv = sqrt(T/(m/L))
v = sqrt(100 N / [(0.100 kg)/(1.0 m)])
v = 31.62 m/s
v = fλ
31.62 m/s = (95 Hz)(λ)
λ = 0.333 m
For every wavelength along a string there will be 2 antinodes.
1.0 m / 0.333 m = 3
3 * 2 = 6 antinodes
6 + 1 = 7 nodes

4 0

2 years ago

Eac of the two Straight Parallel Lines Each of two very long, straight, parallel lines carries a positive charge of 24.00 m C/m.

Cloud [144]

Answer:

The magnitude of the electric field at a point equidistant from the lines is $4.08\times10^{5}\ N/C$

Explanation:

Given that,

Positive charge = 24.00 μC/m

Distance = 4.10 m

We need to calculate the angle

Using formula of angle

$\theta=\sin^{-1}(\dfrac{\dfrac{d}{2}}{2d})$

$\theta=\sin^{-1}(\dfrac{1}{4})$

$\theta=14.47^{\circ}$

We need to calculate the magnitude of the electric field at a point equidistant from the lines

Using formula of electric field

$E=\dfrac{2k\lambda}{r}\times2\cos\theat$

Put the value into the formula

$E=\dfrac{2\times9\times10^{9}\times24.00\times2\times10^{-6}\cos14.47}{2.05}$

$E=408094.00\ N/C$

$E=4.08\times10^{5}\ N/C$

Hence, The magnitude of the electric field at a point equidistant from the lines is $4.08\times10^{5}\ N/C$

6 0

2 years ago

An 80-kg person stands at one end of a 130-kg boat. He then walks to the other end of the boat so that the boat moves 80 cm with

a_sh-v [17]

Explanation:

We assume that length of boat is L cm. And, we suppose to have a coordinate system relative to the bottom of the lake with:

Boat's position initially from x = 0 cm to x = L cm

Let person (S) is initially at x = L cm (right hand end of boat)

and, let boat's center of mass at x = C cm

Therefore, initially center of mass of overall (system's) is at $X_{1}$ where (taking moments about x = 0).

$(80 + 130)X_{1}$

= 130C + 80L ........... (i)

After S walks left to the other end of the boat, the boat moves right 80 cm.

Hence,

Boat's position is now from x = 80 cm to x = (L+ 80) cm

S's position is x = 80 (left end of boat)

Boat's center of mass is at (C + 80).

Therefore, overall (system's) final center of mass is at $X_{2}$ where (taking moments about x = 0).

$(80 + 130)X_{2}$

= $(130)(C + 80) + 80 \times 80$ ............. (ii)

As no external force is acted on the system, the system's center of mass has not moved, so $X_{1} = X_{2}$ .

This means the left hand sides of both equations (i) and (ii) are equal. So, we can equate the right hand sides giving: the equation as follows.

130C + 80L = $(130)(C+80) + 80 \times 80
$

80L = 16800 cm

L = 210 cm

= 2.1 m

Length of boat = 2.1 m

When the person is walking from one end to the other as there was no external force then center of mass of the person-boat system did not move.

6 0

2 years ago