How do we gather most of the information we get about the universe around us?

a uniform rod is hung at onen end and is partially submerged in water. If the density of the rod is 5/9 than of wter, find the f

34kurt

Answer:

$\frac{h_{liquid} }{ h_{body} }$ = 5/9

Explanation:

This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.

B = ρ_liquid g V_liquid

let's write the translational equilibrium condition

B - W = 0

let's use the definition of density

ρ_body = m / V_body

m = ρ_body V_body

W = ρ_body V_body g

we substitute

ρ_liquid g V_liquid = ρ_body g V_body

$\frac{\rho_{body} }{\rho_{liquid} } } = \frac{V_{liquid} }{V_{body} } }$

In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar

V = A h_bogy

Thus

$\frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }$

we substitute

5/9 = $\frac{h_{liquid} }{ h_{body} }$

8 0

2 years ago

Which of these help create radio waves?

inysia [295]

The one that help create radio waves is :
Changing electric and magnetic fields applied at right angles

Radio waves are transverse wave, which means that the oscillations occurring perpendicular to the direction of energy transfer

hope this helps

3 0

3 years ago

How does erosion happen

Zarrin [17]

Erosion happens when rocks and sediments are picked up and moved to another place by ice, water, wind or gravity.

5 0

3 years ago

2. Two long parallel wires each carry a current of 5.0 A directed to the east. The two wires are separated by 8.0 cm. What is th

Gre4nikov [31]

Answer:

The magnitude of magnetic field at given point = $5.33$ × $10^{-5}$ T

Explanation:

Given :

Current passing through both wires = 5.0 A

Separation between both wires = 8.0 cm

We have to find magnetic field at a point which is 5 cm from any of wires.

From biot savert law,

We know the magnetic field due to long parallel wires.

⇒ $B = \frac{\mu_{0}i }{2\pi R}$

Where $B =$ magnetic field due to long wires, $\mu_{0} =$ $4\pi \times10^{-7}$ , $R =$ perpendicular distance from wire to given point

From any one wire $R_{1} =$ 5 cm, $R_{2} =$ 3 cm

so we write,

∴ $B = B_{1} + B_{2}$

$B = \frac{\mu_{0} i}{2\pi R_{1} } + \frac{\mu_{0} i}{2\pi R_{2} }$

$B =\frac{ 4\pi \times10^{-7} \times5}{2\pi } [\frac{1}{0.03} + \frac{1}{0.05} ]$

$B = 5.33\times10^{-5} T$

Therefore, the magnitude of magnetic field at given point = $5.33\times10^{-5} T$

3 0

3 years ago

Two sound waves have equal displacement amplitudes, but wave 1 has two-thirds the frequency of wave 2. What is the ratio of the

zlopas [31]

Answer:

$\dfrac{I_1}{I_2}=\dfrac{4}{9}$

Explanation:

c = Speed of wave

$\rho$ = Density of medium

A = Area

$\nu$ = Frequency

$\nu_1=\dfrac{2}{3}\nu_2$

Intensity of sound is given by

$I=\dfrac{1}{2}\rho c(A\omega)^2\\\Rightarrow I=\dfrac{1}{2}\rho c(A2\pi \nu)^2$

So,

$I\propto \nu^2$

We get

$\dfrac{I_1}{I_2}=\dfrac{\nu_1^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{\dfrac{2}{3}^2\nu_2^2}{\nu_2^2}\\\Rightarrow \dfrac{I_1}{I_2}=\dfrac{4}{9}$

The ratio is $\dfrac{I_1}{I_2}=\dfrac{4}{9}$

8 0

2 years ago