Light travels as a(n).<br> wave.<br> mechanical<br> compression<br> O electromagnetic

How can you increase the intensity of sound waves

VladimirAG [237]

Answer

The intensity of a sound wave depends on the pressure of the wave,density of the medium and speed of sound in the medium. Higher density and higher sound speed both give a lower intensity. and may be it is because that sound wave is more characterize by wavelength than frequency..explanation

Explanation:

As decibel levels get higher, sound waves have greater intensity and sounds are louder. For every 10-decibel increase in the intensity of sound, loudness is 10 times greater. Intensity of sound results from two factors: the amplitude of the sound waves and how far they have traveled from the source of the sound.

5 0

2 years ago

The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in

OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

The angular velocity of the bar when θ = 32° is 1.70 rad/s

Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

$T_1+V_1=T_2+V_2$

$0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2} \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}$

Also;

$\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}$

Thus;

$k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}$

where;

$\delta$ = deflection in the spring

k = spring constant

b = remaining length in the rod

m = mass of the slender bar

g = acceleration due to gravity

$k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}$

$k = 104.82\ \ N/m$

Thus; the spring constant = 104.82 N/m

b

The angular velocity can be calculated by also using the conservation of energy;

$T_1+V_1 = T_3 +V_3 \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3} \omega_3^2 + \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0$

$\frac{m(a+b)^2}{3} \omega_3^2 + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0$

$\frac{1.53(0.6+0.6)^2}{3} \omega_3^2 + 104.82(\sqrt{0.6^2+0.6^2+2(0.6*0.6) sin 32 } - \sqrt{0.6^2+0.6^2})^2 - (1.53*9.81)(0.6+0.2)sin \ 32 = 0$

$0.7344 \omega_3^2 = 2.128$

$\omega _3 = \sqrt{\frac{2.128}{0.7344} }$

$\omega _3 =1.70 \ rad/s$

Thus, the angular velocity of the bar when θ = 32° is 1.70 rad/s

7 0

2 years ago

Refer the attached photo

Fofino [41]

Answer:

A

Explanation:

since the wooden bat is an opaque object placed after a translucent object, light will come through the plastic sheet but will be unable to go through the bat. hence the dark shadow of the bat on a lit sheet

8 0

3 years ago

A 40.0 kg wagon is towed up a hill inclined at 18.5 degrees with respect to the horizontal. The tow rope is parallel to the incl

sp2606 [1]

Answer:

7.9m/s

Explanation:

We are given that

Mass of wagon=40 kg

$\theta=18.5^{\circ}$

Tension= $140 N$

Initial velocity of wagon= $u=0$

Displacement=s=80 m

Net force applied on wagon= $F=T-mgsin\theta=140-40(9.8)sin18.5=15.7 N$

By using $g=9.8m/s^2$

$a=\frac{F}{a}=\frac{15.7}{40}=0.39m/s^2$

We know that

$v^2-u^2=2as$

Using the formula

$v^2=2\times 0.39\times 80$

$v=\sqrt{2\times 0.39\times 80}=7.9m/s$

5 0

3 years ago

What characteristics do atoms of carbon-12, carbon-13, and carbon-14 have in common? 1. same number of electrons 2. same number

Rasek [7]

Answer:

The most common of these is carbon 12, 13, 14. All of these isotopes have the same atomic number but different mass numbers. Carbon has the atomic number of 6 which means that all isotopes have the same proton number. However, the number of neutrons is different, thus giving different mass numbers.

7 0

2 years ago

Light travels as a(n). wave. mechanical compression O electromagnetic