Answer:
option D
Explanation:
given,
Intensity of sound = 20 dB
distance = 15 m
intensity of sound is increased to = 50 dB
distance between the sound level = ?
Using relation
L₁ = 20 dB L₂ = 50 dB r₁ = 15 m r₂ = ?
r₂ = 0.47 m
r₂ = 47 cm
hence, the correct answer is option D
Answer:
Explanation:
en un movimiento sísmico?
As altitude increases, temperature increases.
The stratosphere is the part of the atmosphere that starts in the tropopause and ends in the estratopause. In the troposphere, the air is close to the Earth surface. The air surface can absorb more sunlight energy than the air, so the Earth surface heats the air. As you go higher, the distance to the Earth surface is higher, so the temperature is lower. The troposphere ends in the tropopause, where this trend changes. In the estratopause, there is a lot of ozone, which absorbs the dangerous UV radiation and converts into heat. That heat warms the air. So the air which is close to the estratopause is warm because of the heat released by the ozone reactions. The tropopause is far from the Earth surface and far from the ozone layer, that’s why it is cold. So the tropopause is cold and the estratopause is warm, which means: the air becomes warmer <span>as you rise above the tropopause until you get to the estratopause.</span>
Answer:
Young's modulus (Y) = 3.56×10^11 N/m^2
The speed of sound in tungsten = 6166.4 m/s
Explanation:
Young's modulus (Y) = stress/strain
Stress = force/area
Force = mg = 30×9.8 = 294 N
Area = 1.5 × 2.6 = 3.9 mm^2 = 3.9/10^6 = 3.9×10^-6 m^2
Stress = 294/3.9×10^-6 = 7.54×10^7 N/m^2
Strain = extension/length
Extension = 0.000594 m
Length = 2.8 m
Strain = 0.000594/2.8 = 2.12×10^-4
Y = 7.54×10^7/2.12×10^-4 = 3.56×10^11 N/m^2
Y = h × rho × g
rho = 18.7 g/cm^3 = 18.7 g/cm^3 × 1 kg/1000 g × (100 cm/1 m)^3 = 18,700 kg/m^3
h = 3.56×10^11/(18,700×9.8) = 1.94×10^6 m
From the equations of motion
v^2 = u^2 + 2gh =
Initial speed (u) = 0 m/s
v = sqrt (2×9.8×1.94×10^6)
v = 6166.4 m/s
