Explanation:
Given:
v₀ = 0 m/s
v = 49 m/s
a = 9.8 m/s²
Find: t
v = at + v₀
49 m/s = (9.8 m/s²) t + 0 m/s
t = 5 s
Answer:
The intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².
Explanation:
Given;
intensity of the sound level, dB = 60 dB
The intensity of the sound in W/m² is calculated as;
![dB = 10 Log[\frac{I}{I_o} ]\\\\](https://tex.z-dn.net/?f=dB%20%3D%2010%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5C)
where;
I₀ is threshold of hearing = 1 x 10⁻¹² W/m²
I is intensity of the sound in W/m²
Substitute the given values and for I;
![dB = 10 Log[\frac{I}{I_o} ]\\\\60 = 10 Log[\frac{I}{I_o} ]\\\\6 = Log[\frac{I}{I_o} ]\\\\10^6 = \frac{I}{I_o} \\\\I = 10^6 \ \times \ I_o\\\\I = 10^6 \ \times \ 1^{-12} \ W/m^2 \\\\I = 1\ \times \ 10^{-6} \ W/m^2](https://tex.z-dn.net/?f=dB%20%3D%2010%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5C60%20%3D%2010%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5C6%20%3D%20%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5C10%5E6%20%3D%20%5Cfrac%7BI%7D%7BI_o%7D%20%5C%5C%5C%5CI%20%3D%2010%5E6%20%5C%20%5Ctimes%20%5C%20I_o%5C%5C%5C%5CI%20%3D%2010%5E6%20%5C%20%5Ctimes%20%5C%201%5E%7B-12%7D%20%5C%20W%2Fm%5E2%20%5C%5C%5C%5CI%20%3D%201%5C%20%5Ctimes%20%5C%2010%5E%7B-6%7D%20%5C%20W%2Fm%5E2)
Therefore, the intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².
Answer:
Explanation:
la frecuencia = ω/2π, nada cambio
v(max) = ωA → ω2Α = 2ωA duplicara velocidad máxima
a(max) = ω²Α → ω²2Α = 2ω²Α duplicara la aceleración máxima
la energía total ½kA² → ½k(2Α)² = 4(½kA²) cuatro veces la energía
Answer: higher and lower
Explanation:
charge in an electric field will experience a force in the direction of decreasing potential energy. Since the electric potential energy of a negative charge is equal to the charge times the electric potential the direction of decreasing electric potential energy is the direction of increasing electric potential.
