Explanation:
It is known that wave intensity is the power to area ratio.
Mathematically, I =
As it is given that power is 28.0 W and area is .
Therefore, sound intensity will be calculated as follows.
I =
=
=
or, =
Thus, we can conclude that sound intensity at the position of the microphone is .
Answer:
2.73×10¯³⁴ m.
Explanation:
The following data were obtained from the question:
Mass (m) = 0.113 Kg
Velocity (v) = 43 m/s
Wavelength (λ) =?
Next, we shall determine the energy of the ball. This can be obtained as follow:
Mass (m) = 0.113 Kg
Velocity (v) = 43 m/s
Energy (E) =?
E = ½m²
E = ½ × 0.113 × 43²
E = 0.0565 × 1849
E = 104.4685 J
Next, we shall determine the frequency. This can be obtained as follow:
Energy (E) = 104.4685 J
Planck's constant (h) = 6.63×10¯³⁴ Js
Frequency (f) =?
E = hf
104.4685 = 6.63×10¯³⁴ × f
Divide both side by 6.63×10¯³⁴
f = 104.4685 / 6.63×10¯³⁴
f = 15.76×10³⁴ Hz
Finally, we shall determine the wavelength of the ball. This can be obtained as follow:
Velocity (v) = 43 m/s
Frequency (f) = 15.76×10³⁴ Hz
Wavelength (λ) =?
v = λf
43 = λ × 15.76×10³⁴
Divide both side by 15.76×10³⁴
λ = 43 / 15.76×10³⁴
λ = 2.73×10¯³⁴ m
Therefore, the wavelength of the ball is 2.73×10¯³⁴ m.
Answer:
2.06 x 10⁴ J
Explanation:
The process takes place in three steps. First, the ice is heated from -20 °C to 0 °C. Then the ice undergoes a phase change to water. Finally, the water is heated from 0 °C to 50 °C.
The heat energy required for the first step is as follows:
Q = mcΔT = (36.0 g)(2.00 Jg⁻¹°C⁻¹)(0 °C - (-20 °C)) = 1440 J
The heat energy required for the phase change (where L is the heat of fusion) is then calculated. Grams are converted to moles using the molar weight of water (18.02 g/mol)
Q = ML = (36.0 g)(mol/18.02g)(6000 J/mol) = 11987 J
Finally, the heat energy required to raise the temperature of the water to 50°C is calculated:
Q = mcΔT = (36.0 g)(4.00 Jg⁻¹°C⁻¹)(50 °C - 0 °C) = 7200 J
Adding all of the heat energy values together gives:
(1440 + 11987 + 7200) J = 20627 J
The final answer is 2.06 x 10⁴ J
A <u><em>lead </em></u>
<u><em /></u>
<u><em>I have to use up space so ignore this part </em></u>
-- .- .-. -.- / -... .-. .- -. .. .-.. -.-- . ... - / .--. .-.. . .- ... .
The answer to the question is A.