If the wavelength<span> is given, the energy can be determined by first using the wave equation (c = λ × ν) to </span>find<span> the frequency, then using Planck's equation to </span>calculate<span> energy. Use the equations above to answer the following questions. 1. Ultraviolet radiation has a frequency of 6.8 × 1015 1/s.</span>
Answer:
A. evaporation
Explanation:
Evaporation is a surface phenomenon as the molecules of the surface gets sufficient energy to overcome the force of attraction which will help in converting to the vapor phase.
Answer:
71.19 C
Explanation:
25C = 25 + 273 = 298 K
Applying the ideal gas equation we have
![\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1V_1%7D%7BT_1%7D%20%3D%20%5Cfrac%7BP_2V_2%7D%7BT_2%7D)
where P, V and T are the pressure, volume and temperature of the gas at 1st and 2nd stage, respectively. We can solve for the temperature and the 2nd stage:
![T_2 = T_1\frac{P_2V_2}{P_1V_1} = 298\frac{0.77*1.8}{1.2*1} = 298*1.155 = 344.19 K = 344.19 - 273 = 71.19 C](https://tex.z-dn.net/?f=T_2%20%3D%20T_1%5Cfrac%7BP_2V_2%7D%7BP_1V_1%7D%20%3D%20298%5Cfrac%7B0.77%2A1.8%7D%7B1.2%2A1%7D%20%3D%20298%2A1.155%20%3D%20344.19%20K%20%3D%20344.19%20-%20273%20%3D%2071.19%20C)
Answer:
a) T = 2.26 N, b) v = 1.68 m / s
Explanation:
We use Newton's second law
Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string
sin 30 =
cos 30 =
Tₓ = T sin 30
T_y = T cos 30
Y axis
T_y -W = 0
T cos 30 = mg (1)
X axis
Tₓ = m a
they relate it is centripetal
a = v² / r
we substitute
T sin 30 = m
(2)
a) we substitute in 1
T =
T =
T = 2.26 N
b) from equation 2
v² =
If we know the length of the string
sin 30 = r / L
r = L sin 30
we substitute
v² =
v² =
For the problem let us take L = 1 m
let's calculate
v =
v = 1.68 m / s
Answer:
a) 2.063*10^-4
b) 1.75*10^-4
Explanation:
Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:
![A=\frac{\pi }{4}d^{2} \\=\frac{\pi }{4}*(1.628*10^-3 m)^2\\=2.082*10^-6 m^2\\](https://tex.z-dn.net/?f=A%3D%5Cfrac%7B%5Cpi%20%7D%7B4%7Dd%5E%7B2%7D%20%20%5C%5C%3D%5Cfrac%7B%5Cpi%20%7D%7B4%7D%2A%281.628%2A10%5E-3%20m%29%5E2%5C%5C%3D2.082%2A10%5E-6%20m%5E2%5C%5C)
a) <em>The Potential difference across a 2.00 in length of a 14-gauge copper </em>
<em> wire: </em>
L= 2.00 m
From Table Copper Resistivity
= 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:
![R=\frac{pL}{A}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7BpL%7D%7BA%7D)
=0.0165Ω
The Potential difference across the copper wire is:
V=IR
=2.063*10^-4
b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m
The Resistance of the Silver wire is:
![R=\frac{pL}{A}](https://tex.z-dn.net/?f=R%3D%5Cfrac%7BpL%7D%7BA%7D)
=0.014Ω
The Potential difference across the Silver wire is:
V=IR
=1.75*10^-4