Answer:
Explanation:
Given the following data;
Original volume = V
New volume = V'
Original temperature = T
New temperature = T'
To write an expression for Charles's law;
Charles states that when the pressure of an ideal gas is kept constant, the volume of the gas is directly proportional to the absolute temperature of the gas.
Mathematically, Charles law is given by the formula;
Echo sounding is a type of SONAR used to determine the depth of water by transmitting sound pulses into water. The time interval between emission and return of a pulse is recorded, which is used to determine the depth of water along with the speed of sound in water at the time.
Answer:
The maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J
Explanation:
Given that,
dielectric constant k = 5.5
the area of each plate, A = 0.034 m²
separating distance, d = 2.0 mm = 2 x 10⁻³ m
magnitude of the electric field = 200 kN/C
Capacitance of the capacitor is calculated as follows;
![C = \frac{k \epsilon A}{d} = \frac{5.5*8.85*10^{-12}*0.034}{2*10^{-3}} = 8.275 *10^{-10} \ F](https://tex.z-dn.net/?f=C%20%3D%20%5Cfrac%7Bk%20%5Cepsilon%20A%7D%7Bd%7D%20%3D%20%5Cfrac%7B5.5%2A8.85%2A10%5E%7B-12%7D%2A0.034%7D%7B2%2A10%5E%7B-3%7D%7D%20%3D%208.275%20%2A10%5E%7B-10%7D%20%5C%20F)
Maximum potential difference:
V = E x d
V = 200000 x 2 x 10⁻³ = 400 V
Maximum energy that can be stored in the capacitor:
E = ¹/₂CV²
E = ¹/₂ x 8.275 x 10⁻¹⁰ x (400)²
E = 6.62 x 10⁻⁵ J
Therefore, the maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J
the focal length <span> is much more decent for a concave, and also worse</span><span> for a convex mirror. When the image that is given, distance is good and decent, images are always on the same area of the mirror as the object given , and it is not fake. images distance is </span>never positive <span>, the image is on the oppisite side of the mirror, so the image must be virtual.</span>
A) 750 m
First of all, let's find the wavelength of the microwave. We have
is the frequency
is the speed of light
So the wavelength of the beam is
![\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m](https://tex.z-dn.net/?f=%5Clambda%3D%5Cfrac%7Bc%7D%7Bf%7D%3D%5Cfrac%7B3%5Ccdot%2010%5E8%20m%2Fs%7D%7B12%5Ccdot%2010%5E9%20Hz%7D%3D0.025%20m)
Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:
![y=\frac{m\lambda D}{a}](https://tex.z-dn.net/?f=y%3D%5Cfrac%7Bm%5Clambda%20D%7D%7Ba%7D)
where
m = 1 since we are interested only in the central fringe
D = 30 km = 30,000 m
a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)
Substituting, we find
![y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m](https://tex.z-dn.net/?f=y%3D%5Cfrac%7B%281%29%280.025%20m%29%2830000%20m%29%7D%7B2.0%20m%7D%3D375%20m)
and so, the diameter is
![d=2y = 750 m](https://tex.z-dn.net/?f=d%3D2y%20%3D%20750%20m)
B) 0.23 W/m^2
First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is
![r=\frac{750 m}{2}=375 m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7B750%20m%7D%7B2%7D%3D375%20m)
So the area is
![A=\pi r^2 = \pi (375 m)^2=4.42\cdot 10^5 m^2](https://tex.z-dn.net/?f=A%3D%5Cpi%20r%5E2%20%3D%20%5Cpi%20%28375%20m%29%5E2%3D4.42%5Ccdot%2010%5E5%20m%5E2)
And since the power is
![P=100 kW = 1\cdot 10^5 W](https://tex.z-dn.net/?f=P%3D100%20kW%20%3D%201%5Ccdot%2010%5E5%20W)
The average intensity is
![I=\frac{P}{A}=\frac{1\cdot 10^5 W}{4.42\cdot 10^5 m^2}=0.23 W/m^2](https://tex.z-dn.net/?f=I%3D%5Cfrac%7BP%7D%7BA%7D%3D%5Cfrac%7B1%5Ccdot%2010%5E5%20W%7D%7B4.42%5Ccdot%2010%5E5%20m%5E2%7D%3D0.23%20W%2Fm%5E2)