Answer:
7.78x10^-8T
Explanation:
The Pointing Vector S is
S = (1/μ0) E × B
at any instant, where S, E, and B are vectors. Since E and B are always perpendicular in an EM wave,
S = (1/μ0) E B
where S, E and B are magnitudes. The average value of the Pointing Vector is
<S> = [1/(2 μ0)] E0 B0
where E0 and B0 are amplitudes. (This can be derived by finding the rms value of a sinusoidal wave over an integer number of wavelengths.)
Also at any instant,
E = c B
where E and B are magnitudes, so it must also be true at the instant of peak values
E0 = c B0
Substituting for E0,
<S> = [1/(2 μ0)] (c B0) B0 = [c/(2 μ0)] (B0)²
Solve for B0.
Bo = √ (0.724x2x4πx10^-7/ 3 x10^8)
= 7.79 x10 ^-8 T
It is either a person who studies Linguistics or
a person skilled in foreign languages
Density formula is:
D = m / V
Volume of the matchbox car:
V = 46 ml - 41 ml = 5 ml = 0.005 l = 0.005 dm³
D = 10 g / 0.005 dm³
D = 2000 g / dm³ = 2 kg / dm³ = 2 × 10 ^(-3) kg/m³
Hello,
<span>A police car parked on the side of the highway emits a 1200 Hz sound that bounces off a vehicle farther down the highway and returns with a frequency of 1250 Hz.
How fast is the vehicle going?
Doppler equation formula: </span>ƒL = ƒS(v - vL)/(v - vS)
The wave returns with a frequency of 1250 Hz, the <span>echo frequency is higher; the car must be traveling towards the police car.
</span><span>The wave echo is coming back towards the police car at the same speed as the sound wave travels towards the moving car so t</span><span>he relative speed between the cars is half of the speed of the echo.
* </span><span>speed of sound equals about 337 m/s </span>
2v / 337 = (1250/1200) - 1
<span>2v = 14.04 m/s </span>
<span>v = 7.02 m/s
</span>
Thus, the vehicle is going 7.02 m/s.
Faith xoxo
Answer:
45.8 cm
Explanation:
To solve this, we will use the formula
5 / x² = 7/(1 - x)²
5 / x² = 7 / (1 - 2x + x²)
5 / 7 = x² / (1 - 2x + x²)
x = 0.5 * (√(35) - 5) meters
x = 0.5 * (5.916 - 5)
x = 0.5 * (0.916)
x = 0.458 or x = 45.8
