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beks73 [17]
2 years ago
7

From an electromagnetic wave the electric and magnetic field is what direction toward each other?

Physics
1 answer:
liubo4ka [24]2 years ago
3 0
What are the choices?
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A wave has a frequency of 875 hz and a wavelength of 352 m. At what speed is this wave traveling 
Sonbull [250]

Answer:

308,000 or 30.8×10^3

Explanation:

v=f×lamda

v is ?, f is 875Hz, lamda is 352m

v=875×352

v=308,000

v=30.8×10^3 m/s

5 0
2 years ago
I need guidance pls
seropon [69]
The slowest line is the solid line and the fastest is the dotted line that crosses the solid line
for future reference you just need to find the slope or the line which is traveling most vertical
8 0
2 years ago
For the following statements, choose the word or words inside the parentheses that serve to make a correct statement. Each state
AnnyKZ [126]

Answer:

a) Temperatura, b) Temperature, c)    Constant , d)  None of these , e) Gibbs enthalpy and free energy (G)

Explanation:

a) the expression for ideal gases is PV = nRT

     Temperature

b) The internal energy is E = K T

      Temperature

c)  S = ΔQ/T

In an isolated system ΔQ is zero, entropy  is constant

       Constant

d) all parameters change when changing status

        None of these

e) Gibbs enthalpy and free energy

7 0
2 years ago
.<br> What happens to the average kinetic energy of water molecules as water freezes?
leva [86]

Answer:molecules slow down

Explanation:

because science

6 0
3 years ago
Your friend decides to generate electrical power by rotating a 100,000 turn coil of wire around an axis in the plane of the coil
MakcuM [25]

Answer:

a) I=35mA

b) P=1.73W

Explanation:

a) The max emf obtained in a rotating coil of N turns is given by:

emf_{max}=NBA\omega

where N is the number of turns in the coil, B is the magnitude of the magnetic field, A is the area and w is the angular velocity of the coil.

By calculating A and replacing in the formula (1G=10^{-4}T) we get:

A=\pi r^2 =\pi(0.23m)^2=0.16m^2

emf_{max}=(100000)(0.3*10^{-4}T)(0.166m^2)(140\frac{rev}{s})=69.72V

Finally, the peak current is given by:

I=\frac{emf}{R}=\frac{69.72V}{1400\Omega}=49.8mA

b)

we have that

I_{rms}=\frac{I}{\sqrt{2}}=\frac{0.0498A}{\sqrt{2}}=0.035A

P_{rms}=I^2{rms}R=(0.035A)^2(1400\Omega)=1.73W

hope this helps!!

6 0
2 years ago
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