Two point charges are separated by 25.0 cm (see (Figure 1)). Assume that q1 = -7.50 nC and q2 = -10.5 nC.

The energy flow per unit time per unit area (S) of an electromagnetic wave has an average value of 310 mW/m2. The maximum value

Aleksandr-060686 [28]

Answer:

5.096*10^-8

Explanation:

Given that

The average value of the electromagnetic wave is 310 mW/m²

To find the maximum value of the magnetic field the wave is closest to, we say

Emax = √Erms

Emax = √[(2 * 0.310 * 3*10^8 * 4π*10^-7)]

Emax = √233.7648

Emax = 15.289

Now, with our value of maximum electromagnetic wave gotten, we divide it by speed of light to get our final answer

15.289 / (3*10^8) = 5.096*10^-8 T

Suffice to say, The maximum value of the magnetic field in the wave is closest to 5.096*10^-8

6 0

2 years ago

A thermometer is placed into a flask of water with pieces of ice and reads 273.15. Which temperature scale is being used? A. Kel

zzz [600]

A. Kelvin. It is 32 degrees fahrenheit

6 0

2 years ago

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When one person shouts at a football game, the sound intensity level at the center of the field is 58.4 dB. When all the people

Tanzania [10]

Answer:

The number of people at game are approximately 22909

Explanation:

Given data

When one person shout $\beta _{1}=58.4dB$

When n number of person shout together $\beta _{n}=102dB$

The sound intensity level during one person shout is given by:

$\beta _{1}=10log(\frac{I_{1}}{I_{o}} )\\58.4=10log(\frac{I_{1}}{I_{o}} )\\5.84=log(\frac{I_{1}}{I_{o}} )\\\frac{I_{1}}{I_{o}} =10^{5.84}\\I_{1}=10^{5.84}*I_{o}$

The sound intensity level during n number of person shout is given by:

$\beta _{n}=10log(\frac{I_{n}}{I_{o}} )\\102=10log(\frac{I_{n}}{I_{o}} )\\10.2=log(\frac{I_{n}}{I_{o}} )\\\frac{I_{n}}{I_{o}}=10^{10.2}\\I_{n}=10^{10.2}*I_{o}$