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
Answer:An infrared sensor system is set up so that two sensors start timing when the infrared ... Determine the average acceleration during each time interval. c. ... the ground with a an initial velocity of 169 m/s at an angle of 23.0 O above the horizontal; a. ... A 4.00 kg mass is sitting at rest on a horizontal surface.
Explanation:
Answer:
Explanation:
In an electric field E force on charge q
F = Eq , acceleration a = Eq / m
a = 664 x 1.6 x 10⁻¹⁹ / 1.67 x 10⁻²⁷
= 636.16 x 10⁸ m /s²
b )
initial velocity u = 0
final velocity v = 1.46 x 10⁶ m/s
v = u + at
1.46 x 10⁶ = 0 + 636.16 x 10⁸ x t
t = 2.29 x 10⁻⁵ s
c )
s = ut + 1/2 a t²
= 0 + .5 x 636.16 x 10⁸ x ( 2.29 x 10⁻⁵ )²
= 1668 x 10⁻²
= 16.68 m
d )
Kinetic energy = 1/2 m v²
= .5 x 1.67 x 10⁻²⁷ x ( 1.46 x 10⁶ )²
= 1.78 x 10⁻¹⁵ J .
Answer:
B.The value k is constant for each of the eight planets in our solar system.
Explanation:
Kepler's third law states that the square of orbital period of the planet is proportional to the cube of semi-major axis.
P² = k A³
k is constant for all the planets in the solar system.
Where, G is the gravitational constant and M is the mass of the Sun.
The orbital period is measured in units of time, and the semi-major axis is measured in units of distance.
With increase in length of the semi-major axis, the orbital period increases.
The value of k can be found by: P²÷A³.
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