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
Explanation:
Make a table, listing the x and y coordinates of each square's center of gravity and its mass. Multiply the coordinates by the mass, add the results for each x and y, then divide by the total mass.

The x-coordinate of the center of gravity is 15/14 a.
The y-coordinate of the center of gravity is 47/42 a.
Answer:
a) The centripetal acceleration of the car is 0.68 m/s²
b) The force that maintains circular motion is 940.03 N.
c) The minimum coefficient of static friction between the tires and the road is 0.069.
Explanation:
a) The centripetal acceleration of the car can be found using the following equation:

Where:
v: is the velocity of the car = 51.1 km/h
r: is the radius = 2.95x10² m

Hence, the centripetal acceleration of the car is 0.68 m/s².
b) The force that maintains circular motion is the centripetal force:

Where:
m: is the mass of the car
The mass is given by:

Where P is the weight of the car = 13561 N

Now, the centripetal force is:

Then, the force that maintains circular motion is 940.03 N.
c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:



Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.
I hope it helps you!