Answer:
B = 1.353 x 10⁻³ T
Explanation:
The Magnetic field within a toroid is given by
B = μ₀ NI/2πr, where N is the number of turns of the wire, μ₀ is the permeability of free space, I is the current in each turn and r is the distance at which the magnetic field is to be determined from the center of the toroid.
To find r we need to add the inner radius and outer radius and divide the value by 2. Hence,
r = (a + b)/2, where a is the inner radius and b is the outer radius which can be found by adding the length of a square section to the inner radius.
b = 25.1 + 3 = 28.1 cm
a = 25.1 cm
r = (25.1 + 28.1)/2 = 26.6 cm = 0.266m
B = 4π x 10⁻⁷ x 600 x 3/2π x 0.266
B = 1.353 x 10⁻³ T
The strength of the magnetic field at the center of the square cross section is 1.3 x 10⁻³ T
W = Fd
W = 1225 N x 10 m = 12250
T is in seconds (s)
<span>2pi is dimensionless </span>
<span>L is in meters (m) </span>
<span>g is in meters per second squared (m/s^2) </span>
<span>so you can write the equation for the period of the simple pendulum in its units... </span>
<span>s=sqrt(m/(m/s^2)) </span>
<span>simplify</span>
<span>s=sqrt(m*s^2*1/m) cancelling the m's </span>
<span>s=sqrt(s^2) </span>
<span>s=s </span>
<span>therefore the dimensions on the left side of the equation are equal to the dimensions on the right side of the equation.</span>
I can not solve the problem if I do not have the mass.
Answer:
n1 sin θ1 = n2 sin θ2 Snell's Law (θ1 is the angle of incidence)
sin θ2 = n1 / n2 * sin θ1
sin θ2 = 2.4 / 1.33 * sin θ1
sin θ2 = 1.80 * .407 = .734
θ2 = 47.2 deg
