Answer:
Radius of the loop is 0.18 m or 18 cm
Explanation:
Given :
Current flowing through the wire, I = 45 A
Magnetic field at the center of the wire, B = 1.50 x 10⁻⁴ T
Number of turns in circular wire, N = 1
Consider R be the radius of the circular wire.
The magnetic field at the center of the current carrying circular wire is determine by the relation:
Here μ₀ is vacuum permeability constant and its value is 4π x 10⁻⁷ Tm/A.
Substitute the suitable values in the above equation.
R = 0.18 m
"The proton and neutron have nothing to do with the isotope little billy"
It must be a virtual image, because this is the only kind of image it can produce.
During upward projection the final velocity is zero, and the gravitational acceleration is -10 m/s² (against the gravity).
Therefore; using the equation;
S = 1/2gt² + ut
Where s is the height h, g is gravitational acceleration, and t is the time and u is the initial velocity u, is 16 ft/s.
Thus; h= 1/2(-10)t² + 16t
We get; h = -5t² + 16t
Therefore; the quadratic equation is 5t² - 16t + h =0
Answer:
I think D sorry if I'm wrong
