Answer:
Yes
Explanation:
Yes because Electromagnetic waves differ from mechanical waves in that they do not require a medium to propagate. This means that electromagnetic waves can travel not only through air and solid materials, but also through the vacuum of space.
Answer:
(a). = 5.43
(b). 13
(C) 0
Explanation:
part A:
φ(outside) = ∬B(outside) dS
note that;
φ(sum) = 2[φ(outside) + φ(inside).
then, we say
φ(outside) = ∫ μI/ 2πr dr (taking boundaries at R - d/2 and d/2.
μI/2πr dr= μI/2πr ln 2R- d/d
= 1.26× 1) ^-6 ×10 ln 40-2.5/2.5.
= 5.43 μWb.
magnetic flux in the conductor can be calculated from magnetic induction as integral of d/2 to R
please note that dS= 1. dr
φ (inside) = B(inside) dS
∫2μIr/ π d^2 dV.
μI/πd2 r^2 (at boundary d/2 and 0)
μI/4π
= 1.26×10^-6 ×10/4π.
= 1.0032 μWb .
where B(inside) = μI(inside)/ 2πr and I= Ir^2/ (d/2) ^2.
φ(sum)= φ(outside + inside)
=2(5.42 + 1.0032
= 13μWb
(c). is zero
Answer:
it will be zero degree
the angle of reflection is zero on hemisphere sides and angle of i is equal to angle of r
Answer:
a= 4.14 m/s²
Explanation
We calculate the weight component parallel to the displacement of the block:
We define the x-axis in the direction of the inclined plane , 25° to the horizontal.
W= m*g : Total block weight
Wx= W*sen25°= m*g* sen25°
We apply Newton's second law :
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Problem development
We apply the formula (1) to calculate the acceleration of the block:
∑Fx = m*a
Wx = m*a
m*g* sen25° = m*a : We divide by m on both sides of the equation
g* sen25° = a
a = g* sen25° = 9.8* sen25° = 4.14 m/s²