Answer:
16∠45° Ω
Explanation:
Applying,
Z = V/I................... Equation 1
Where Z = Impedance, V = Voltage output, I = current input.
Given: V = 120cos(10t+75°), = 120∠75°, I = 7.5cos(10t+30) = 7.5∠30°
Substitute these values into equation 1
Z = 120cos(10t+75°)/7.5cos(10t+30)
Z = 120∠75°/ 7.5∠30°
Z = 16∠(75°-30)
Z = 16∠45° Ω
Hence the impedance of the linear network is 16∠45° Ω
Answer:
Gas Bubbles Appear,
Formation of a Precipitate,
Color Change,
Temperature Change,
Production of Light,
Volume Change,
Change in Smell or Taste
Explanation:
<span> the tangent function of 2mi/5mi = 21.8 degrees
cosine of that angle = cos (21.8 degrees)
=0.928
multiply 0.928 with 570 = 570*0.928
Closing velocity = 529.23 mph
</span><span>As the plane gets closer to radar station, tangent function becomes infinity and the cosine function becomes zero.....
</span>0*570 = 0
Complete Question: A +10 nC charge is located at the origin. What is the electric field at the position (x2,y2)= (-5.0cm, 0cm)? Write electric field vector in component form.
E2x, E2y=? N/C
Answer:
Ex = -3.6*10⁴ N/C Ey=0
Explanation:
As the charge producing the field is positive, the direction of the field, which is the one that would take a positive test charge located in the point of interest, will be away from this point, pointing to the left.
If we choose the positive direction to be to the right, the electric field component along the x-axis will be negative.
The magnitude of the field can be obtained applying the electric field definition, and the Coulomb's Law to the charge in the origin, as follows:
E = k*q/r² = 9*10⁹N*m²/C²*10⁻⁸C/(0.05)² m² = 3.6*10⁴ N/C
As the electric field follows the same line as the electric force, it has only component on the x axis, so:
Ex = -3.6*10⁴ N/C
Ey = 0
