#7 Please answer question correctly showing work for thumbs up.
#7 Please answer question correctly showing work for thumbs up. If a-2.1 mm, b = 3.1...
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11 Looking back at the figure fr problem §13.29, determine the electric eld at the location of charge q. Then, get the force on the charge q using F qE. What will the force on q be if it were -24 nC, or 200 nC? The field calculation makes it easy to get the force on any charge placed at that position 9. If a 3.0 mm, b 4.0 mm, Q1 60 nC, Q2-80...
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13.2 Electric Force. Coulomb's Law Coulomb's Lau Firi a F12= 41 o 2f12-Note that Fi2 is the force on l by 2; and 12-71-72 points from 2 to 1.fi 7 is the unit vector in the direction of In general, for multiple di rete charges, net one on char e 1, F -4 (Pairwise vector sum; superposition principle). For convenience, ke = 9 x 109 appropriate units. Think How is N3L applicable here? Σ in...
Homework #2 Problem 20.46-MC 5 of 7 Part B Review Constants A negative point-charge Q1 is 10 cm direcly below a positive point-charge Q2 as shown in (Figure 1). If Q1 =-5.0 nC and Q2 = 1.5 nC , what is the magnitude of the net electric field created by the two charges at the point P? Express your answer using three significant figures. View Available Hint(s) nel N/C Submit Figure 1 of 1 X Incorrect; Try Again 5 cm...
Homework #2 Problem 20.46-MC 5 of 7 Part B Review Constants A negative point-charge Q1 is 10 cm direcly below a positive point-charge Q2 as shown in (Figure 1). If Q1 =-5.0 nC and Q2 = 1.5 nC , what is the magnitude of the net electric field created by the two charges at the point P? Express your answer using three significant figures. View Available Hint(s) nel N/C Submit Figure 1 of 1 X Incorrect; Try Again 5 cm...
1) Please answer full question (a-c) showing work. Ensure answer is correct for thumbs up. a) If the FFT of an 1000-Hz square wave was measured to be approximately -1.8 dBV. What is the corresponding rms value? b) The third harmonic of the square wave was measured to be approximately -11.34 dBV. What is the corresponding rms value? c) The output voltage across an 8-Ω speaker was measured as 1-kHz sinusoidal signal with peak amplitude 7 Vp. Calculate the value...
QUESTION 8 A particle with a charge of 1.5 u Cand a mass of 2.5 x 10-6 kg is released from rest at point A and accelerates towa point B, arriving there with a speed of 42 m/s. What is the potential difference V B-V A between A and B? a.-2.1 103 V b.-4.1 x 103 V O c.0.9 103 v d. +1.5x 103 V O e. +4.1 103 V QUESTION 1 In vacuum, two particles have charges of q...
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Question 1 5 pts As shown in the figure, an unknown charge is located at the origin and an electron is located at x = +0.2 nm. If the net electric force on a small positive test charge is zero at x = -1.5 nm, what is the magnitude and sign of the unknown charge? unknown charge test charge q electron e X=-1.5 nm x=0 x=+0.2 nm 0 -0.78e 0 +0.78e 0 -0.018e +0.018e +28e The...
Homework 2 Chapter 17 Do not put any work or answers on this page. You must show your work to receive full credit for a correct answer. Answers MUST be in correct form. Unknown Variable calculated value & correct units Review Deadline 1 pm EST Thursday Jan 24 2019 Due Date 1 pm EST Tuesday Jan 29 2019 1) An electric field of magnitude 1.80 (10)6 N exerts a force of 9.00 N on a test charge placed in the...
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Question 5 2 pts Three charges 91.92 and 23 are fixed on the three points shown below. Charge 9: = + 9.0x10C, charge 92 = - 4.0x10C, and charge 03 = -3.2 x 10-6c. Determine the magnitude and direction of the net electrostaticforce on charge 91 due to the other two charges. 120 9 Equatione F-F4, F- Facos, + cose F = F., F., - . sin 0, +F, sin , Wallal i World o Netforce -...
Question 7 (3 Points) This assignment is set up for sequential assessment. Complete each question and submit the answer before moving on to the next question. Correct answers for each question will be made available once the maximum number of attempts have been made for each submission. Capacitance 3 - Mixed C=K Formulas: Capacitors with dielectrics (with dielectric) (without dielectrie) EA d d E = EA AV = Ed For ALL capacitors Q=CAV + -Q+Q2 -Q2 +Q3 -Q3 Step 1...