Determine the magnitude of the force F2, if the resultant force (FR) is directed along the positive y axis. F1 = 1,317 lb θ1=θ1=33° θ2=θ2=46° Magnitude of Force F2 = ____ lb
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Determine the magnitude of the force F2, if the resultant force (FR) is directed along the positive y axis.
F1 = 1,317 lb
θ1=θ1=33°
θ2=θ2=46°
Magnitude of Force F2 = ____ lb
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- Calculate the speed of an electron (in m/s) after it accelerates from rest through a potential difference of 140 V. 7.02e25 m/s The speed of the electron v can be found using the kinetic energy equation, K = 1/2(mv^2). The change in kinetic energy is equal to the work done on the electron, K = W = -U = -qV. Using those two equations, solve for the speed v in terms of the potential V and constants. Remember that the mass of an electron is 9.1e-31 kg and its charge is 1.6e-19 C. Submit Answer Incorrect. Tries 1/2 Previous TriesQT T e cnet ensry ot sty shanges rom 12,1150 Jdus o th s of a s o N on an et atmass 4. Thaworkdon by b o s 0 244 0364 O 484 O 60JSuppose an electron (q= -e = -1.6 x 10^-19 C, m = 9.1 x 10^-31 kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for the final speed of the electron. Express numerical answer in two significant figures. Use the template in the attached photo to solve for the problem.
- A) Through what potential difference does an electron have to be accelerated, starting from rest, to achieve a speed of 0.965 cc? Express your answer in volts. B) What is the kinetic energy of the electron at this speed? Express your answer in joules. Express your answer in joules. C) What is the kinetic energy of the electron at this speed? Express your answer in electronvolts. Express your answer in electronvolts.An electron is accelerating under a potential difference of 105 volts. (m,=9,108x10 31) a) What is the kinetic energy of the electron?What is the speed of an electron after being accelerated from rest through a 2.5×107 V potential difference? Express your answer as a fraction of cc.
- ITCE 250, ITCE 112 - Google X A Meet - ITCE112,250 - Digita x G Google Account O Introduction to Number SysX O (7) WhatsApp G Google Search - pzjio X Content .edu.bh/ultra/courses/_24054 1/cl/outline Remaining Time: 15 minutes, 24 seconds. * Question Completion Status: QUESTION 6 A long coaxial cable is shown in the figure where the radii of the conducting concentric cylinders are indicated. The linear charge densities of the inner and the outer cylinders are la=3.000 nC/m and Ab =-7.000 nC/m, respectively. Find the magnitude of the electric field (in N/C) at 31.77 mm from the center. S1mm a 30mm Click Save and Submit to save and submit. Click Save All Annoers to save all annvers Ps Au Ae Save All Answ 28 mm2. A particle of mass m moves in a force field whose potential in spherical coordinates is V =- (K cos 0)/r². Obtain the canonical equations of motion.In large CRT televisions, electrons are accelerated from rest by a potential difference of 23.88 kV and shot onto a phosphorescent screen to produce an image. What is the speed of the electrons when they reach the screen? (g. = 1.602 x 10-19C ;me = 9.11 x 10 -31 kg) Answer: x10' m (express your answers in tenths place or one decimal digit only)
- ... Path of trajectory AV w An electron is fired at a speed v¡ = 3.4 x 106 m/s and at an angle 0; = 30.5° between two parallel conducting plates as shown in the figure. If s = 1.5 mm and the voltage difference between the plates is AV = 98 V, determine how close, w, the electron will get to the %3D %3D bottom plate. Put your answer in meters and include at 6 decimal places in your answer. Do not include units. The x-axis of the coordinate system is in the middle of the parallel plate capacitor. Round your answer to 6 decimal places.The answer for this problem would be "C" but could you explain me why? The reason for this is to understand this kind of problems in the future not only this one.Suppose an electron (q = -e = -1.6 x 10¬9 C,m=9.1 x 10¬3' kg) is accelerated from rest through a potential difference of Vab = +5000 V. Solve for %3D the final speed of the electron. Express numerical answer in two significant figures. The potential energy U is related to the electron charge (-e) and potential Vab is related by the equation: U = Assuming all potential energy U is converted to kinetic energy K, K+U = 0 K = -U 1 Since K mv and using the formula for potential energy above, we arrive at an equation for speed: 2 v = ( 1/2 Plugging in values, the value of the electron's speed is: x 107 m/s V=