In this example, suppose a proton (mp 1.67 x 10 27 kg) is released from rest at the positive plate. What is its speed just before it reaches the negative plate? Express your answer with the appropriate units. PA v=2.40.105 m S HAMO ?

2 A train of mass 20,000 kg is travelling along a horizontal surface at 35 ms-1 when the driver sees a man tied to the tracks ahead, disengages the engine and applies the brakes. When braking the total resistive force on the train is (4000 + 500v) N, where v is the speed of the train. How far does the train travel before coming to rest?

= Two point particles have equal charge q 1100 esu and different masses, m₁ = 25 gr and m₂ = 65 gr, are confined to move freely along the x axis. Note that esu is the unit of electric charge when working with units of mass in grams and distance in centimeters. = The electrical force between the particles is a repelling force which is derived from the potential energy function U(x1, x2) of the two-particles system. U(x1, x2) depends only on the distance of the particles and is given by U(x1, x2) - q2 | x2 − x1| At t = 0) the particle of mass m2 is at rest at the origin (x2 = 0), and the particle of mass m₁ begins to move from point x1 = 76 cm with velocity vo 134 cm/s towards particle m2. = a. What is the location of the center of mass (in cm) at t - 0? -21. b. What is the location of the center of mass (in cm) at t = 0.4 sec? -6.2 c. What will be the minimum distance (in cm) between the two particles? 6.46 X d. Once the distance between the particles is minimal, what will be the speed…

What is the electric potential 4.6×10−11 m from a proton (charge +e)? Let V=0 at r=∞.Express your answer using two significant figures.   What is the potential energy of an electron at this point?

so the distance is from -3 to 5 cm which is 8cm or .08m. I did parts a-e and need help for parts f-h. This is what I have so far: a: V1=1000V b: 3.2e^-16 J c. potential energy; no work needed to move particle as it leaves electric field  d. 3.2e^-16 J e. m1<m2<m3<m4, v1>v2>v3>v4 f. distance =.08m -time takes particle 1 hit detector: .98e^-6 sec ---> .08m/.98e^-6 s= 8.16e^4 m/s -time takes particle 2 hit detector: 1.4e^-6 s --> .08m/1.4e^-6 s= 5.71e^4 m/s -time takes particle 3 hit detector: 1.7e^-6 s --> .08m/1.7^-6 s= 4.7053^4 m/s -time takes particle 4 hit detector: 1.96^-6 s --> .08m/1.96^-6 s= 4.082e^4 m/s   I am not sure if i Made any mistakes so please let me know if i DO!

11. An electron in the picture tube of an old television set is accelerated from rest through a potential difference Vba = +5000V. a. What is the change in potential energy of the electron? (ans: -8.0 x 1010 J) b. What is the speed of the electron as a result of this acceleration? (ans: 4.2 x 10' m/s) c. What is the speed of a proton that is accelerated through a potential difference Vba = - 5000 V? (ans: 9.8 x 10° m/s) 020-12-11 Page 4 of 17

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=

Help with the following question, please write the answers on a piece of paper

3. Calculate the gravitational field due to a homogeneous cylinder at exterior points on the axis of the cylinder. Perform the calculation by a) computing the force directly and b) computing the potential first.

An electron moving parallel to the x axis has an initial speed of 3.40 x 106 m/s at the origin. Its speed is reduced to 1.98 x 105 m/s at the point x = 2.00 cm. (a) Calculate the electric potential difference between the origin and that point. Volts (b) which point is at the higher potential? O the point x = 2.00 cm ● the origin O both have the same potential