What force is required to accelerate an electron to 1.00 x 10^2 m/s?

106 MeV) is Extra: A very rare but occasionally observed decay of a muon µ (rest energy = into an electron e (rest energy 0.511 MeV) and a photon y (rest energy decays at rest by this mode, what is the kinetic energy of the electron and photon combined after the decay? (1M V = 10°eV = 1.6 x 10-13 J) zero). If a muon

Extra: You have two disks lying flat on ice, one has a string wrapped around the outside edge which unravels as you pull, the other has a string attached to a peg in the center. The discs have equal masses and dimensions and are initially at rest. If you start pulling both strings at the same time with the same constant force, at some time t later which of the two disks has moved further? Or have they both moved the same distance? Explain.

Please help with explanation. I will upvote

The mass of an electron is 9.11 x 10 ^-31 kg. The mass of a proton is 1.67 x 10 ^-27 kg. They are about 5.3×10^ −11 m apart in a hydrogen atom. What is the gravitational force between these two particles in the hydrogenatom?

A body of mass m is attracted toward a 10.7 kg mass, 27.6 cm away, with a force of magnitude 6.72 x 10-8 N. Find m.

We will use differential equations to model the orbits and locations of Earth, Mars, and the spacecraft using Newton’s two laws mentioned above. Newton’s second law of motion in vector form is: F^→=ma^→ (1) where F^→ is the force vector in N (Newtons), and a^→ is the acceleration vector in m/s^2,and m is the mass in kg. Newton’s law of gravitation in vector form is: F^→=GMm/lr^→l*r^→/lr^→l where G=6.67x10^-11 m^3/s^2*kg is the universal gravitational constant, M is the mass of the larger object (the Sun), and is 2x10^30 kg, and m is the mass the smaller one (the planets or the spacecraft). The vector r^→ is the vector connecting the Sun to the orbiting objects. Step one ) The motion force in Equation(1), and the gravitational force in Equation(2) are equal. Equate the right hand sides of equations (1) and (2), and cancel the common factor on the left and right sides. Answer: f^→=ma^→        f=Gmm/lr^→l^2         a^→=Gmm/lr^→l^2 x r^→/lr^→l        r^→=r^→/lr^→l * Gmm Could you please…

An electron travels with v =5.90×10^6i^m/s through a point in space where E =(2.20×10^5i−2.20×10^5j^)V/m and B =−0.180k^T. What is the force on the electron? Express vector F in the form Fx, Fy, where the x and y components are separated by a comma.

Consider the oscilloscope deflection mechanism as seen in the figure next. The speed of the electron entering the area of deflation (2.3cm long) is 9.6 x 10^8 cm/s with a constant acceleration of 9.4 x 10^6 cm/s?. The vertical distance when leaving the area of deflection is +Ed d Vox 3D Id L 0.53cm 0.29cm 0.27cm 0.16cm None of the choices O O Screen

You are the pilot of a spacecraft intended for travel at very high speeds. Before leaving you measure the spacecraft to be 31.4 m long and have a mass of 5.28 x 104 kg. During your travel, you pass a planet and exchange information with an observer on the planet. You are told that your spacecraft has been measured to be 28.6 m long. a. How fast is your spacecraft travelling with respect to the planet? b. You are told there is a nearby planet that is stationary with respect to the first planet. The observer on the first planet says it will take you 25 s to reach the nearby planet. How far away does your co-pilot on the spacecraft say the nearby planet is? c. How much energy was required to accelerate your spacecraft to this speed?