Engineering, 29.06.2021 01:00 leaving2020asap
The potential difference across the plates will be zero. The accelerating potential difference will be av. When the electrons enter the region between the plates, they will be deflected by the magnetic field set up by the Helmholtz coils. The electrons will be deflected along a circular arc. This arc is determined by the condition that the magnetic force acting on the electrons provides the centripetal acceleration necessary for the circular motion, i. e.. evB = mv2/R). (1) where e is the charge of an electron, 1.60 x 10 C. Ris the radius of the circular path, Bis the magnetic field, and v is the speed of the electrons. The electrons originate at the cylindrical end of the tube and are accelerated through a potential difference AV acquiring a kinetic energy given by mv = AV Solving equation (2) for v, and substituting it into equation (1). we obtain (RBP = (2me) AV (3) The magnetic field is given by the relation B = B1, (4) where 3 = 0.00465 and l is the current in amperes. Substituting BI for B into Eq.(3) and rearranging the variables, we have (RIY = (2m'eß-) AV. This is the basic equation (5) Physica en 2 Despacio Detection The radius of the path can be determined by knowing a point on the path with coordinates (x, y) and using R = 1* Cos-15* + y4y2y. (6) For each accelerating potential, Eq. (6) allows us to calculate R. Now that we know R for each accelerating potential, we can graph (RIY vs. AV, and the slope of the line will equal Zm/e32 (Note: This comes from the fact that the general form of a straight line is y=mx +b. So if (RI) is plotted on the y axis and AV is plotted on the x axis, then the slope (m) will be the coefficient of x which in this case is (2m/elsa). Now we can calculate the mass of an electron Procedures: There are four different High Voltage Power Supplies. Wire your particular High Voltage Power Supply apparatus according to the diagrams at the end of this Magnetic Deflection lab. Your lab instructor must check your connections before you turn on the power supplies
Answers: 3
Engineering, 04.07.2019 19:10, ndaha
An external consultant recommends that a plant installs a bank of capacitors for power factor correction. this will reduce the peak electrical demand charges by an average of 93 kw every month. the plant current pays $13 per kw in peak demand charges. the capacitor bank will include 223 kw of fixed capacitors, and 183 of variable capacitors. the fixed capacitors cost $59 per kw, and the variable capacitors will cost $65 per kw. the consultant charges 21% of the equipment costs to install the capacitors. because this project will reduce the demand for the electric utility, they are prepared to provide a one-time rebate of $42 per kw of reduced demand. what is the simple payback period for this project (in years)?
Answers: 2
Engineering, 04.07.2019 19:10, shayshay7874
An electric kettle is made out of stainless steel, weighs two pounds (when empty) and is equipped with a heating element that consumes 2 kw of electricity. assuming that the water and the kettle are at the same uniform temperature at any moment of time, calculate the shortest possible time to bring 2 quarts of water from room temperature to the onset of boiling
Answers: 2
The potential difference across the plates will be zero. The accelerating potential difference will...
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