Microwaves in a microwave oven are produced by electrons circling in a magnetic field at a frequency of 2.4 ghz. (a) what’s the magnetic field strength? (b) the electrons’ motion takes place inside a special tube called a magnetron. if the magnetron can accommodate electron orbits with maximum diameter 2.5 mm, what’s the maximum electron energy?

6.3^2 + 12^2 = thirteen.6^2. Left hand area balances suitable hand area! The triangle with factors 6.3 cm, 12 cm and thirteen.6 cm is nice-angled. Makes issues easier. Draw a triangle XYZ with XY = 6.3 cm, YZ = 12 cm and XZ = thirteen.6 cm (the hypotenuse) --- on a sheet of paper. be conscious that XY and YZ has a suitable perspective between them. Circle X and Y. that's the place there are 2 currents of length 15.0 A coming into the web site frequently. The magnetic field located at factor located at a distance R far flung from an prolonged straight away cutting-edge I is given by way of B = (mu-nought)*(I) / (2*pi*R) “mu-nought” is (4*pi*10^-7), so B = (2I)/R the two currents make magnetic fields B1 and B2 at Z, so as that they are at suitable angles with XZ and YZ respectively. the attitude between B1 and B2 is theta, the place theta is likewise the attitude at Z in triangle XYZ. Theta is rather discovered through fact all 3 factors of XYZ are prevalent, making use of easy sin, tan or cos. Draw a parallelogram with 2 arrows for factors, of length B1 and B2, with the attitude theta between them. locate the diagonal of the parallelogram between B1 and B2, and the attitude it makes with B1. for this reason you be responsive to that's the dimensions and path of the ensuing field. ok, right this is the place physics ends, something is arithmetic. It’s as much as you presently to the two draw to scale or calculate.

graph slopes whilst gas cools, then it is flat whilst conensation to liquid happens, the coolong liquid then graphs as a slope when it solidifies graph is flat

latent heat

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards