更新1:
according to your wer for question 1
the question doesn't mention the diameter of the table
leg distance from the edge and the position of the Mass relative to the legs. I just copied the question right away.....
1. You have to draw a geometrical diagram as follows: Join the three legs of the table together to form a large triangle. From the centre of the table (where the centre of gravity is located)
draw lines joining the table centre with the three legs. There will then be three *** aller isosceles triangles forming. The apex of each triange is 360/3 degrees = 120 degrees. Take any one triangle
drop a perpendicular line from the apex (i.e. the table centre) to the base of the triangle and produce it to meet the rim of the table at a point
say point A. This perpenicular line would divide the triangle into o identical ones
each with apex angle 120/2 deg. = 60 degrees. The required mass (M
say)is placed at A to topple the table. The pivot is the line joining the nearest o legs from A (i.e. the base line of the triangle) Distance of the table centre from the pivot = R.cos(60) where R is the radius of the table Moment given by the table mass = 36 x R.cos(60) Distance of A from the pivot = R-R.cos(60) Hence
moment given by the mass at A = M.[R-R.cos(60)] For equilibrium
36 x R.cos(60) = M.[R-R.cos(60)] M = 36.cos(60)/(1-cos(60)) kg = 36 kg ------------------------------------------ 2. Let the end nearest to the piano be A and the other end be B. The reactions at A and B are R and R' respectively hence
R + R' = (160 + 300)g N = 460g N where g is the acceleration due to gravity i.e. R = 460g - R' Take moment about A R'.L = 160g(L/2) + 300g(L/4) where L is the length of the beam R = (160/2 + 300/4)g N = 155g N = 1550 N (take g = 10 m/s2) thus
R' = (460-155)g N = 305g N = 3050 N
1. Not sufficient information. To calculate it
need following information: a. diameter of the table. b. leg distance from the edge. c. position of the Mass relative to the legs. 2. At the end closer to the piano: (1/2)160+(3/4)300 = 305kg At another end: (1/2)160+(1/4)300 = 155kg 2009-01-26 22:12:41 补充： 002
How about placing the Mass right above one of the legs? What assumptions you have made?
更新1:
as the value of V in the equation is the Voltage of the capacitor is this repersented that we can't prove I=(Io)e^[- t/(CR)] by my method. if it can
please continue at this step (dQ/dt=V(dC/dt)=C(dV/dt))
更新2:
how to solve A charged capacitor of capacitance 47F is connected to a voltmeter. The voltmeter reading is found to drop from 8V to 2V in 7.8s Find the resistance of the voltmeter As the voltmeter must connected to the capacitor in parallel so the equation I=(Io)e^[- t/CR] can not be applied?
Your derivation is mathematically correct
but the sign is incorrect. Consider a circuit consisting of a cell with e.m.f. E and a resistor and capacitor connected in series
then by Q=CV dQ/dt =C(dV/dt) But note that for the notation V
it represents the voltage across the resistor
so V = IR is not valid. Instead
E = IR + V
so V = E - IR
and dV/dt = dE/dt - d(IR)/ dt
but since E
the e.m.f. of the cell
is assumed to be constant
dE/dt = 0 and dV/dt = -d(IR)/dt)
and dQ/dt = - C(dIR/dt) I/(CR)= - dI/dt -t/(CR)= [ln(I)] (upper=I and lower=Io) =ln(I)-ln(Io) =ln(I/Io) I=(Io)e^[-t/(CR)]
as required. Also
if a resistor is connected in parallel to the capacitor
the circuit with capacitor has no resistance at all. The capacitor is charged up very quickly and the equation for I doesn't apply (substituting R = 0 in the equation and you will see why) Yet in the example: A charged capacitor of capacitance 47F is connected to a voltmeter. The voltmeter reading is found to drop from 8V to 2V in 7.8s Find the resistance of the voltmeter Note that the capacitor is not being charged up
instead
it is discharging. For discharging capacitor with a resistor in the circuit
we have the equation: V = V0(e^[-t/(CR)])
where V is the voltage across the resistor. Applying the equation and you can get the wer.
参考: /
Your equation is wrong. By Q=CV dQ/dt=V(dC/dt)=C(dV/dt) =C(dIR/dt) < ---- this is wrong
V is not equal to IR.It is because V is the p.d. across the capacitor
but IR is the p.d. across the resistor. They are not the same. When the resistor is connected in parallel with the capacitor
the capacitor will be charged almost immediately to its full capacity after connected to the battery. The relevant equation for series connection could not apply.
