Physics Calculation Questions Detailed Process Thank you

1) Heat absorbed by water Q absorption = cm T = A 20 ) =;

The electrical energy consumed w=q sucked.

Answer: Energy consumption;

2) q put = vq = ;

Q suck q put.

Answer: The thermal efficiency of the gas stove is;

3) When the user needs the same heat, the cost of using the induction cooker to heat is yuan degree = yuan;

The cost of heating with a gas stove is $3;

Use an induction cooker** low; Then the use of induction cooker is safe, convenient and fast; Compared with gas stoves, it is less polluting to the environment

There are two types of questions to be asked about:

The first one is as shown in the picture

The particle vibrates downward from the equilibrium position O point I draw, reaches the M point after 3s, and then passes through the M point for the second time after 2S, according to the symmetry of the simple harmonic motion, it can be seen from the figure that the particle vibrates downward from the equilibrium position O point I draw.

The period of the trough (between 1 and 2) 1 4 is 4s, and then from 2 to 3 is the third time to pass the m point, according to the symmetry of the simple harmonic motion, the vibration time is a period less than 2s, so the particle point passes through 4*4 -2 = 14s for the third time to pass the m point.

The second type is as shown in the second picture

The particle vibrates downward from the equilibrium position O point I draw, reaches the M point after 3S, and then passes through the 2S second time through the M point, as can be seen from the figure, the particle vibrates downward from the equilibrium position O point I drew to the second time through the M point has not passed a cycle, according to the symmetry of the simple harmonic motion, a period difference (3 - 2) 4 s, in this case a period is:

3+2+ 1 4 = 21 4 s, and then from 2 to 3 is the third time to pass the m point, according to the symmetry of the simple harmonic motion, the vibration time is a period less than 2s, so the particle passes through 21 4 -2 = 13 4 s for the third time to pass the m point.

16s 4*(3+2:2) is a period of vibration.

1) 10 ohms: vmax=12 vmin=(12 (10+40))*10=

The voltmeter varies from volts to 12 volts;

40 ohms: vmax=12 vmin=(12 (40+40))*40=6

The voltmeter varies from 6 volts to 12 volts .

2) When the 10 ohm resistor is connected to the circuit, the voltmeter change range is more obvious.

When R1 is 10 ohms, the maximum resistance in the circuit is 10+40=50 ohms, so the minimum indication of the voltmeter is: 12 (10+40)*10= The maximum indication is the power supply voltage of 12V. When R1 is 40 ohms, the maximum resistance of R2 is 40 ohms, so the minimum indication of the voltmeter is 6V, and the maximum indication is the power supply **12V.

When the R1 circuit is connected and disconnected, the voltmeter variation range is more obvious.

Solution: (4) When r2=0, the voltage representation number umin=0v, ammeter imax=u source r1=15 24=

When R2=20, the current representation imin=u, the source r=15 (20+24)=, and the voltage representation umax=iminr2=

Then the change range of the voltmeter is: 0, and the change range of the ammeter:

5) The range of the ammeter is 0, then the maximum circuit current is Imax=, then the voltage at both ends of R1 is U1Max=ImaxR1=, then the voltage at both ends of the voltage representation number R2 is minimum U2min=U source-U1Max=15-9=6V

The minimum value of R2 is R2min=U2Min, iMax=6

When R2 is the maximum value Rmax=20, R total Max=R1+R2Max=30, then the minimum current is: imin=u source RTotal Max=15 30=

The voltage represents the number R2, and the voltage at both ends is the maximum U2max=iminr2max=

Then the value range of R2 resistance value is: 10 20, the voltage number range is: 6V 10V, and the current number range is:

R2 is at its maximum:

i=u/(r1+r2)=15/(24+20)=u2=i*r2=

The rheostat is the minimum withered (r2=0).

i'=u/r1=15/24=

u2=0 Conclusion: voltmeter is not Lachang: ; Ammeter:

4) r1 = 24 ohms.

i most ruined whisper large = u r1 = 15 fiber biyun nian 24 = a u1 = 0i min = u (r1 + rz) = 15 (24 + 20) = u2 = irz = u( i (

5) i1= i1=u (r1+rz) rz=15 ohm u1=i1r1=9v u2=6v

rz>=25 ohms i( u(6---15V).