初三物理綜合應(yīng)用專(zhuān)題

6767五用[2018·]1,15℃(1)[

=1.0×10kg/m,c水=4.2×10

J/(kg·℃(2)?q

=4.2×107)4

J,,100kg20℃40℃:[3

J/(kg·℃)](1)(2)5(3)η(4),(煤J/kg,)

33[2018·]T5-110s20m/s,(1)0~10s,(2)f10~20s?(3)80%,(已5×10

7

[2018·],T5-215m

Nm,,(

2323gN/kg)(1)(2)(3)500cm,2×10Pa,.[2018·]T5-3,0.05m/sFt:(g10(1)A?(2)A,?(.01%)(3)V,?(不

)[2018·],,,,A22

20cm,cmcmN,g10N/kg(1)(2)(3)(4))

K?

11.[2018·]T5-5

0

“V6WL,,8Ω,SW:

,,LT55(1)R

0

(2)100s,(3)(4)

[2018·]T5-6,V12V“1A0~0.6A,0~15VLL12(1)L(2)(3)L,

1aa2bb01aa2bb0[2018·]L62.4”A,0~15V160

,a,=40Ω,UV,Ia

;SbR,I

II∶4,R∶ab∶2,,,a)(1)(2)

0

U(3)S

1

,

3-33036667吸5=6=0133參考答案3-33036667吸5=6=0133:m=

V=1.0×10

kg/m

×10×10

=10kg,m(t-t

)=4.2×10J/(kg·℃℃℃)=3.57×10(2)

=3.57×106

=

放

=

=085m

3天然氣:Q

=cm=4.2×103

℃)×100℃-20℃)=8.4×10JE=4.2×10J/h×5(3)

吸

×100

×100%=(4)Q

,m===.kg:0~10sv===m/s(2)10~20s,F=f=4000N,ts=10s,s'=v't'm,10~20sW=Fs'N×200J(3)10~20放==1×10J,m=

放

==0kg:

=V

,

=

=×

kg/m

×10(2)n,η=

有用

×100%=

-

浮

×%總

-421Δ-414443-421Δ-414443F=312.5,F=(G-

G拉,N=(2000N-1000

),GN(3)F

=2×10

=100F=FN,F+F=G,人F=G-FN/kg-100人支4=G輪

F

=2G

4F=2×125N+2000N=250N,Δp

浮

=

浮

-

浮

=。.:=×N,浮=G-F=×10N-×10N=×N,AV=V=

浮

==1

。水(2)η

有用

×=

×=

×100%=

×100%=80%,總

動(dòng)

動(dòng)

動(dòng)G=5000NAη=1

有用

×100%=

×100%=

×=

×≈71%總

動(dòng)

動(dòng)(3)P===Fv=(G+G

)·2v=(G+G)v=×104

N+N)×05=W,P=UI=V×A=1900W,P=P-P=W-1750=150W,

熱

==Ω

3-2-33-3-2-42-4-2-33-2-33-3-2-42-4-2-33-3L1LL:p=

kg/mN/kg×0.2(2)V=S

h=100×10m×12×10mV=1.2×10

F浮=水排=1.0×10kg/m3N/kg×1.2×10N(3)3+F=GF=浮

浮

=

=N,,,=FN(4)

,'=(1)

×12N=3'=

浮

=

=NF'=2'=2×7N=14N,F=FN-8N=6N;Fx·2x=Fcm/N;6N6N×0.5cm/N=3cm,L=2cm,=S容L+S-S)h-4m2×2×10)××12×10mm=水3kg/m×1.3×10m.:R

0

2V2RR0

0

==

=10Ω(2)100s,:Q=Pt=6W×100J(3)SS,L,LW,U

=V,I=05:U=IR=0.5A×8Ω=4V,L

LLL=I'=LLLaaaa02abbLLL=I'=LLLaaaa02abbU=U+U=V+4V=6V(4)P=

,,SS,12,R'=0Ω,L0L+P=PL+=6W+=9.6W:I==0.5A(2)==

=24,,A,=

滑

=

=Ω,V3V,I'=L

=R=

滑

=

=120,12~120Ω(3)P

滑

)2=

滑

滑

滑

=

-

滑

滑

=滑-

滑滑,R,,I'=

==A總U

″=I'R=Ω=9VP滑大=U″I'=(U-UL″)I'=(18V-9W.:U

=6V,P=2.4R額

=L

額

=

=15Ω額(2)

1

,U0

=12V,=40Ω,I==0.3U=I(R+R)=0.3R①;SL,:II=34,I=I×0.3則