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关注微信公众号Laboratory measurement of the pure rotational spectrum of vibrationally excited
Title:Laboratory measurement of the pure rotational spectrum of vibrationally excited HCO(+) (nu2 = 1) by far-infrared laser sideband spectroscopy
Authors:; ; ; ;
Affiliation:AA(California, University, Berkeley), AB(California, University, Berkeley), AC(California, University, Berkeley), AD(California, University, Berkeley), AE(California, University, Berkeley)
Publication:Astrophysical Journal, Part 2 - Letters to the Editor (ISSN X), vol. 316, May 1, 1987, p. L45-L48. Research supported by the University of California. ()
Publication Date:05/1987
Category:Astrophysics
NASA/STI Keywords:Formyl Ions, Ground State, Infrared Lasers, Interstellar Matter, Laser Spectroscopy, Rotational Spectra, Far Infrared Radiation, Frequency Modulation, Semiconductor Diodes, Semiconductor Lasers
Bibliographic Code:
Laboratory observations of the pure rotational spectrum of HCO(+) in its
lowest excited bending state are reported. Because of their severe
excitation requirements, such vibrational satellites and the high-J
ground-state lines, also measured here, sample only hot, dense regions
of matter in active molecular cloud cores and circumstellar envelopes.
As the HCO(+) abundance is tied directly to the gas fractional
ionization, it is probable that the vibrationally excited formyl ion
transitions will provide high-contrast observations of shocked molecular
material, rather than the more quiescent, radiatively heated gas
surrounding stellar sources detected with the few vibrationally excited
neutral species observed to date.
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arXiv e-prints在三角形ABC中,a^2+b^2=2b^2,其中abc分别是角A,B,C对边长,求证B小于等于60度,若B=45度,A是
在三角形ABC中,a^2+b^2=2b^2,其中abc分别是角A,B,C对边长,求证B小于等于60度,若B=45度,A是钝角,求A
a^2+c^2=2b^2余弦定理cosB=(a^2+c^2-b^2)/2ac=(a^2+c^2)/4ac a^2+c^2>=2ac>=2ac/4ac=1/2 y=cosx在(0°,90°)内是减函数∴B
与《在三角形ABC中,a^2+b^2=2b^2,其中abc分别是角A,B,C对边长,求证B小于等于60度,若B=45度,A是》相关的作业问题
cosA=根号6/3那么sinA=根号3/3即tanA=[根号3/3]/[根号6/3]=根号2 /2即tan2A=2tanA/(1-tan^2A)=(根号2)/(1-1/2)=2根号2
1.由题意知： A+C=2B,A+B+C=180得：B＝60°且A＋C＝120°?∴tan（A＋C）＝tan120°＝－√3＝(tanA+tanC)/tanAtanC??又∵tanAtanC＝2＋√3 ∴tanA＋tanC＝tan（A＋C)（1－tanAtanC）＝tan120°(1－2－√3 ）＝－√3（－1－√3
快好了,请稍等 再问： 好的 再答： 证明： ∵AD＝CD，BD＝CD ∴∠A＝∠ACD，∠B＝∠BCD ∴∠ACB＝∠ACD+∠BCD＝∠A+∠B ∵∠A+∠B+∠ACB＝180 ∴2∠ACB＝180 ∴∠ACB＝90 ∵∠BDC＝∠A+∠ACD＝2∠A，DF平分∠BDC ∴∠BCF＝∠BDC/2＝∠A ∴DF∥AC
由题：2b^2=a^2+c^2余弦定理：b^2=a^2+c^2-2ac*cosBb^2=2ac*cosB=根号2倍ac由正弦定理：ac/（sinAsinC）=b^2/(sinB)^2则 根号2倍sinAsinC=（sinB）^2即 根号2倍sinAsin（A+B）=1/2展开：（sinA）^2+sinAcosA=0.5
1、cosB=(a²＋c²－b²)/(2ac)=b²/(2ac).2b²=a²＋c²≥2ac,则：b²≥ac,从而cosB≥1/2,则0°
因为 三角形ABC中,AB=AC所以 三角形ABC为等腰三角形因为 AD平分角BAC所以 AD垂直于BC所以 三角形ABD和ACD为直角三角形因为 点E、F分别为AB、AC中点所以 DF=1/2AC,DE=1/2AB因为 AB=AC所以 ED=FD=AF=AE所以 四边形AEDF为菱形
第一小问 ∵A+C=2B ∴B=60° A+C=120°tanA * tanC=2+√3可得 tanA *tan（120°—A)=tanA *{(tan120—tanA)/(1+ tanA *tan120)【这里应用的是正切公式】即可求出A C 的度数
证明：∵AB=BC=AC，∴△ABC为等边三角形，∴∠B=∠BAC=∠ACB=60°．在△ABE和△CAD中BE＝AD∠B＝∠BACAB＝CA，∴△ABE≌△CAD（SAS）．∴∠BAE=∠ACD．∵∠CPE=∠ACD+∠PAC．∴∠CPE=∠BAE+∠PAC=60°．
1.根据余弦定理：c^2=a^2+b^2-2abCOSC代入数据得,c^2=12所以 c=2倍根号3根据正弦定理：c/sinc=a/sinA所以 2倍根号3/二分之根号3=2/sinA故 sinA= 二分之一 所以 A=30度根据三角形内角和定理：所以B=90度2.（1）根据余弦定理,b^2=a^2+c^2-2acCO
在△ABC中，由a=4，b=5，以及S=12abosinC=53，可得sinC=32，故C=60°或&1200．当C=60°&时，由余弦定理 c2=a2+b2-2abocosC，求得c=21．当C=120°&时，由余弦定理 c2=a2+b2-2abocosC，求得c=61．综上可得，边c的
（1）易知cosC=0不满足条件，因此cosC≠0，由不等式x2cosC+4xsinC+6≥0对一切实数x恒成立，∴△=16sin2C-24cosC≤0，cosC＞0，化为2cos2C+3cosC-2≥0，解得cosC≥12，又0＜C＜π，当cosC=12时，角C取得最大值π3．（2）角C取得最大值时为π3，∵a=2b
证明：∵∠ACB＝90,D是AB的中点∴AD＝BD＝CD （直角三角形中线特性）∴∠B＝∠DCB∵∠ADC＝∠B+∠DCB∴∠ADC＝2∠B∵DE平分∠ADC∴∠ADE＝∠ADC/2∴∠ADE＝∠B∴DE∥BC又∵AD＝CD∴∠A＝∠DCA∵∠BDC＝∠A+∠DCA∴∠BDC＝2∠A∵DF平分∠BDC∴∠BDF＝∠BD
COSA=根号6/3.则sinA=根号(1-6/9)=根号3 /3那么tanA=(根号3/3)/(根号6/3)=根号2/2所以,tan2A=2tanA/(1-tan^2A)=根号2 /(1-1/2)=2根号2或者cos2A=2cos^2A-1=1/3tan2A=2√2
证：AH⊥BC交EF于G点∵D,E,F,分别是边BC,CA,AB的中点∴AE=EC,AF=FB,BD=DC根据三角形的中位线定理,可得FH=1/2AC,EF=1/2BC,DE=1/2ABFH‖AC,EF‖BC,DE‖AB∵AH⊥BC∴AH⊥EF∴Rt△AGE≌△Rt△HGE∴AE=EH=FD同理,AF=FH=DEEF为
因为AB=AC,且∠A=120°,所以∠B=30°,又因为MN⊥AB,所以在直角△BNM中,MN=½BM(直角三角形中,30°所对的直角边等于斜边的一半),请采纳,谢谢.
1.,a^2+c^2=2b^2,即SinA^2+CosC^2=2*1/2=1所以SinA=CosC,即C+π/2=A,又A+C=3π/4所以A=2π/3,C=π/62.a^2+c^2=2b^2>=2ac,即b^2>=ac由余弦定理得,b^2=a^2+c^2-2acCosB,即b^2=2acCosB所以2acCosB>=
证明：作CG//AB ,交DF于G则⊿FCG∽⊿FBD,＝＞BF：CF=BD：CG ⊿CGE∽⊿ADE,＝＞AE：EC=AD：CG∵AD=BD∴BF：CF=AE：EC
证明：作CH∥AB.交DF于点H则△FCH∽△FBD∴BF:FC=BD:CH易证△ECH∽△EAD∴AE：EC=AD：CH∵AD=BD∴BF:CF=AE:EC
证：∵△ABC中,D、E、F是BC、AC、AB的中点（已知）∴DF、DE是△ABC的中位线（中位线定义）∴DF=1/2AC,DE=1/2AB（三角形中位线定理）又∵AH⊥BC于点H（已知）∴△ABH和△ACH都是直角三角形（直角三角形定义）又∵F为AB中点,E为AC中点（已知）∴HF是Rt△AHB的斜边上的中线,HE是Multiple glomus tumors of the urinary bladder in a cow associated with bovine papillomavirus type 2 (BPV-2) infection.
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):39-42. doi: 10.1354/vp.45-1-39.Multiple glomus tumors of the urinary bladder in a cow associated with bovine papillomavirus type 2 (BPV-2) infection.1, , , , , , .1Department of Pathology and Animal Health, Faculty of Veterinary Medicine, Naples University Federico II, Naples, Italy. sante.roperto@unina.itAbstractWe report a case of multiple glomus tumors associated with bovine papillomavirus type 2 (BPV-2) infection in the urinary bladder of a 13-year-old cow suffering from severe chronic enzootic hematuria. Macroscopically, multiple submucosal reddish nodules were seen swelling the vesical mucosa. Histologically, neoplastic proliferation was characterized by the presence of numerous blood vessels. These were lined by normal endothelial cells surrounded by round epithelioid cells with central nuclei, prominent nucleoli, acidophilic cytoplasm, and well-defined cytoplasmic borders. Tumor cells were distributed around open vascular lumina and in perivascular spaces. They were immunohistochemically positive for actin and vimentin and negative for cytokeratins, desmin, and factor VIII-related antigen. On the basis of these findings, this tumor was diagnosed as glomus tumor, a neoplasm not previously reported in cattle and exceedingly rare in animals. BPV-2 DNA was amplified from the formalin-fixed, paraffin-processed tissue specimens obtained by laser capture microdissection. This report widens the spectrum of mesenchymal tumors of the bovine urinary bladder. Finally, the microscopic pattern of tumor described here shares striking morphologic and immunohistochemical similarities with the angiomatous form of glomus tumor known to occur in man.PMID:
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