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salas1985
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路lu橙

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物理就在我们身边 物理是一门以理解为主的学科,对于许多学生来说,学起来比较吃力,究其原因关键是学习物理的方法和氛围不够,总是把物理想象的多么多么的深奥,其实物理就在我们身边。要想走出物理难学的怪圈,应该从以下三个方面入手。 第一、物理身边有。1、物理无处不在。生活中的好多物体都包含有物理知识,例如我们学习用的台灯包含的物理知识就有:杠杆、摩擦力、变阻器、导体和绝缘体、光的反射、能量变化、物态变化以及化学知识等等。另外,在娱乐时也有物理知识,电视中的《奇思妙想》、《绝对挑战》都包含了许多物理知识,甚至电视广告中也有物理知识:广告露露饮料需要加热饮用时,在冰天雪地里怎么加热,后来主人公用冰块磨成冰透镜会聚太阳光解决了这一难题;学生从物理学习中找到自信和快乐。2、物理中深奥的道理尽量用身边的事例来说明。你像《气体压强和流速的关系》一节中讲解飞机升天的原因时,学生大都不知道怎么回事,学起来有一定困难,讲解时我首先用两张纸做了简单的实验,得出气体流速越大压强越大,接着再用学生比较感兴趣的足球里的“贝氏弧线”进一步理解这一关系,然后再讲解飞机升天的问题,这样学生就很容易理解。3、物理实验尽量用身边的器材来完成。证明大气压存在的实验器材可以是:啤酒瓶、热水、盆子、玻璃杯、硬纸片、挂衣钩、吸管、注射器;研究《气体压强和流速的关系》时用的是两张纸;研究重力势能的大小因素时用的书本、橡皮等等都是我们身边再熟悉不过的器材,这样学生随时随地都可以做自己感兴趣的物理实验。 第二、实验重探究。我听说过这样一句话:如果美国人提出一个实验课题,美国人自己需要两年完成,而中国人只需半年就可以完成,但是关键是中国人提不出问题。由此说明中国人的探究能力的欠缺,所以作为教师的我们有责任提高学生的探究能力。探究实验教学是提高学生探究能力的重要途径,所谓探究实验教学,就是首先对某一问题提出猜想,然后再设计实验验证或者否定自己猜想的教育理念,由于现实条件的限制,我们在课堂上不可能都实验探究,但是我们至少可以按照这个思路教育学生、讲解知识,这样学生慢慢的有了解决实际问题的能力和兴趣,回家后他们会想办法去做课堂上无法完成的事情。记得讲完《压强》后,有一学生看到电视上公安部门利用犯罪分子留下的脚印来破案,于是他就想留有的痕迹的大小跟什么因素有关呢?通过观察他猜想可能与物体的质量、下落的高度、物体与地面的接触面积以及地面的松软情况有关,于是他就找来一把刻度尺、一些沙子、两个大小不同的球、两个质量不同的球来做探究实验,验证了自己的猜想,得出了正确的结论。这样课堂上学过的探究方法又在同学的日常生活中得到运用。 只要留意身边处处都有物理知识,只要用心随时都能做物理实验,那么不久的将来学生的探究精神和综合素质会有较大的提高。

大学物理论文800字数量

107 评论(15)

zouyuehong

Here we present the derivation of the new set of equations termed, Lorentz transformations, and all the subsequent LORENTZ TRANSFORMATIONSWe consider two coordinate systems (frames of reference) one stationary S and one moving at some velocity v relative to S, then according to the two postulates of Relativity, stated in the main text, the displacement in both frames is of the same Therefore, we have (A-1) (A-2)We should note here that in the old Galilean transformations these equations would be (A-3)which is in direct contradiction to Postulate 2, a firm experimental Equations (A-1) and (A-2) can be written as(A-4) (A-5)That is, (A-6)We are interested in finding and in terms of x and That is, = (x, t) (A-7) = (x, t) (A-8)This is accomplished via the formation of two linear simultaneous equations as follows:(A-9) (A-10)where a11, a12, a21, and a22 are constants to be It is required that the transformations are linear in order for one event in one system to be interpreted as one event in the other system; quadratic transformations imply more than one event in the other Solution of problems involving motion begins with an assumption of their initial conditions; , where does the problem begin?The classical assumption is to set = 0 at = Therefore, according to S, the system appears to be moving with a velocity v, so that x = We can obtain this from E (A-9) by writing it in the form = a11(x - vt) so that, when = 0, x = Therefore, we conclude that a12 = - We can write Equations (A-9) and (A-10) as (A-11) (A-12)Substituting and into Equation (A-6) and rearranging, we get (A-13)Since this equation is equal to zero, all the coefficients must That is,(A-14) (A-15) (A-16)Solving these equations we obtain(A-17) (A-18)where β = v/c and Thus, substituting these values in E (A-11) and (A-12) we obtain the famous Lorentz coordinate transformation equations connecting the fixed coordinate system S to the moving coordinate system :(A-19) (A-20)We may also obtain the inverse transformations (from system to S) by replacing v by –v and simply interchanging primed and unprimed This gives,(A-21) (A-22)VELOCITY TRANSFORMATIONSAs a direct consequence to these new transformations, all the other mathematical operations and physical variables follow For example, the velocity equations (though still the derivatives of the displacement) assume a new form, so the Lorentz form of the velocities is:From E (A-19) and (A-20) we have: (A-23) (A-24)Therefore:(A-25)ENERGY CONSIDERATIONSConsider a particle of rest mass m0 being acted by a force F through a distance x in time t and that it attains a final velocity The kinetic energy attained by the particle is defined as the work done by the force F The applicable equations are,(A-26)We note thatand thatSubstituting d(γv) in E (45) and integrating, we obtain (A-27)That is, (A-28)This says that K = (m – m0)c2 and finally one sees that the total energy is equal to the sum of the kinetic energy K and the rest energy E0 = , E = K + Eo = γm0c2 = γE0, (A-29) where E0 = m0c2 and E = 给分吧
117 评论(8)

老卢

兄弟……800字……自己写吧……
324 评论(8)

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