Two masses,m1=18.0kg and m2=26.5kg,are connected by a rope that hangs over a pully.The pully is an uniform cylinder of radius 0.260m and mass 7.50kg.Initially,m1 is on the ground and m2 rests 3.00m above the ground.If the system is now released,use c

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Two masses,m1=18.0kg and m2=26.5kg,are connected by a rope that hangs over a pully.The pully is an uniform cylinder of radius 0.260m and mass 7.50kg.Initially,m1 is on the ground and m2 rests 3.00m above the ground.If the system is now released,use c
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Two masses,m1=18.0kg and m2=26.5kg,are connected by a rope that hangs over a pully.The pully is an uniform cylinder of radius 0.260m and mass 7.50kg.Initially,m1 is on the ground and m2 rests 3.00m above the ground.If the system is now released,use c
Two masses,m1=18.0kg and m2=26.5kg,are connected by a rope that hangs over a pully.The pully is an uniform cylinder of radius 0.260m and mass 7.50kg.Initially,m1 is on the ground and m2 rests 3.00m above the ground.If the system is now released,use conservation of energy to determine the speed of m2 just before it strikes the ground.Assume the pully if frictionless.
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Two masses,m1=18.0kg and m2=26.5kg,are connected by a rope that hangs over a pully.The pully is an uniform cylinder of radius 0.260m and mass 7.50kg.Initially,m1 is on the ground and m2 rests 3.00m above the ground.If the system is now released,use c
机械能守恒
有得到的能量=失去的能量
得到的是m1gh 1/2*m1v2 1/2*m2v2 与滑轮的动能
失去的是m2gh
有m1gh+1/2*m1v2+1/2*m2v2+转动动能=m2gh
转动动能=1/2*Jw2=1/2*1/2mR2*w2

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用能量守恒定律,重力势能转换为动能,详细过程不想写、、

Two masses,m1=18.0kg and m2=26.5kg,are connected by a rope that hangs over a pully.The pully is an uniform cylinder of radius 0.260m and mass 7.50kg.Initially,m1 is on the ground and m2 rests 3.00m above the ground.If the system is now released,use c a mass of =masses of吗? 如图,两质量分别为m1=1kg和m2=4kg小球 英语翻译two masses,m1 and m2,connected to each other and to a central post by cords,rotate about the post at frequency f (revolutions per second) on a frictionless horizontal surface at distances r1 and r2 from the post.derive an algebraic expres m1=75kg,m2=168.5kg,斜面与m2,m2与m1间的摩擦系数均为0.04,求m2刚要移动时的角度. two boxes,m1=1.0kg with a coefficient of kinetic friction of 0.10,and m2=2.0kg with a coefficient of 0.20,are placed on a plane inclined at Θ=30.what acceletation does each box experience?If a taut string is connected to the boxes,with m2 initially 如图所示,物体A的质量是m1=2kg,长木板B的质量为m2=1kg.. 关于滑轮的物理题求受力、分解m1=6.4kgm2=3.4kg摩擦力忽略不计 一道高中物理题,求解,求详细答案 如图所示,m1重47kg,m2重35kg,m1和斜面的摩擦系数一道高中物理题,求解,求详细答案如图所示,m1重47kg,m2重35kg,m1和斜面的摩擦系数为μs=0.42 μk=0.19a)如果小明按 一道万有引力计算题求具体数字 计算机里面算不出来 Two 203-kg masses are separated by a distance of 0.96 m.Using Newton's law of gravitation,find the magnitude of the gravitational force exerted by one mass on the other. 某店将甲乙两种糖果混合在一起卖,单价为(a1*m1+a2*m2)/(m1+m2),m1、m2分别为甲乙糖的质量,a1、a2分别是甲乙的单价,a1=20¥/kg,a2=16¥/kg,现将10kg乙和一箱甲均匀混合,卖出5kg后,再加入5kg乙,这时单价 m1=3Kg,m2=2Kg,m3=1Kg.求m1下降加速度.忽略悬挂线和滑轮的质量、轴承摩擦和阻力,线不可伸长. 力学 运动学m1=2.5 kg,m2=1.5 kga=30°,b=45°忽略摩擦力求m1 m2 如图16-45所示,固定在轻质弹簧的两端质量分别是m1=0.5kg,m2=1.49kg的两个物体置于固定在轻质弹簧的两端质量分别是m1=0.5kg,m2=1.49kg的两个物体置于光滑水平面上,m1靠在光滑竖直墙上.现有一颗m=0.01 动量守恒之碰撞1,2两球在光滑水平面上同向而行,动量分别是P1=5kg*m/s,P2=7kg*m/s,1球从后面和2球发生碰撞,碰后2球动量变为10kg*m/s,则两球质量M1,M2之间关系可能为A,M1=M2 B,2*M1=M2C,4*M1=M2 D,6*M1=M2 已知m1=4kg,m2=1kg,u1=0.3,u2=0.1.m1可视为质点,m2上表面的长L=1m,g=10m/s^21)F=20N,求m1,m2之间的摩擦力f=?2)F=60N,求m1从m2的左端到右端的时间t=? 如图所示,m1=10kg 、m2=50kg,m2与墙壁间用一轻绳相连.m1 与m2 间、m1与地面间的动摩擦因数均为μ=0.5,θ=53o .现要将m1匀速拉出且m2保持不动.求:(1)绳对m2的拉力大小.(2)拉力F的大小(sin53°=0.8,cos53° 20、在光滑水平面上,有一质量m1=20kg的小车,通过一根几乎不可伸长的轻绳与我不明白为什么μm3gs1=1/2(m1+m2)(v1^2-v2^2) 关m1什么事?20、在光滑水平面上,有一质量m1=20kg的小车,通过一根几乎