3t3kg等于多少t，一固定斜面长5m高3m斜面上物体A的质量为m5kg物体与斜面
来源：整理 编辑：亚灵电子网 2023-02-26 00:00:37
本文目录一览
1，一固定斜面长5m高3m斜面上物体A的质量为m5kg物体与斜面
2，3t3kg 303t最大的是哪个
3.03t=3030kg3t3kg=3003kg所以3.03t大再看看别人怎么说的。
3，3吨5kg用小数怎样写
3吨5kg用小数表示是3.005吨解析：kg表示千克，5kg表示5千克1吨=1000千克，5÷1000=0.005（吨）3吨5千克=3吨+5千克=3吨+0.005吨=3.005吨解：1000kg=1t3kg=3÷1000=0.003t3t3kg=3+3÷1000=3.003t760÷1000=0.760t3t3kg换小数等于3.003t，760kg换小数等于0.760t
4，如图所示在倾角37的固定斜面上放置一质量M1kg长度L15m
由牛顿第二定律对滑块：ma1=mgsinθ-μ1mgcosθ可得a1=4m/s^2对平板：Ma2=Mgsinθ+μ1mgcosθ-μ2(m+M)gcosθ可得a2=1m/s^2设滑块运动到平板的下端B用时t1,则L=（a1-a2)t1^2/2,可得t1=1s在这段时间内，平板的位移为X1=a2t1^2/2=0.5m在t1末，滑块速度为v1=a1t1=4m/s平板速度为v2=a2t1=1m/s滑块离开平板后，二者加速度变为一样，设为a3,则a3=g(sinθ-cosθ)=2m/s^2设滑块与平板下端到达地面C所用时间分别为t2,t3,则X剩=v1t2+a3t2^2/2 可得t2=(6^0.5-2)sX剩=v2t3+a3t3^2/2 可得t3=1s故，△t=t3-t2=(3-6^0.5)s
5，环境风险评价章节中源项分析中源强计算中的热量蒸发估算 沸点高于环
环境风险评价中的泄漏液体蒸发通常分为三种:闪蒸蒸发、热量蒸发和质量蒸发。通常计算蒸发总量为三种蒸发之和,作为保守计算的源项。
对于液态物质的蒸发途径的源项分析,保守处理都是这样计算蒸发量的,无论是HCl,还是别的液体。
1 闪蒸量的估算
过热液体闪蒸量可按下式估算
Q1=F·WT /t 1
式中:
Q1——闪蒸量,kg/S;
WT——液体泄漏总量,kg;
t1——闪蒸蒸发时间,s;
F ——蒸发的液体占液体总量的比例;按下式计算
F=Cp×(TL-Tb)/H
式中:
Cp——液体的定压比热,J/(kg·K);
TL——泄漏前液体的温度,K;
Tb——液体在常压下的沸点,K;
H ——液体的气化热,J/kg。
2 热量蒸发估算
当液体闪蒸不完全,有一部分液体在地面形成液池,并吸收地面热量而气化称为热量蒸发。热量蒸发的蒸发速度Q2按下式计算:
Q2=λ×S×(T0-Tb)/[H(παt)^0.5]
式中:
Q2——热l量蒸发速度,kg/s;
T0——环境温度,k;
Tb——沸点温度;k;
S ——液池面积,m2;
H——液体气化热,J/kg;
λ——表面热导系数(见表A2-1),W/m·k;
α——表面热扩散系数(见表A2-1),m2/s;
t——蒸发时间,s。
3 质量蒸发估算
当热量蒸发结束,转由液池表面气流运动使液体蒸发,称之为质量蒸发。
质量蒸发速度Q3计算公式可以查阅专著,或咨询我。
液池最大直径取决于泄漏点附近的地域构型、泄漏的连续性或瞬时性。有围堰时,以围堰最大等效半径为液池半径;无围堰时,设定液体瞬间扩散到最小厚度时,推算液池等效半径。
4 液体蒸发总量的计算
Wp=Q1t1+Q2t2+Q3t3
液体泄漏估算中密度的单位通常是kg/m3
液体泄漏估算中密度的单位 是kg/m3,这是通用的单位。
盐酸是否存在热量蒸发?
闪蒸蒸发、热量蒸发和质量蒸发是在三个温度范围的三种状态,是由盐酸含有的热能决定的。当热量蒸发结束,转由液池表面气流运动使液体蒸发,就是质量蒸发。
6，一个关于求物体的位移的问题
解：动磨擦力Ft=MgU=3*10*0.2=6 N, 方向向右的水平推力F1, F1-Ft=Ma1,15-6=3a1, 得a1=3(10(m/s^2)), 下面的略去单位。6s之后的速度，v1=a1*t1=3*6=18, s1=1/2 *a1*t1^2=1/2 *3*6^2=54,
持续作用一段时间设为 t2,加速度设为a2, 则有， F2+Ft=Ma2, 得 a2=6，在t=v1/a2=3S时，上面的式子都是成立的。 一个大小仍为12牛，方向向右持续作用的水平推力F3时，向右的加速度为 a3, a3=(F3-Ft)/m=2，由静止到，18m/s需要9S，6+3+9=18显然是大于14S的，那么在F2时，作用时间，t2,末速度为，V2，可得，
方程式t=t1+t2+t3=6+t2+t3=14, V2=v1-a2*t2=18-6*t2, v3=V2+a3*t3=V2+2*t3=18
解得，t2=2, t3=6, v2=6, 则S2=V1*t2-a2*t2*t2/2=18*2-6*2*2/2=24, S3=v2*t3+a3*t3*t3/2=6*6+2*6*6/2=72
S=S1+S2+S3=54+24+72=150(米) （题中的物体一直是向右运动的）
我做得比较详细吧.从侧面证明了楼主说的是正确的,这下放心了吧.解：前六秒对物体受力分析可求得其加速度a1=3m/s2 设向右为正方向
第6s末其速度为v1=a1*t1=18m/s
第二阶段：在-12N的作用下对物体受力分析（此时摩擦力仍向左）可求得其加速度a2=-6m/s2
设作用了t2 则在第（6+t2）s时 v2=18-6t2
第三阶段：在12N的作用下对物体受力分析（假设此时速度仍向右，则摩擦力向左）可求得其加速度a3=2m/s2
设作用了t3 则第14s末v3=v2+a3t3=18-6t2+2t3=18m/s 化简：t3=t2
又因为 6+t2+t3=14 所以解得t2=2s t3=6s v2=6m/s
画出v-t图（我无法将其在电脑上画出，不过你可以根据上面的解答过程画出）求出面积即为物块在14s内发生的位移 答案为150m
7，2011年买的华为et127 是否可以用win7
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3t3kg等于多少t等于 多少 固定
