皮皮学，免费搜题
登录
搜题
【多选题】
在液力变出油道上常设有一个回油阀，其作用是A.
只有在液力变内的油压高于一定值时才打开，让受热液压进入液压油散热器
B.
当进入液力变的液压油压过，限压阀开启，让部分液压油泄回到油底壳，以防止液力变中的油压过导致油封漏油
C.
不但可以防止液力变油压过低而影响动力传递，而且可以降低液压油散热器油压，使之低于196kP，以防止油压过高造成耐压能力较低的散热器及油管漏油或破裂
D.
让一部分冷却液压油流回到齿轮变速器，对齿轮变速器中的轴承和齿轮进行润滑
拍照语音搜题，微信中搜索"皮皮学"使用
参考答案：
参考解析：
知识点：
皮皮学刷刷变学霸
举一反三
【简答题】完形填空。 Thomas Edison was a famous American scientist. He was l in 1847. When he was a child, he liked to 2 out how things worked. He was in school 3 only three months. He asked his teacher a lot of str...
【单选题】下列语句段中不是死循环的是
A.
k=0; do{ ++k; } while k<=0);< /div>
B.
i=100; while (1){ i=i%100+1; if (i==20) break;}
C.
s=3379; while (s++%2+3%2) s++;
D.
for (i=1;;i++) sum=sum+1;
【简答题】阅读理解。 Thomas Edison was a famous American scientist. He was born in 1847. When he was a child, he liked to find out how things worked. He was in school for only three months. He asked his teacher a lo...
【单选题】下列语句段中不是死循环的是
A.
i=100; while (1) { i=i%100+1; if (i==20) break; }
B.
for (i=1;;i++) sum=sum+1;
C.
k=0; do { ++k; } while (k<=0);< /div>
D.
s=3379; while (s++%2+3%2) s++;
【简答题】The inventor Thomas Edison was regarded as a(n) d____ boy when he was young because he couldn't work out easy mathematical calculations.
【简答题】When Thomas Edison was born in the small town of Milan, Ohio, in 1847, America was just beginning its great _____1______ development. The time in which he lived was an age of invention, filled with hu...
【简答题】下列语句段中不是死循环的是A. i=100; while(1) { i=i%100+1; if(i==20) break; }B. for(i=1;;i++) sum=sum+1;C. k=0; do { ++k; } while (k>=0);D. s=3379; while(s++%2+3%2) s++;
【单选题】the financial means to remain independent, Thomas Edison was compelled to seek employment as a night telegraph operator.
A.
He was deprived of
B.
Depriving of
C.
Although he was deprived of
D.
Deprived of
相关题目：
【单选题】下列语句段中不是死循环的是
A.
k=0; do{ ++k; } while k<=0);< /div>
B.
i=100; while (1){ i=i%100+1; if (i==20) break;}
C.
s=3379; while (s++%2+3%2) s++;
D.
for (i=1;;i++) sum=sum+1;
【单选题】下列语句段中不是死循环的是
A.
i=100; while (1) { i=i%100+1; if (i==20) break; }
B.
for (i=1;;i++) sum=sum+1;
C.
k=0; do { ++k; } while (k<=0);< /div>
D.
s=3379; while (s++%2+3%2) s++;