作业帮英语翻译原题是英文 中文是个人翻译,不标准 所以附上英文原版.1.How long would it take a je
来源：学生作业帮 编辑：作业帮 分类：英语作业 时间：2024/05/14 13:03:28
英语翻译
原题是英文 中文是个人翻译,不标准 所以附上英文原版.
1.How long would it take a jet travelling at Mach 1.8 to fly a distance of 10500 km?The air temperature at this elevation is -40°F (about -4.444°C).
1.一架喷射机在1.8马赫情况下飞行10500千米需要多长时间?飞机所在环境 -40华氏度（-4.444摄氏度左右）.
2.As a car is driving towards you,the sound coming from its engine has a frequency of 1310 Hz.After it passes you,the engine sound has a frequency of 1100 Hz.If the air temperature is 15°C,what was the speed of the car?
2.一辆汽车在向你驶来时,它引擎所发出的声音在 1310 Hz,当他经过你以后的声音为 1100 Hz.如果当天气温为15摄氏度,汽车的速度为多少?
3.A bullet of mass 5 g is travelling at a constant speed of 120 m/s when it hits and lodges into a pendulum bob of mass 1 kg.What maximum height will the pendulum bob move to AND what will be its displacement angle?
3.一个5克的子弹以120 m/s 速度射入一个1千克的静止钟摆；该钟摆能上升的最高高度为多少?上升角度为多少?
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懒得算也好歹写一下思路!
悬赏分肯定会在50分以上,但是现在不能给,要不然没有满意答案就拿不回来了（你们懂得~）
原题是英文 中文是个人翻译,不标准 所以附上英文原版.
1.How long would it take a jet travelling at Mach 1.8 to fly a distance of 10500 km?The air temperature at this elevation is -40°F (about -4.444°C).
1.一架喷射机在1.8马赫情况下飞行10500千米需要多长时间?飞机所在环境 -40华氏度（-4.444摄氏度左右）.
2.As a car is driving towards you,the sound coming from its engine has a frequency of 1310 Hz.After it passes you,the engine sound has a frequency of 1100 Hz.If the air temperature is 15°C,what was the speed of the car?
2.一辆汽车在向你驶来时,它引擎所发出的声音在 1310 Hz,当他经过你以后的声音为 1100 Hz.如果当天气温为15摄氏度,汽车的速度为多少?
3.A bullet of mass 5 g is travelling at a constant speed of 120 m/s when it hits and lodges into a pendulum bob of mass 1 kg.What maximum height will the pendulum bob move to AND what will be its displacement angle?
3.一个5克的子弹以120 m/s 速度射入一个1千克的静止钟摆；该钟摆能上升的最高高度为多少?上升角度为多少?
——————————————————————————————
懒得算也好歹写一下思路!
悬赏分肯定会在50分以上,但是现在不能给,要不然没有满意答案就拿不回来了（你们懂得~）
（1）音速 u=331.3+(0.606c)m/s ,在-4.444摄氏度时,u=328.6m/s
1.8马赫即为1.8倍音速
t=10500/（1.8*328.6）=17.75s
(2)根据多普勒效应,观察者不动,波源相对介质以速度v接近观察者,观察者接收到的频率
f'=[u/(u-v)]·f
若其远离观察者,则观察者接受到的频率
f'=[u/(u+v)]·f
u是波传递的速度,为音速340m/s,v是波源运动的速度,待求,则上下一比
1310/1100=(340+v)/(340-v),v=29.63m/s
（3）根据能量守恒1/2m子弹v^2=（m钟+m子弹）gh,所以上升高度h约为3.6m.
1.8马赫即为1.8倍音速
t=10500/（1.8*328.6）=17.75s
(2)根据多普勒效应,观察者不动,波源相对介质以速度v接近观察者,观察者接收到的频率
f'=[u/(u-v)]·f
若其远离观察者,则观察者接受到的频率
f'=[u/(u+v)]·f
u是波传递的速度,为音速340m/s,v是波源运动的速度,待求,则上下一比
1310/1100=(340+v)/(340-v),v=29.63m/s
（3）根据能量守恒1/2m子弹v^2=（m钟+m子弹）gh,所以上升高度h约为3.6m.
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