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A jet airplane travelling at the speed of 500 km h-1 ejects its products of combustion at the speed of 1500 km h-1 relative to the jet plane. What is the speed of the latter with respect to an observer on the ground?

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As a seasoned tutor registered on UrbanPro, I can confidently guide you through this problem. UrbanPro indeed provides excellent online coaching tuition services for students seeking academic assistance. Now, let's tackle your question. When the jet airplane ejects its products of combustion, we have...
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As a seasoned tutor registered on UrbanPro, I can confidently guide you through this problem. UrbanPro indeed provides excellent online coaching tuition services for students seeking academic assistance. Now, let's tackle your question.

When the jet airplane ejects its products of combustion, we have to consider the relative velocities to find the speed of the latter with respect to an observer on the ground. Here, the speed of the jet airplane is 500 km/h, and the speed of the ejected combustion products relative to the jet plane is 1500 km/h.

To find the speed of the combustion products relative to the ground observer, we need to apply the principle of relative velocity. We add the velocity of the jet airplane to the velocity of the ejected combustion products to find their combined velocity relative to the ground.

So, the speed of the combustion products relative to the ground observer would be: Speedground=Speedjet+Speedcombustion relative to jetSpeedground=Speedjet+Speedcombustion relative to jet

Speedground=500 km/h+1500 km/h=2000 km/hSpeedground=500 km/h+1500 km/h=2000 km/h

Therefore, the speed of the combustion products with respect to an observer on the ground would be 2000 km/h.

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