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Problem 13 Medium Difficulty

ssm mm A rocket of mass $4.50 \times 10^{5} \mathrm{kg}$ is in flight. Its thrust is directed at an angle of $55.0^{\circ}$ above the horizontal and has a magnitude of $7.50 \times 10^{6} \mathrm{N}$ . Find the magnitude and direction of the rocket's acceleration. Give the direction as an angle above the horizontal.

Answer

$|a|=10.3067 \mathrm{m} / \mathrm{sec}^{2} \quad \theta=21.9495^{\circ}$

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Video Transcript

this question. We have to determine where it's the magnitude and the angle that the acceleration factor makes with the horizontal axis. We begin by drawing our two reference access the white access on the X axis, which coincides. If there is Aunt Alexis, then he proceeded to decompose both forces the weight and the trust force on the axe. And why components? How can we do that? Let me call this half on. These were for the weight. We have nobody and remember that the weight is equals to you. The mass times gravity. Then we have 4.5 times Stand to the fifth times, 9.8. Let's keep it like this for a while now we proceed to that. Compose the force half on both access. How can he do that? Remember that for the X component. We have magnitude times the co sign off the angle between the force on the horizontal axis. For the white component. We have the magnitude times this sign off the off the angle between the force and horizontal axis to reform. F Component X is equal to 7.5 time. Stand to the sixth time's The cursed sign off 55 degrees, which gives us approximately 4.3 time stan to the sixth new terms now for do I. Component. We have 7.5 times stand to the sixth times the sign off 55 degrees, which gives us approximately 6.1 time stan to this 60. No, we knew the components off the force F note that we already know the component off the weight force. It only has one component that points towards the negative. Why direction? The reform? We can write it like that Free X component off the weak force is that goes to zero and then vertical component is a question minus 4.5 Time standard 50 times 9.8, which is approximately 44.1 times 10. The fifth new clothes. Now, let me summarize these results. Now we can use Newton's second law to complete What are the accent? Why components off the acceleration for the axe component we have? The following the net force on the X direction is because to the mass times the acceleration on X direction, then the net force index direction is he goes to the X component off the force F. So 4.3 times 10 to the sixth is equals to 4.5 times 10 to the fifth times acceleration over Dex access. Then the X component off the acceleration is it goes to 4.3 times 10 to the sixth, divided by 4.5 times 10 to the fifth. These is equals to 4.3 times 10 to the sixth, minus five divided by 4.5. And this gives us approximately 9.6 meters per second squared for Dex component off acceleration for the white component. We have the following net force on the Y direction. Is it close to the mass times acceleration on the wind direction? The net force in the wind direction is it close to 6.1 times 10 to the sixth, minus 44.1 times 10. So the fifth and this is equals to 4.5 times stand to the 50 times they y component off acceleration Then the white component off acceleration is Eco's truth 5.6 point one times 10 to the sixth miners 44.1 times 10 to the fifth, divided by 4.5 times 10 to the fifth and this is approximately 3.8 meters per second squared. Now let me organize the results. The magnitude of the resulting acceleration is equal to the square it off. It's X component squared. Plus it's why components squared. It's just the category Imperium, and he's requested a square it off 9.6 squared plus 3.8 square, which is approximately then 0.3 meters per second squared. So this is the magnitude of the resulting acceleration. Now, what is the angle that the resulting acceleration makes with the horizontal access? Well, we know that the tangent off that angle is equals. Two. The white component off acceleration divided by the X component off the acceleration. And these equals two 3.8 divided by 9.6, which is approximately zero point for me, then the angle is ICO student inverse tangent off 0.4, which is approximately 21.8 degrees. So this is the angle between the horizontal. Under resulting acceleration, a drawing will be the following year is the result of acceleration. Its magnitude is 10.3 meters per second squared, and the angle it makes with the horizontal axes is 21.8 degrees