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mestny [16]
3 years ago
5

Robert has just bought a new model roket, and is trying to measure its flight characteristics. The rocket engine package claims

that it will maintain a constant thrust of 11.4 N until the engine is used up. Robert launches the rocket on a windless day, so that it travels straight up, and uses his laser range-finder to measure that the height of the rocket when the engine cuts off is 19.0 m. He also measures the rocket's peak height, which is 23.5 m. If the rocket has a mass of 0.613 kg, how much work is done by the drag force on the rocket during its ascent?
Physics
1 answer:
rewona [7]3 years ago
3 0
At the peak of its flight ALL the energy given to the rocket is potential energy (its velocity is zero) and that is calculated as mgh So Energy given to rocket = mgh Energy expended by engine = F x D (D= height where engine stops) Energy 'lost' to drag is the difference between the two values. please if this helped mark it as the brainiest answer.
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3 years ago
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A monkey running through the jungle goes 0.198 km straight to the East, then turns 15.8° deflection from straight East toward th
inn [45]

Answer:

|\vec r|=339.82\ m

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Explanation:

<u>Displacement </u>

It's a vector magnitude that measures the space traveled by a particle between an initial and a final position. The total displacement can be obtained by adding the vectors of each individual displacement. In the case of two displacements:

\vec r=\vec r_1+\vec r_2

Given a vector as its polar coordinates (r,\theta), the corresponding rectangular coordinates are computed with

x=rcos\theta

y=rsin\theta

And the vector is expressed as

\vec z==

The monkey first makes a displacement given by (0.198 km,0°). The angle is 0 because it goes to the East, the zero-reference for angles. Thus the first displacement is

\vec r_1==\ km=\ m

The second move is (145 m , -15.8°). The angle is negative because it points South of East. The second displacement is

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The total displacement is

\vec r=\ m+\ m

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In (magnitude,angle) form:

|\vec r|=\sqrt{337.52^2+(-39.48)^2}=339.82\ m

\boxed{|\vec r|=339.82\ m}

\displaystyle tan\theta=\frac{-39.48}{337.52}=-0.1169

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5 0
3 years ago
A student pushes on a crate with a force of 100 Newtons directed to the right.
romanna [79]

Answer:

100N

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For every action, there is an equal and opposite reaction.

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2 years ago
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You would get 13.7 
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