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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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SOVA2 [1]
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2 years ago
Fluid originally flows through atube at a rate of 200 cm3/s. Toillustrate the sensitivity of the Poiseuille flow rate to various
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Answer:

Q_{2}=1200cm^{3}/s

Explanation:

Given data

Q₁=200cm³/s

We know that:

F=n\frac{vA}{l}\\

can be written as:

ΔP=F/A=n×v/L

And

Q=ΔP/R

As

n₂=6.0n₁

So

Q=ΔP/R

Q=\frac{nv}{lR}\\ \frac{Q_{2}}{n_{2}}= \frac{Q_{1}}{n_{1}}\\ Q_{2}=\frac{Q_{1}}{n_{1}}*(n_{2})\\Q_{2}=\frac{200}{n_{1}}*6.0n_{1}\\ Q_{2}=1200cm^{3}/s

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2 years ago
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7 0
3 years ago
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AVprozaik [17]

Answer:

103239.89 days

Explanation:

Kepler's third law states that the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

                               a³ / T² = 7.496 × 10⁻⁶  (a.u.³/days²)

where,

a is the distance of the semi-major axis in a.u

T is the orbit time in days

Converting the mean distance of the new planet to astronomical unit (a.u.)

                       1 a.u = 9.296 × 10⁷ miles        

                                      \frac{4004 * 10^{6}}{9.296 * 10^{7}}  =  43.07\ a.u.

Substituting the values into Kepler's third law equation;

                                    \frac{(43.07)^{3}}{T^{2}}  =  7.496 * 10^{-6}  

                                    T^{2} = \frac{(43.07)^{3}}{7.496 * 10^{-6}} (days)²

                                    T^{2} = \sqrt{\frac{(43.07)^{3}}{7.496 * 10^{-6}}}

                                    T = 103239.89 days

An estimate time T for the new planet to travel around the sun in an orbit is 103239.89 days

7 0
2 years ago
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laiz [17]

Answer:

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Explanation:

Given data

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Work

Solution

The work done on the charge is equal to difference in potential energy

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4 0
3 years ago
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