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frez [133]
3 years ago
10

A spacecraft on its way to Mars has small rocket engines mounted on its hull; one on its left surface and one on its back surfac

e. At a certain time, both engines turn on. The one on the left gives the spacecraft an acceleration component in the x direction of ax = 5.10 m/s2, while the one on the back gives an acceleration component in the y direction of ay = 7.30 m/s2. The engines turn off after firing for 645 s, at which point the spacecraft has velocity components of vx = 3780 m/s and vy = 4470 m/s. What was the magnitude and the direction of the spacecraft's initial velocity before the engines were turned on? Express the magnitude as m/s and the direction as an angle measured counterclockwise from the +x axis.
Physics
1 answer:
lozanna [386]3 years ago
4 0

Answer:

v=545.41 \frac{m}{s}

β=-25.93

Explanation:

Give the acceleration in 'x' and 'y' also the time can find the initial velocity using equation of uniform acceleration motion

For axis x

a_{x}=5.10\frac{m}{s^{2}} \\v_{fx}=3780\frac{m}{s} \\v_{fx}=v_{ox}+a_{x}*t\\v_{ox}=v_{fx}-a_{x}*t\\v_{ox}=3780 \frac{m}{s}-5.1\frac{m}{s^{2} }*645s\\v_{ox}=490.5\frac{m}{s}

For axis y

a_{y}=7.30\frac{m}{s^{2}} \\v_{fy}=4470\frac{m}{s} \\v_{fy}=v_{oy}+a_{y}*t\\v_{oy}=v_{fy}-a_{y}*t\\v_{oy}=4470\frac{m}{s}-7.30\frac{m}{s^{2} }*645s\\v_{oy}=-238.5\frac{m}{s}

Maginuted

v=\sqrt{v_{xo}^{2} +v_{yo}^{2} } \\v=\sqrt{490.5x^{2} +(-238.5)^{2} }\\v=545.41 \frac{m}{s}

The  direction is knowing when find the angle so

\beta =tan^{-1}*\frac{v_{yo} }{v_{xo}}\\\beta =tan^{-1}*\frac{-238.5}{490.5}\\\beta =tan^{-1}*-0.48\\\beta =-25.93

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bagirrra123 [75]

Answer:

500 J

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We know that mass = 10 kg and velocity = 10 m/s.

KE = (m × v × v) ÷ 2

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KE = 500 J

The kinetic energy of the object is 500 J (joules). Hope this helps, thank you !!

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A student of mass 60.0 kg, starting at rest, slides down a slide 20.0 m long, tilted at an angle of 30.0° with respect to the ho
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Answer:

Explanation:

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One of the most important properties of materials in many applications is strength. Two of the qualitative measures of the stren
katrin2010 [14]

To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.

In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.

By definition we know that the tensile strength is defined as

\sigma = \frac{F}{A}

Where,

\sigma =Tensile strength

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