1 points OSColPhys2016 3.4.WA.034 My Notes Ask Your Teacher A spacecraft on its

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1 points OSColPhys2016 3.4.WA.034 My Notes Ask Your Teacher 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 surface. At a certain time, both engines turn on. The one on the left gives the spacecraft an acceleration component in the x direction of ay 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 620 s, at which point the spacecraft has velocity components of Vx -3785 m/s and vy4446 m/s. What was the magnitude and direction of the spacecraft's initial velocity, before the engines were turned on? Express the direction as an angle measured counterclockwise from the +x axis. magnitude direction m/s o counterclockwise from the +x-axis Additional Materials Reading -1 points OSColPhys2016 3.4.WA.035. Older televisions display a picture using a device called a cathode ray tube, where electrons are emitted at high speed and collide with a phosphorescent surface, causing light to be emitted. The paths of the electrons are altered by magnetic fields. Consider one such electron that is emitted with an initial velocity of 1.95 x 107 m/s in the horizontal direction when magnetic forces deflect the electron with a vertically upward acceleration of 5.40 x 1015 m/s2. The phosphorescent screen is a horizontal distance of 8.6 cm away from the point where the electron is emitted. My Notes Ask Your Teacher (a) How much time does the electron take to travel from the emission point to the screen? (b) How far does the electron travel vertically before it hits the screen? cm Additional Materials Reading

Explanation / Answer

Vx = ax*t = 5.10*620 = 3162 m/sec

Vox = Vx-Vx = 3785-3162 = 623 m/sec

Vy = ay*t = 7.30*620 = 4526 m/sec

Voy = Vy-Vy = 4446-4526 = -80 m/sec

Magnitude: V0 = sqrt (623^2 + (-80)^2) = 628.12 m/s

= arctan (80/623) = 7.32°

This is the angle below +x

So the required angle = 360 – 7.32 = 352.68°

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