Bam!— Apollo 15 Lands on the Moon The first word spoken on the surface of the Moon after Apollo 15 landed on July 30, 1971, was “Bam!” This was James Irwin’s involuntary reaction to their rather bone-jarring touchdown. “We did hit harder than any of the other flights!” says Irwin. “And I was startled, obviously, when I said. Bam!’" The reason for the “firm touchdown” of Apollo 15, as pilot David Scott later characterized it, was that the rocket engine was shut off a bit earlier than planned, when the lander was still 4.30 ft above the lunar surface and moving downward with a speed of 0.500 ft/s. From that point on the lander descended in lunar free fall, with an acceleration of 1.62 m/s 2 . As a result, the landing speed of Apollo 15 was by far the largest of any of the Apollo missions In comparison, Neil Armstrong s landing speed on Apollo 11 was the lowest at 1.7 ft/s—he didn’t shut off the engine until the footpads were actually on the surface. Apollos 12 , 14 , and 17 all landed with speeds between 3.0 and 3.5 ft/s. To better understand the descent of Apollo 15, we show its trajectory during the final stages of landing in Figure 2-47 (a) In Figure 2-47 (b) we show a variety of speed-versus-time plots. (a) (b) FIGURE 2-4 7 Problems 108, 109, 110, and 111 109. •• What was the impact speed of the lander when it touched down? Give your answer in feet per second (ft/s), the same units used by the astronauts. A. A. 2.41 ft/s B. B 6.78 ft/s C. C. 9.95 ft/s D. D. 10.6 ft/s
Bam!— Apollo 15 Lands on the Moon The first word spoken on the surface of the Moon after Apollo 15 landed on July 30, 1971, was “Bam!” This was James Irwin’s involuntary reaction to their rather bone-jarring touchdown. “We did hit harder than any of the other flights!” says Irwin. “And I was startled, obviously, when I said. Bam!’" The reason for the “firm touchdown” of Apollo 15, as pilot David Scott later characterized it, was that the rocket engine was shut off a bit earlier than planned, when the lander was still 4.30 ft above the lunar surface and moving downward with a speed of 0.500 ft/s. From that point on the lander descended in lunar free fall, with an acceleration of 1.62 m/s 2 . As a result, the landing speed of Apollo 15 was by far the largest of any of the Apollo missions In comparison, Neil Armstrong s landing speed on Apollo 11 was the lowest at 1.7 ft/s—he didn’t shut off the engine until the footpads were actually on the surface. Apollos 12 , 14 , and 17 all landed with speeds between 3.0 and 3.5 ft/s. To better understand the descent of Apollo 15, we show its trajectory during the final stages of landing in Figure 2-47 (a) In Figure 2-47 (b) we show a variety of speed-versus-time plots. (a) (b) FIGURE 2-4 7 Problems 108, 109, 110, and 111 109. •• What was the impact speed of the lander when it touched down? Give your answer in feet per second (ft/s), the same units used by the astronauts. A. A. 2.41 ft/s B. B 6.78 ft/s C. C. 9.95 ft/s D. D. 10.6 ft/s
The first word spoken on the surface of the Moon after Apollo 15 landed on July 30, 1971, was “Bam!” This was James Irwin’s involuntary reaction to their rather bone-jarring touchdown. “We did hit harder than any of the other flights!” says Irwin. “And I was startled, obviously, when I said. Bam!’"
The reason for the “firm touchdown” of Apollo 15, as pilot David Scott later characterized it, was that the rocket engine was shut off a bit earlier than planned, when the lander was still 4.30 ft above the lunar surface and moving downward with a speed of 0.500 ft/s. From that point on the lander descended in lunar free fall, with an acceleration of 1.62 m/s2. As a result, the landing speed of Apollo 15 was by far the largest of any of the Apollo missions In comparison, Neil Armstrong s landing speed on Apollo 11 was the lowest at 1.7 ft/s—he didn’t shut off the engine until the footpads were actually on the surface. Apollos 12, 14, and 17 all landed with speeds between 3.0 and 3.5 ft/s.
To better understand the descent of Apollo 15, we show its trajectory during the final stages of landing in Figure 2-47 (a) In Figure 2-47 (b) we show a variety of speed-versus-time plots.
(a)
(b)
FIGURE 2-47
Problems 108, 109, 110, and 111
109. •• What was the impact speed of the lander when it touched down? Give your answer in feet per second (ft/s), the same units used by the astronauts.
For each of the actions depicted below, a magnet and/or metal loop moves with velocity v→ (v→ is constant and has the same magnitude in all parts). Determine whether a current is induced in the metal loop. If so, indicate the direction of the current in the loop, either clockwise or counterclockwise when seen from the right of the loop. The axis of the magnet is lined up with the center of the loop. For the action depicted in (Figure 5), indicate the direction of the induced current in the loop (clockwise, counterclockwise or zero, when seen from the right of the loop). I know that the current is clockwise, I just dont understand why. Please fully explain why it's clockwise, Thank you
A planar double pendulum consists of two point masses \[m_1 = 1.00~\mathrm{kg}, \qquad m_2 = 1.00~\mathrm{kg}\]connected by massless, rigid rods of lengths \[L_1 = 1.00~\mathrm{m}, \qquad L_2 = 1.20~\mathrm{m}.\]The upper rod is hinged to a fixed pivot; gravity acts vertically downward with\[g = 9.81~\mathrm{m\,s^{-2}}.\]Define the generalized coordinates \(\theta_1,\theta_2\) as the angles each rod makes with thedownward vertical (positive anticlockwise, measured in radians unless stated otherwise).At \(t=0\) the system is released from rest with \[\theta_1(0)=120^{\circ}, \qquad\theta_2(0)=-10^{\circ}, \qquad\dot{\theta}_1(0)=\dot{\theta}_2(0)=0 .\]Using the exact nonlinear equations of motion (no small-angle or planar-pendulumapproximations) and assuming the rods never stretch or slip, determine the angle\(\theta_2\) at the instant\[t = 10.0~\mathrm{s}.\]Give the result in degrees, in the interval \((-180^{\circ},180^{\circ}]\).
What are the expected readings of the ammeter and voltmeter for the circuit in the figure below? (R = 5.60 Ω, ΔV = 6.30 V)
ammeter
I =
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