Apollo 14 astronaut Alan Shepard famously took two golf shots on the Moon where it’s been estimated that an expertly hit shot could travel for 70.0 s through the Moon’s reduced gravity, airless environment to a maximum range of 4.00 km (about 2.5 miles). Assuming such an expert shot has a launch angle of 45.0°, determine the golf ball’s (a) kinetic energy as it leaves the club, and (b) maximum altitude in km above the lunar surface. Take the mass of a golf ball to be 0.045 0 kg and the Moon’s gravitational acceleration to be g moon = 1.63 m/s 2 .
Apollo 14 astronaut Alan Shepard famously took two golf shots on the Moon where it’s been estimated that an expertly hit shot could travel for 70.0 s through the Moon’s reduced gravity, airless environment to a maximum range of 4.00 km (about 2.5 miles). Assuming such an expert shot has a launch angle of 45.0°, determine the golf ball’s (a) kinetic energy as it leaves the club, and (b) maximum altitude in km above the lunar surface. Take the mass of a golf ball to be 0.045 0 kg and the Moon’s gravitational acceleration to be g moon = 1.63 m/s 2 .
Solution Summary: The author explains the kinetic energy of the golf ball as it leaves the club.
Apollo 14 astronaut Alan Shepard famously took two golf shots on the Moon where it’s been estimated that an expertly hit shot could travel for 70.0 s through the Moon’s reduced gravity, airless environment to a maximum range of 4.00 km (about 2.5 miles). Assuming such an expert shot has a launch angle of 45.0°, determine the golf ball’s (a) kinetic energy as it leaves the club, and (b) maximum altitude in km above the lunar surface. Take the mass of a golf ball to be 0.045 0 kg and the Moon’s gravitational acceleration to be gmoon = 1.63 m/s2.
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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