Now we have a rod-shaped space station of length 1296 m and mass 1.43 x 10^6 kg, which can change its length (kind of like an old-fashioned telescope), without changing its overall mass. Suppose that the station is initially rotating at a constant rate of 1.83 rpm. If the length of the rod is reduced to 1.83 m, what will be the new rotation rate of the space station?
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- A planet rotates through one complete revolution every 27 hours. Since the axis of rotation is perpendicular to the equator, you can think of a person standing on the equator as standing on the edge of a disc that is rotating through one complete revolution every 27 hours. Find the angular velocity of a person standing on the equatorA space station consists of two donut-shaped living chambers, A and B, that have the radii shown in the figure. As the station rotates, an astronaut in chamber A is moved 369 m along a circular arc. How far along a circular arc is an astronaut in chamber B moved during the same time if r_a = 266 m and r_b = 165 m? Note: Express your answer in whole numbers. No unit is required for the final answer. Set your calculator in radians. ra A rB B.Consider a non-rotating space station in the shape of a long thin uniform rod of mass 6.76 x 10^6 kg and length 1131 meters. Rocket motors on both ends of the rod are ignited, applying a constant force of F = 5.27 x 10^5 N to each end of the rod as shown in the diagram, causing the station to rotate about its center. If the motors are left running for 1 minutes and 43 seconds before shutting off, then how fast will the station be rotating when the engines stop? 1.22 rpm 1.46 rpm 0.81 rpm 0.57 rpm
- a) What is the period of rotation of Venus in seconds? (The period of rotation of Venus in hours is 5,832.5 hr.) Answer in seconds (b)What is the angular velocity (in rad/s) of Venus? (Enter the magnitude.) Answer in rad/s (c) Given that Venus has a radius of 6.1 ✕ 106 m at its equator, what is the linear velocity (in m/s) at Venus's surface? (Enter the magnitude of the linear velocity at the equator.) Answer on m/sArtificial gravity is a must for any space station if humans are to live there for any extended length of time. Without artificial gravity, human growth is stunted and biological functions break down. An effective way to create artificial gravity is through the use of a rotating enclosed cylinder, as shown in the figure. Humans walk on the inside of the outer edge of the cylinder, which has a diameter of ?=2735 m that is large enough such that its curvature is not readily noticeable to the inhabitants. (The space station in the figure is not drawn to scale.) Once the space station is rotating at the necessary angular speed ? to create an artificial gravity of 1 g, how many minutes would it take the space station to make one revolution?Suppose that humans have created a colony outside of our solar system on a planet called Wfirst21. Wfirst21 has a mass of 2.65×1025 kg2.65×1025 kg and a day that lasts 23.7 h23.7 h (which defines the rotational period of the planet). The colony is located on the planet's equator. The colonists set up a communications satellite which orbits Wfirst21. The satellite has a circular orbit that keeps it positioned directly above the colony. Calculate the radius ?r of the satellite's orbit in kilometers.
- Consider a non-rotating space station in the shape of a long thin uniform rod of mass 9.84 x 10^6 kg and length 1286 meters. Rocket motors on both ends of the rod are ignited, applying a constant force of F = 4.77 x 10^5 N to each end of the rod as shown in the diagram, causing the station to rotate about its center. If the motors are left running for 2 minutes and 49 seconds before shutting off, then how fast will the station be rotating when the engines stop? 0.73 rpm 0.29 rpm 0.88 rpm 1.32 rpmYour roommate is working on his bicycle and has the bike upside down. He spins the 56.0 cmcm -diameter wheel, and you notice that a pebble stuck in the tread goes by three times every second. What is the pebble's acceleration? Express your answer with the appropriate units. The answer is NOT 110 49.7 49.8 or 0.63A wheel (of radius 0.5 meters) spins with an angular velocity of 2 radians per second (rad/s). If this wheel were rolling on the ground without sliding, how fast would the wheel be traveling? O 1 m/s O 0.5 m/s It's not this one. O 2 m/s
- ФЫ В А ПРОЯВАПРОЛДЗА ПРОЛДЖЭ ЯЧСМИТНЧСМИТЬБНСМИТЬБЮ SOLVE STEP BY STEP IN DIGITAL FORMAT 39.- A small sphere of lead with a mass of 25g is connected to the origin by a thin rod of 74cm and negligible mass. The rod rotates around the z axis in the (x,y) plane. A constant force of 22N in the "y" direction acts on the sphere. a) Assuming that the sphere is a particle, what will be the rotational inertia about the origin? b) If the rod makes an angle of 40° with the positive x-axis, calculate its angular acceleration. 40.- Three particles are connected to a thin rod one meter long and of negligible mass that rotates around the origin in the (x,y) plane. Particle 1 (mass 52g) is attached at a distance of 27cm from the origin, particle 2 (35g) is at 45cm and particle 3 (24g) is at 65cm. a) What is the rotational inertia of the system? b) if instead the rod rotated around the center of mass of the system, what would its rotational inertia be? ИЦУКЕНГШУКЕНІ ШЩКЕНТ ШщзхьAn amusement park ride has a platform that rotates at 1.5 rad/s. The car (C) is mounted to the platform and can rotate. The car rotates at 2.0 rad/s with respect to the platform. The center of the car is 3 meters from the center of the platform O. The rider B is 0.75 meters from the center of the car. Find Vr and Ar.Now we have a rod-shaped space station of length 1402 m and mass 8.50 x 10^6 kg, which can change its length (kind of like an old-fashioned telescope), without changing its overall mass. Suppose that the station is initially rotating at a constant rate of 1.51 rpm. If the length of the rod is reduced to 1.51 m, what will be the new rotation rate of the space station? 5.08 rpm 2.54 rpm 3 8.13 rpm 4 1.52 rpm