A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases at a rate [dM (t)] / dt = bv2 (t), where v(t) is its instantaneous velocity. The instantaneous acceleration of the satellite is: 1) –bv3(t) 2) –bv3 / M(t) 3) –2bv3 / M(t) 4) –bv3 / 2M(t)
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A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases at a rate [dM (t)] / dt = bv2 (t), where v(t) is its instantaneous velocity. The instantaneous acceleration of the satellite is:
- 1) –bv3(t)
- 2) –bv3 / M(t)
- 3) –2bv3 / M(t)
- 4) –bv3 / 2M(t)
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- A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases at a rate [dM (t)] / dt = bv2 (t), where v(t) is its instantaneous velocity. The instantaneous acceleration of the satellite is: 1) –bv3(t) 2) –bv3 / M(t) 3) –2bv3 / M(t) 4) –bv3 / 2M(t)An object is moving with ordinary speed 0.5c in S reference frame along a direction making an angle (tan 9 = 2) with respect to the positive x - axis. Find the zeroth component of the proper velocity.If ever the human race would need to leave Earth, the Trappist-1 system might make a good destination – the star there has seven temperate terrestrial planets, a few of which are close to Earth’s size and mass. It’s 39.6 light years away, though, so it could take a long time to get there. (a) If the spaceship had a speed of 0.95c, how long would the journey take from the Earth’s reference frame? (b) How long would the journey be for the passengers on the spaceship?
- An astronaut wishes to visit the Andromeda galaxy, making a one-way trip that will take 25.3 years in the space-ship's frame of reference. Assume the galaxy is 2.00 million light years away and his speed is constant. (a) How fast must he travel relative to Earth? The following approximation will prove useful: z 1 1 + x for x << 1. 2 - (Complete the equation for your answer.) (1-| C (b) What will be the kinetic energy of his spacecraft, which has mass of 1.08 x 10° kg? (c) What is the cost of this energy if it is purchased at a typical consumer price for electric energy, 13.0 cents per kWh?A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases at a rate [dM (t)] / dt = bv2 (t), where v(t) is its instantaneous velocity. The instantaneous acceleration of the satellite is: 1) –bv3(t) 2) –bv3 / M(t) 3) –2bv3 / M(t) 4) –bv3 / 2M(t)A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases at a rate [dM (t)] / dt = bv2 (t), where v(t) is its instantaneous velocity. The instantaneous acceleration of the satellite is: 1) –bv3(t) 2) –bv3 / M(t) 3) –2bv3 / M(t) 4) –bv3 / 2M(t)
- Suppose a cosmic ray colliding with a nucleus in the Earth's upper atmosphere produces a muon that has speed v = 0.99c. The muon then travels at constant speed and lives 1.5 μs as measured in the muon's frame of reference. (You can imagine this as the muon's internal clock.)Randomized Variablesv = 0.99 ct = 1.5 μs Part (a) How many kilometers does the muon travel according to an Earth-bound observer? Part (b) How many kilometers of the Earth pass by as viewed by an observer moving with the muon? Base your calculation on its speed relative to the Earth and its lifetime (proper time).Consider a cosmic ray colliding with a nucleus in the Earth's upper atmosphere that produces a muon which has a velocity v = 0.916c. The muon then travels at constant velocity and lives 1.52 µs as measured in the muon's frame of reference. (You can imagine this as the muon's internal clock.) (a) How far (in km) does the muon in travel according to an Earth-bound observer? (b) How far (in km) does it travel as viewed by an observer moving with it? Base your calculation on its velocity relative to the Earth and the time it lives (proper time).What are (a) K, (b) E, and (c) p (in GeV/c) for a proton moving at speed 0.990c? What are (d) K, (e) E, and (f) p (in MeV/c) for an electron moving at speed 0.990c?
- The relative velocity of frame Y with respect to frame Z is Byz = 0.5. The relative velocity of frame X with respect to Y is Bxy = -0.25. The relative velocity of frame W with respect to frame X is Bwx = 0.125. Finally, the relative velocity of frame V with respect to frame W is Byw = -0.0625. What is the relative velocity of frame V with respect to frame Z (By/z)? A. 0.3125 B. 0.3425 C. 0.4375 D. 0.5000Suppose an astronaut, travels so fast that γ = 30.00 . (a) She travels from the Earth to the nearest star system, Alpha Centauri, 4.300 light years (ly) away as measured by an Earthbound observer. How far apart are the Earth and Alpha Centauri as measured by the astronaut? (b) In terms of c , what is her velocity relative to the Earth? You may neglect the motion of the Earth relative to the Sun.An unstable high-energy particle is created in the laboratory, and it moves at a speed of 0.992c. Relative to a stationary reference frame fixed to the laboratory, the particle travels a distance of 1.05 × 10-3 m before disintegrating. What is (a) the proper distance and (b) the distance measured by a hypothetical person traveling with the particle? Determine the particle's (c) proper lifetime and (d) its dilated lifetime. (a) Number Units (b) Number Units (c) Number Units (d) Number Units