1) A particle has a rest mass of 6.95×10−27 kg and a momentum of 5.15×10−18 kg⋅m/s.
Determine the total relativistic energy of the particle.E=
JJ
Find the ratio of the particle's relativistic kinetic energy to its rest energy.
??rest=
2)
Estimate the difference Δtdiff in a 15000-s time interval as measured by a proper observer and a relative observer traveling on a commercial jetliner.
Δ?diff=
s
3)
Suppose that you have found a way to convert the rest energy of any type of matter directly to usable energy with an efficiency of 57.0%. How many liters of water would be sufficient fuel to very slowly push the Moon 2.30 mm away from the Earth? The density of water is ?water=1.00kg/liter, the Earth's mass is ?earth=5.97×1024 kg, the Moon's mass is ?moon=7.36×1022 kg, and the separation of the Earth and Moon is ?E,M=3.84×108 m
water:
Liters
1) A particle has a rest mass of 6.95×10−27 kg and a momentum of 5.15×10−18 kg⋅m/s....
Suppose that you have found a way to convert the rest energy of any type of matter directly to usable energy with an efficiency of 73.0%. How many liters of water would be sufficient fuel to very slowly push the Moon 2.70 mm away from the Earth? The density of water is Pwater 1.00kg/liter, the Earth's mass is Mearth 5.97 x 1024 kg, the Moon's mass is 7.36 x 1022 kg, and the separation of the Earth and Moon is...
Suppose that you have found a way to convert the rest energy of any type of matter directly to usable energy with an efficiency of 81.0%. How many liters of water would be sufficient fuel to very slowly push the Moon 2.10 mm away from the Earth? The density of water is Pwater = 1.00kg/liter, the Earth's mass is Mearth = 5.97 x 10 kg, the Moon's mass is Mmoon = 7.36 x 1022 kg, and the separation of the...
Suppose that you have found a way to convert the rest energy of any type of matter directly to usable energy with an efficiency of 45.0%. How many liters of water would be sufficient fuel to very slowly push the Moon 2.30 mm away from the Earth? The density of water is pwater = 1.00kg/liter, the Earth's mass is Mcarth = 5.97 x 1024 kg, the Moon's mass is Mimon = 7.36 x 1022 kg, and the separation of the...
A particle has a rest mass of 6.95×10^-27 and a momentum of 4.73×10^−18 kg⋅m/s. Determine the total relativistic energy of the particle. E = ______________ J Find the ratio of the particle's relativistic kinetic energy to its rest energy. ?? rest = _________________
A particle of mass 1.67Ý10-27 kg moves with a speed of 2.6Ý108 m/s relative to an observer. What is the kinetic energy of the particle in the rest frame of the observer?
A proton (rest mass is 1.673 times 10^-27 kg, rest energy is 938.3 MeV) has kinetic energy of 2500 MeV. find its momentum (in kg-m/s) (use relativistic relations) find its wavelength if one measures the proton's position to an uncertainty delta x of + l-5.60 times 10^-14 m. find the minimum possible uncertainty in the proton's momentum A panicle of mass 6.646 times 10^-27 kg is confined to a one-dimensional box of length 3.0 times 10^-14 m. What wavelength photon...
What is the rest mass of a particle with a relativistic momentum of 5.32 x 10^-19 kg*m/s and is moving at 0.75 c (c is the speed of light)?
1. A 30.00 kg mass moving with the velocity of 10 m/s. The momentum of the body will be A. 144.2 kg m/s. B. 187.8 kg m/s. C. 320.0 kg m/s. D. 442.4 kg m/s. E. None of the above 2. Out of the following which one is not a vector quantity? A. Force B. Impulse C. Mass D. velocity 3. Impulse is the product of which two quantities? A. Mass and Time B. Mass and Momentum C. Force and...
1. An unstable particle with mass 2.81 x 10-27 kg is initially at rest. The particle decays into two fragments that fly off along the x axis with velocity components u, 0.987c and u, --0.898c. From this information, we wish to determine the masses of fragments 1 and 2 (a) Is the initial system of the unstable particle, which becomes the system of the two fragments, isolated or nonisolated? [i.e. Are there external forces on the system?] (b) Based on...
Earth Mass/Radius: 1 M= 5.97 x 10^24 kg 1 R= 6.38 x 10^6 m Compare th centripetal force of s 75 kg person standing on the equator of the earth to that of the gravitational force due to the earths rotation. In other words, compute the ratio ( F gravitational/ F centripetal). How many revolutions per day would the earth have to turn to make these forces equal to one another (to makes this ratio exactly 1)?