HW #3-1-solutions

HW #3-1-solutions - pasha(sep635 HW#3-1 Antoniewicz(56445...

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pasha (sep635) – HW #3-1 – Antoniewicz – (56445) 1 This print-out should have 23 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 of 4) 3.0 points Each part oF this problem will state an ex- ample oF objects interacting via some Force. Choose the Fundamental interaction that is responsible in each case. Aneu tronou ts ideanuc leusdecaysin toa proton, electron, and antineutrino. 1. Strong 2. Electromagnetic 3. Gravitational 4. Weak correct Explanation: Notice the presence oF an antineutrino in this decay. Neutrinos (and their antiparticle partners) only interact via the weak Force, so this must be a weak interaction. 002 (part 2 of 4) 2.0 points Protons and neutrons attract each other in anuc leus. 1. Gravitational 2. Strong correct 3. Electromagnetic 4. Weak Explanation: The Force that holds the nucleus together must be strong enough to overcome the repul- sion between protons due to the electromag- netic Force. This is a strong interaction. 003 (part 3 of 4) 2.0 points The Earth pulls on the Moon. 1. Gravitational correct 2. Strong 3. Weak 4. Electromagnetic Explanation: This is a gravitational interaction, in which massive bodies attract each other. 004 (part 4 of 4) 2.0 points Protons in a nucleus repel each other. 1. Weak 2. Strong 3. Gravitational 4. Electromagnetic correct Explanation: This is an example oF an electromagnetic in- teraction, in which particles with like charges repel. 005 (part 1 of 2) 5.0 points The mass oF the Sun is 2 × 10 30 kg, and the mass oF Mercury is 3 . 3 × 10 23 kg. The distance From the Sun to Mercury is 4 . 8 × 10 10 m. ±irst, calculate the magnitude oF the gravi- tational Force exerted by the Sun on Mercury. Use G =6 . 67 × 10 - 11 m 3 kg · s 2 . Correct answer: 1 . 91068 × 10 22 N. Explanation: The magnitude oF the gravitational Force between two objects is given by ± ± ± ~ F gr ± ± ± = ± ± ± ± - Gm 1 m 2 r 2 ± ± ± ± = Gm 1 m 2 r 2 . We just need to plug in the given constants to fnd the answer: ± ± ± ~ F gr ± ± ± = Gm 1 m 2 r 2 = ( G )(2 × 10 30 kg)(3 . 3 × 10 23 kg) (4 .

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pasha (sep635) – HW #3-1 – Antoniewicz – (56445) 2 = 1 . 91068 × 10 22 N . where G =6 . 67 × 10 - 11 m 3 kg · s 2 . 006 (part 2 of 2) 4.0 points Calculate the magnitude of the gravitational force exerted by Mercury on the Sun. Correct answer: 1 . 91068 × 10 22 N. Explanation: By Newton’s third law, and by simply look- ing at the formula for the gravitational force, it is clear that the force exerted by Mercury on the Sun will be the same as that exerted by the Sun on Mercury. 007
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HW #3-1-solutions - pasha(sep635 HW#3-1 Antoniewicz(56445...

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