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- There is relatively little space between atoms in solids and liquids, so that the average density of an atom is about the same as matter on a macroscopic scale - approximately 10^3 kg/m^3. The nucleus of an atom has a radius of about 10^-5m of that of the entire atom, and conatins nearly all mass of the atom. a) What is the approximate density, in kilograms per cubic meter, of a nucleus. b) One possible remnant of a supernova, called a neutron star, can have the density of a nucleus, while being the size of a small city. What would be the raduis, in kilometers, of the neutron star with a mass 10 times that of the Sun? The radius of the Sun is 7*10^8 m and its mass is 1.99*10^30 kgS Can 6 PAR к Торс K Unit K In x K Moti = Cop K Unit S Spee S Topo S Math Micr eb.kamihq.com/web/viewer.html?source-filepicker&document_identifier=137VZR5BZOVSAIMOA55WU555_CvJ9NacO + 100 P e Interpreting Graphs Answer the questions following the graphs on each side Dietance va. Time 2 7 10 11 12 13 14 Time in soconds 1. From 1 second to 2 seconds, how fast is the object traveling. (Take the difference in distance and divide it by the time in between the 2 distances) 2. Is the object going as fast between 9 and 12 seconds as it is between 1 and 4 seconds? How can you tell? 3. What is the motion of the object between 4 and 6 seconds? acerthe average density of an atom is approximately 103 kg/m3. The nucleus of an atom has a radius about 10-5 times that of the entire atom, and contains nearly all the mass of the atom. What is the approximate density, in kilograms per cubic meter, of a nucleus?
- What is the square root of 1023? (Hints: 10 is an integer number. Integer numbers are exact, they don't have any uncertainty. The result however is going to be a real number. Computers approximate real numbers using floating point numbers. Floating point numbers have a given number of significant figures or digits. LON-CAPA typically expects 3, 4 or 5 digits. 2 or 1 is not enough, 6 or more is too many. Very large or very small numbers are expressed in scientific or exponential notation like the number above. Numbers in scientific notation are entered to computer software - like LON- CAPA as 1.23E15 or 1.23e15, for example; and 1.23E-15 or 1.23e-15, if the exponent is negative. If both the number and the exponent are negative, then you type -1.23E- 15 or -1.23e-15. On calculators the button is usually labelled as 'EE', but 'E' or 'EEX' or 'Exp' are also possible.)Problem 1 Use the properties of Gamma functions to solve these expressions by hand. a program such as a T[ ] × [4] b C 3 — 92 xr 5 9 T[-2] × [2] Г xr 4 4Please Asap
- 4 Su24 on Question 1 (Mandatory) (10 points) ✓ Saved The correct formula to calculate a Percent Error between experimental value X and accepted value A is O al X-A 100% X+A b) IX-AI 100% Ob) A Oc) IX-AI 100% (X+A) 1/2 d) IX+AI 100% (X-A) acer ) 0:16:50 remainA valid, but probably useless, dimensionless group is givenby (μT0g)/ (YLα) , where everything has its usual meaning,except α . What are the dimensions of α ?( a ) θL-1T-1 , ( b ) θL-1T-2 , ( c ) θML-1 , ( d ) θ-1LT-1 ,( e ) θLT-1calculate: (a) the average of fex) = Cos (3x) from II to TT (い th average of f(x) = x² from (-10) to (l0) (c) th average of frxl = 3x from (-3) to (3)
- one hundred milli- (100 times one one-thousandth): 102 ✕ 10−3 = 10(2 + [−3]) = 10(2 − 3) = 10−1 = 0.1 (100 ÷ 1,000 = 0.1) one hundred micro- (100 times one one-millionth): 102 ✕ 10−6 = 10(2 + [−6]) = 10(2 − 6) = 10−4 How do we write one hundred nano- (100 times one-billionth)? 102 ✕ 10−9 = 10 We write ten micrometers as follows. 10 ✕ 10−6 m = 10−5 m How do we write ten nanoseconds? 10 ✕ 10−9 s = 10 sIf v→1=34i^+17j^, v→2x=80, and v→1 X v→2= 28k^ then : v2y = v2z=