A few days ago
Please help me answer my Physics homework..a step by step solution..thanks,..?
The radius of a collapsing star destined to eventually become a pulsar decreases by 10 percent while at the same time 15 percent of its mass escapes. Calculate the percent change in density.
Top 2 Answers
A few days ago
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If the radius of the star is r, then assuming it is spherical, its volume v is proportional to r^3.
v = kr^3 …(1)
If the radius decreases by 10%, it becomes 90% = 0.9 of what it was. It is multiplied by 0.9.
If V is the new volume, then:
V = k(r*0.9)^3
V = kr^3 * (0.9)^3 …(2)
Dividing (2) by (1):
V / v = 0.9^3
V = 0.9^3 * v …(3)
Its mass decreases by 15%.
If the original mass is m and the new mass M, then:
M = 0.85m …(4)
Mass / volume = density.
Dividing (4) by (3):
M / V = 0.85m / 0.9^3 v
= 0.85 / (0.9^3) (m/v).
If D is the new density and d the old, then:
D / d = 0.85 / (0.9^3)
= 1.167 to 3 dec. pl.
The increase in density is 16.7%.
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4 years ago
mass m = 1500 kg velocity v = 20 m/s Kinetic power Ek = 0.5 x m x v^2 = 0.5 x 1500 kg x 4 hundred m^2/s^2 = 3 hundred,000 Joule = 3 hundred kJ distance s = a hundred m very final velocity vt = 10 m/s decceleration a vt^2 = 2 x a x s (10 m/s)^2 = 2 x a x a hundred m a = -0.5 m/s^2 (minus skill a decceleration) uncomplicated rigidity F = m x a = 1500 kg x -0.5 m/s^2 = -750 Newton (minus skill that the rigidity is utilized against the vehicle’s circulate)
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