Yhuihoifjhh <span>F = Gm1m2 / r^2
if the masses are doubled then the force is increased by a factor of 4
if the distance is doubled the force is decreased by a factor of 1/ 2^2
the net result is no change in force</span>
Answer:
x₁ = 0.1878 m
Explanation:
For this exercise we will use conservation of energy
starting point. Highest point
Em₀ = U = m g h
final point. Lowest point with fully compressed spring
Em_f = K_e + U
Em_f = ½ K x² + m g x
energy is conserved
Em₀ = Em_f
m g h = ½ K x² + m g x
½ K x² + mg (x- h) = 0
let's substitute
½ 7.3 x² + 0.030 9.8 (x- 0.25) = 0
3.65 x² + 0.294 (x- 0.25) = 0
x² + 0.080548 (x- 0.25) = 0
x² - 0.020137 + 0.080548 x = 0
x² + 0.080548 x - 0.020137 = 0
let's solve the quadratic equation
x = [0.080548 ±√ (0.080548² + 4 0.020137)] / 2
x = [0.080548 ± 0.29502] / 2
x₁ = 0.1878 m
x₂ = -0.1072 m
These are the compression and extension displacement of the spring
Answer:
Pu - 239 have the smaller critical mass.
Explanation:
Critical mass is the smallest amount of certain element of mass that is needed to achieve a nuclear chain reaction early . Since Pu - 239 releases an average of 2.7 neutrons per fission as compared to U - 235 that releases 2.5 neutrons per fission. So, Pu - 239 has smaller critical mass, because Pu - 239 has a higher probability for fission and produces a large no. of neutrons per fission event. Infact of all the basic nuclear fuels, Pu - 239 has smallest critical mass. Critical mass depends on the nuclear properties of elements undergoing fission reaction. Hence, as Pu - 239 produces large no. of neutrons per fission than U - 235 and Pu - 239 has smaller critical mass.
Answer:
15.065ft
Explanation:
To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.
By definition the drag force is expressed as:
Where
is the density of the flow
V = Velocity
= Drag coefficient
A = Area
For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3
For second Newton's Law the Force is also defined as,
Equating both equations we have:
Integrating
Here,
Replacing:
Answer:
2041.66666667
Explanation:
35 N → 350 METERS for 6 seconds
force distance
350 M / 6s = 58.3
58.3 x 35 N = 2041.66666667 work
