If we let
p as the directed multigraph that has no isolated vertices and has an Euler circuit
q as the graph that is weakly connected with the in-degree and out-degree of each vertex equal
The statement we have to prove is
p ←→q (for biconditional)
Since
p → q (assuming that p is strongly connected to q)
q ← p (since p is strongly connected to q)
Therefore, the bicondition is satisfied
Newton is the correct unit of force
Answer:
the coin does not slide off
Explanation:
mass (m) = 5 g = 0.005 kg
distance (r) = 15 cm = 0.15 m
static coefficient of friction (μs) = 0.8
kinetic coefficient of friction (μk) = 0.5
speed (f) = 60 rpm
acceleration due to gravity (g) = 9.8 m/s^{2}
lets first find the angular speed of the table
ω = 2πf
ω = 2 x π x 60 x 
ω = 6.3 s^{-1]
Now lets find the maximum static force between the coin and the table so we can get the maximum velocity the coin can handle without sliding
static force (Fs) = ma
static force (Fs) = μs x Fn = μs x m x g
Fs = 0.8 x 0.005 x 9.8 = 0.0392 N
Fs = ma
0.0392 = 0.005 x a
a = 7.84 m/s^{2}
= a x r
= 7.84 x 0.15
Vmax = 1.08 m/s
ωmax = 
ωmax =
= 7.2 s^{-1}
now that we have the maximum angular acceleration of the table, we can calculate its maximum speed in rpm
Fmax = 
Fmax =
= 68.7 rpm
since the table is rotating at a speed less than the maximum speed that the static friction can hold coin on the table with, the coin would not slide off.
Answer:
m=ρV
V=4/3 * pi * r3
V=1.3 * 3.14 * 3.9^3
V=242.14 cm^3
m=7.58 * 242.14
m=1.8 kG
Explanation:
1. We calculate volume for sphere.
2. Then we calculate mass of sphere.