Answer:
0.65812
Explanation:
m = Mass of suitcase = 16 kg
= Incline angle = 40°
g = Acceleration due to gravity = 9.81 m/s²
a = Acceleration of the block = 1.36 m/s²
f = Frictional force
The normal force is given by

In x direction

Frictional force is given by

The coefficient of kinetic friction between the suitcase and the ramp is 0.65812
In order to answer this, we would need to know
the half-life of ⁴⁰K. Perhaps when you look that up,
you'll be able to answer this on your own.
Answer:
Explanation:
Given that
Initial velocity wo=0.210rev/s
Then, 1rev=2πrad
wo=0.21×2πrad/s
wo=0.42π rad/s
Given angular acceleration of 0.9rev/s²
α=0.9×2πrad/s²
α=1.8π rad/s²
Diameter of blade
d=0.75m,
Radius=diameter/2
r=0.75/2=0.375m
a. Angular velocity after t=0.194s
Using equation of angular motion
wf=wo+αt
wf=0.42π+ 1.8π×0.194
wf= 0.42π + 0.3492π
wf=1.319+1.097
wf= 2.42rad/s
If we want the answer in revolution
1rev=2πrad
wf= 2.42/2π rev/s
wf=0.385 rev/s
b. Revolution traveled in 0.194s
Using angular motion equation
θf - θi = wo•t + ½ αt²
θf - 0= 0.42π•0.194 + ½ × 1.8π•0.194²
θf = 0.256 + 0.106
θf = 0.362rad
Now, to revolution
1rev=2πrad
θf=0.362/2π=0.0577rev
Approximately θf= 0.058rev
c. Tangential speed? At time 0.194s
Vt=?
w=2.42rad/s at t=0.194s
Using circular motion formulae, relationship between linear velocity and angular velocity
V=wr
Vt=wr
Vt= 2.42×0.375
Vt=0.9075 m/s
Vt≈0.91m/s
d. Magnitude of resultant acceleration
Tangential Acceleration is given as
at=αr
at=1.8π× 0.375
at=2.12rad/s²
Now, radial acceleration is given as
ar=w²r
ar=2.42²×0.375
ar=2.196 m/s²
Then, the magnitude is
a=√ar²+at²
a=√2.196²+2.12²
a=√9.3171
a=3.052m/s²
a≈ 3.05m/s²
The answer is : B. allow for bonding metals to be stable. Metals, nonmetals all bond to be stable. Metallic bonding is the force of attraction between valence electrons and the metal ions.The electrons and the positive ions in the metal have a strong attractive force between them.