Answer:
<em>1.228 x </em>
<em> mm </em>
<em></em>
Explanation:
diameter of aluminium bar D = 40 mm
diameter of hole d = 30 mm
compressive Load F = 180 kN = 180 x
N
modulus of elasticity E = 85 GN/m^2 = 85 x
Pa
length of bar L = 600 mm
length of hole = 100 mm
true length of bar = 600 - 100 = 500 mm
area of the bar A =
=
= 1256.8 mm^2
area of hole a =
=
= 549.85 mm^2
Total contraction of the bar =
total contraction =
==>
= <em>1.228 x </em>
<em> mm </em>
Answer:
Explanation:
a ) Time period T = 2 s
Angular velocity ω = 2π / T
= 2π / 2 = 3.14 rad /s
Initial moment of inertia I₁ = 200 + mr²
= 200 + 25 x 2.5²
=356.25
Final moment of inertia
I₂ = 200 + 25 X 1.5 X 1.5
= 256.25
b ) We apply law of conservation of momentum
I₁ X ω₁ = I₂ X ω₂
ω₂ = I₁ X ω₁ / I₂
Putting the values

ω₂ = 4.365 rad s⁻¹
c ) Increase in rotational kinetic energy
=1/2 I₂ X ω₂² - 1/2 I₁ X ω₁²
.5 X 256.25 X 4.365² - .5 X 356.25 X 3.14²
= 684.95 J
This energy comes from work done against the centripetal pseudo -force.
Normal Force = 54 N
acceleration = 1.2 m/s^2
For Normal Force:
According to the force diagram, we can come up with the equation (all up and down forces):
10 sin 30 + Normal Force - 58.8 = 0
Normal Force = 53.8 N = 54 N
For acceleration:
According to the force diagram, we can come up with the equation (all left and right forces):
10 cos 30 - 1.5 = (6.0) (Acceleration)
Acceleration = 1.19 m/s^2 = 1.2 m/s^2
Answer:
6.02×10²³
Explanation:
Mole measures the number of particles in a specific substance. The numeric value of a mole for atom or molecules is approximately 6.02×10²³ atoms or molecules.