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:
E) 6.5 A
Explanation:
Given that
L = 40 m H
C= 1.2 m F
Maximum charge on capacitor ,Q= 45 m C
The maximum current I given as
I = Q.ω
ω =angular frequency
By putting the values
ω = 144.33 rad⁻¹
Maximum current
I = 45 x 10⁻³ x 144.33 A
I= 6.49 A
I = 6.5 A
E) 6.5 A
Answer:
<h2>2.2 m/s²</h2>Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
f is the force
m is the mass
From the question we have
We have the final answer as
<h3>2.2 m/s²</h3>Hope this helps you