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>
The given question is incomplete. The complete question is as follows.
In a nuclear physics experiment, a proton (mass kg, charge +e =
C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed
m/s. The proton comes momentarily to rest at a distance
m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are
m apart?
Explanation:
The given data is as follows.
Mass of proton = kg
Charge of proton =
Speed of proton =
Distance traveled =
We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.
=
where,
U =
Putting the given values into the above formula as follows.
U =
=
=
Therefore, we can conclude that the electric potential energy of the proton and nucleus is .