Answer:
0.29 m
Explanation:
9 mm = 0.009 m in diameter
Cross-sectional area 
Let the tensile modulus of Nickel
.
The elongation of the rod can be calculated using the following formula:

Answer:

Explanation:
Given that,
The mass of the paperclip, m = 1.8 g = 0.0018 kg
We need to find the energy obtained. The relation between mass and energy is given by :

Where
c is the speed of light
So,

So, the energy obtained is
.
Answer:
- Fx = -9.15 N
- Fy = 1.72 N
- F∠γ ≈ 9.31∠-10.6°
Explanation:
You apparently want the sum of forces ...
F = 8.80∠-56° +7.00∠52.8°
Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...
f∠α = (-f·cos(α), -f·sin(α))
This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.
= -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))
≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)
≈ (-9.15309, 1.71982)
The resultant component forces are ...
Then the magnitude and direction of the resultant are
F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)
F∠γ ≈ 9.31∠-10.6°
Explanation:
The general equation of an AC current is given by :

Where
I₀ is the peak value of current
is angular frequency

So,

We know that,

So, the frequency is 50 Hz and the maximum rms value of current is 14.14 A.