Answer:
The percentage change in resistance of the wire is 69%.
Explanation:
Resistance of a wire can be determined by,
R = (ρl) ÷ A
Where R is its resistance, l is the length of the wire, A its cross sectional area and ρ its resistivity.
When the wire is stretched, its length and area changes but its volume and resistivity remains constant.
= 1.3l, and 
= 
So that;
= (ρ) ÷ = (ρ × 1.3l) ÷ ()
= (1.3lρ) ÷ ()
= 
× [(ρl) ÷ A]
= 1.69R (∵ R = (ρl) ÷ A)
= 1.69R
Where is the new resistance, is the new length, and is the new area after stretching the wire.
The change in resistance of the wire = - R
= 1.69R - 1R
= 0.69R
The percentage change in resistance = 
× 100
= 0.69 × 100
= 69%
The percentage change in resistance of the wire is 69%.
Thomson experiment he calculated the charge to mass ratio just be passing the fundamental charge through a tube
He calculated the charge to mass ratio just by finding the deflection of charge while it is passing through the constant electric field
so here we will use the deflection as following
let say it passes the field of length "L"
so here we have
now in the same time if it deflect by some distance
now by solving this equation we can find e/m ratio
so here correct answer will be
the electron's charge-to-mass ratio
I think it will be muscle contraction. Hope this helps! Mark brainly please!
Answer:
restoring force is 2 N
Explanation:
given data
angle pulled = 5°
force = 1 N
pulled = 10°
to find out
restoring force
solution
we know here force
force = m×g×sinθ ..........1
so here θ is very small so sinθ = θ
1 = mg(5)
mg = 1 /5 ..................2
and
now for 10 degree
we know here θ is small so sinθ = θ
so from equation 1
force = m×g×sinθ
put equation 2 here
force = 1/5 × 10
force = 2 N
so restoring force is 2 N
F = M (V^2/r) = 20 x 100 /.15 =13333.33 N<span>convert units as needed to wind up with Newtons Kg-m/s^2 hope it helps :)</span>
