I got you kid It’s A- 2.5m/sb
Answer:
At 81. 52 Deg C its resistance will be 0.31 Ω.
Explanation:
The resistance of wire =
Where 
=Resistance of wire at Temperature T
= Resistivity at temperature T ![=\rho_0 \ [1 \ + \alpha\ (T-T_0\ )]](https://tex.z-dn.net/?f=%3D%5Crho_0%20%5C%20%5B1%20%5C%20%2B%20%5Calpha%5C%20%28T-T_0%5C%20%29%5D)
Where 
l=Length of the wire
& A = Area of cross section of wire
For long and thin wire the resistance & resistivity relation will be as follows
![\frac{0.25}{0.31}=\frac{1}{[1+\alpha(T-20)]}](https://tex.z-dn.net/?f=%5Cfrac%7B0.25%7D%7B0.31%7D%3D%5Cfrac%7B1%7D%7B%5B1%2B%5Calpha%28T-20%29%5D%7D)
T = 81.52 Deg C
'A biased oppinion'
When you like something and right a report on it your opinion will come through about that subject, the same thing occurs if you dislike something.
:) Hope this helps x
Answer:
-2040 m/s²
Explanation:
Taking toward the wall to be positive, the initial velocity is 10.1 m/s and the final velocity is -8.3426 m/s.
Average acceleration is the change in velocity over change in time.
a = Δv / Δt
a = (-8.3426 m/s − 10.1 m/s) / 0.00905 s
a = -2040 m/s²
Answer:
Work done = 4584.9 J
Explanation:
given: q1=3.0 mC = 3.0 × 10⁻³ C, r = 20 cm = 0.20 m, q1 = 34μC = 34 × 10⁻⁶ C
Solution:
Formula for the potential difference at the center of the circle
P.E = K × q1 q2 /r (Coulomb's constant k= 8.99 × 10⁹ N·m² / C²)
P.E = 8.99 × 10⁹ N·m² / C² × 3.0 × 10⁻³ C × 34 × 10⁻⁶ C / 0.20 m
P.E = 4584.9 J = Work done
