The electrical resistance of a wire varies directly as its length and inversely as the square of its diameter. a wire with a len
1 answer:
The electrical resistance of the wire is proportional to its length L and inversely proportional to the square of its diameter, 
.
The length of the wire in the problem is changed from 200 inches to 500 inches, so the new length is 2.5 times the initial length:
the diameter of the wire is not changed, so the new electrical resistance must be 2.5 times the original value:
The length of the wire in the problem is changed from 200 inches to 500 inches, so the new length is 2.5 times the initial length:
the diameter of the wire is not changed, so the new electrical resistance must be 2.5 times the original value:
You might be interested in
Answer:
Below
Explanation:
First draw the vectors that represent both electric fields.
E1 is the elictric field created by q1, E2 is the one created by q2.
● q1 is negative so E1 will point from P.
● q2 is positive so E2 will point out of P
(Picture below)
■■■■■■■■■■■■■■■■■■■■■■■■■■
The resulting electric field is equal to the sum of the two fields since both vectors are colinear.
Let E be the total field.
● E = E1 + E2
The formula of the electric field intensity is:
● E = K ×(q/d^2)
-K is Coulomb's constant
-d is the distance between the charge and the object ( here P)
-q is the charge
■■■■■■■■■■■■■■■■■■■■■■■■■■
● E1 = K × (q1/d1^2)
The distance between q1 and P is the qum of 0.15 m 0.25 m. (0.4 m)
Coulombs constant is 9×10^9 m^2/C^2
● E1 = 9×10^9 ×[-6.39 × 10^(-9)/ 0.4^2]
● E1 = -359.43 N/C
■■■■■■■■■■■■■■■■■■■■■■■■■■
● E2 = K ×(q2/d^2)
The distance between q2 and P is 0.25 m.
● E2 = 9×10^9×[3.22×10^(-9) /0.25^2]
● E2 = 463.68 N/C
■■■■■■■■■■■■■■■■■■■■■■■■■■
● E = E1 + E2
● E = -359.43+463.68
● E = 105.25 N/C
Complete Question
A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.
If the acceleration of the projective is : a = c/s m/s2
Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?
Answer:
The value of the constant is
Explanation:
From the question we are told that
The acceleration is
The initial position of the projectile is s= 1.5m
The final position of the projectile is
The velocity is
Generally
and acceleration is
so
=>
integrating both sides
Now for the limit
a = 200 m/s
b = 0 m/s
c = s= 3 m
d == 1.5 m
So we have
=>