Resistance
R
of a conducting wire of length
l
and of area of cross section
A
is given by the expression
R
=
ρ
l
A
where
ρ
is resistivity or specific resistance of the material of the conductor.
Inserting given values in the expression in SI units we get
34
=
(
1.7
×
10
−
8
)
l
π
(
0.1
1000
)
2
4
⇒
l
=
34
1.7
×
10
−
8
×
π
4
(
0.1
1000
)
2
⇒
l
=
15.71
m
, rounded to two decimal places.
B. of attraction between two objects
Answer: 5.5 m/s
Explanation:
As we are told the runner accelerates at a constant rate, we can use the following equation:
Where:
is the runner's final velocity
is the runner's initial velocity
is the runner's constant acceleration
is the time it takes the runner to reach the final velocity
Finally:
D=Vot+1/2at^2
In this case, there is no initial y velocity so the term Vot=0 so d=1/2at^2
acceleration=acceleration due to gravity=-9.8m/s^2
It falls - 22cm or -0.22m
We have - 0.22=1/2(-9.8)t^2
t^2=(-0.44)/(-9.8)
t=sqrt[0.44/9.8]