Answer:
(D)
Explanation:
Given :
l=3.5 m


Resistance can be calculated as :


Resistance of the wire will be 1.1×
ohms
Option D is correct
Answer:
it should be 8,000
Explanation:
because if you multiply the mass value by 1000, you will get 8,000
:):):):)
Answer:
Time = 6 years
Explanation:
First, we will calculate the no. of half life periods required to reduce the mass of Protactinium to the given value:

where,
n = no. of half-life periods = ?
m = initial mass = 86.3 g
m' = remaining mass = 10.8 g
Therefore,

Since the bases are the same. Therefore equating powers:
n = 3
Now we calculate the time:

<u>Time = 6 years</u>