actually the e is a variable with the value 1.56x10^20
So, I have to find X in [1.56x10^20]^(0.50/X)
How do i find X in e^(0.50/x) on a scientific calc?
do I use LN, and if so, could you pls tell me how I can plug it in?
5 answers
To find an unknown (x), you need an equation.
What has been supplied is an expression containing x. Does it equal some value?
What has been supplied is an expression containing x. Does it equal some value?
its actually a physics question
q=(2 coul)[1.56x10^20]^([0.50/sec][t])
t=time in sec.
q=(2 coul)[1.56x10^20]^([0.50/sec][t])
t=time in sec.
The full question:
The charge (in coul.) on an object increases according to q=(2 coul)e^[(o.05/sec)(t)], where t=time in sec. At how many sec. is the object deficient 1.56x10^20 electrons (from neutral)?
A 0 B 12.56 C 21.94 D 28 E 42.21 F 50.5
The charge (in coul.) on an object increases according to q=(2 coul)e^[(o.05/sec)(t)], where t=time in sec. At how many sec. is the object deficient 1.56x10^20 electrons (from neutral)?
A 0 B 12.56 C 21.94 D 28 E 42.21 F 50.5
One coulomb = 6.24150965(16)×10^18 electrons.
1.56*10^20 electrons
= 1.56*10^20/6.24150965(16)×10^18 coulombs
=25 coulombs
So the equation becomes:
25 coulombs = 2 coulombs * e^(0.05t)
e^(0.05t) = 25/2=12.5
take ln on both sides:
0.05t = ln 12.5 = 2.526
t = 2.526/0.05 = 50.5 sec.
1.56*10^20 electrons
= 1.56*10^20/6.24150965(16)×10^18 coulombs
=25 coulombs
So the equation becomes:
25 coulombs = 2 coulombs * e^(0.05t)
e^(0.05t) = 25/2=12.5
take ln on both sides:
0.05t = ln 12.5 = 2.526
t = 2.526/0.05 = 50.5 sec.
3 answers