Answer:
<em>0.45 mm</em>
Explanation:
The complete question is
a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?
A) 0.45 mm
B) 0.63 mm
C.) 0.68 mm
D) 0.91 mm
Current in the fuse is 1.0 A
Current density of the fuse when it melts is 620 A/cm^2
Area of the wire in the fuse = I/ρ
Where I is the current through the fuse
ρ is the current density of the fuse
Area = 1/620 = 1.613 x 10^-3 cm^2
We know that 10000 cm^2 = 1 m^2, therefore,
1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2
Recall that this area of this wire is gotten as
A = 
where d is the diameter of the wire
1.613 x 10^-7 = 
6.448 x 10^-7 = 3.142 x 
=
d = 4.5 x 10^-4 m = <em>0.45 mm</em>
Answer:
Explanation:
As we know by force equation that force along the inclined planed due to gravity is given as
so the acceleration due to gravity along the plane is given as
now we have
now we know that
Answer:
Hello there use something that looks like this
Explanation:
This is an accurate representation of something you are working on!
As you can see the wire and the core are represented on the left and is showing how it can be represented on your right hand and how they are similar!
Answer:
She passes through a loud spot at x = 19.5m from the first speaker. Constructive interference occurs here.
At the quiet spot destructive interference occurs. The minimum distance for this to occur is 20.25 from the first speaker.
Explanation:
Answer:
(a) 
(b) 
Explanation:
Hello.
(a) In this case, since the initial volume is 18.5 dm³ and the final volume is 21 dm³ (18.5 +2.5), we can compute the work at constant pressure as shown below:
Which is negative as it expands against the given pressure.
(b) Moreover, of the process is carried out reversibly, the pressure can change, therefore, we need to compute the work via:
Whereas the moles are computed from the given mass of argon:
Thus, the work is:
Regards.
