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:
Decreased by a factor of 4.5
Explanation:
"We have Newton formula for attraction force between 2 objects with mass and a distance between them:
where is the gravitational constant on Earth.
are the masses of the object and Earth itself. and R distance between, or the Earth radius.
So when R is tripled and mass is doubled, we have the following ratio of the new gravity over the old ones:
Since and
So gravity would have been decreased by a factor of 4.5
(a) 12.0 V
In this problem, the capacitor is connected to the 12.0 V, until it is fully charged. Considering the capacity of the capacitor, , the charged stored on the capacitor at the end of the process is
When the battery is disconnected, the charge on the capacitor remains unchanged. But the capacitance, C, also remains unchanged, since it only depends on the properties of the capacitor (area and distance between the plates), which do not change. Therefore, given the relationship
and since neither Q nor C change, the voltage V remains the same, 12.0 V.
(b) (i) 24.0 V
In this case, the plate separation is doubled. Let's remind the formula for the capacitance of a parallel-plate capacitor:
where:
is the permittivity of free space
is the relative permittivity of the material inside the capacitor
A is the area of the plates
d is the separation between the plates
As we said, in this case the plate separation is doubled: d'=2d. This means that the capacitance is halved: . The new voltage across the plate is given by
and since Q (the charge) does not change (the capacitor is now isolated, so the charge cannot flow anywhere), the new voltage is
So, the new voltage is
(c) (ii) 3.0 V
The area of each plate of the capacitor is given by:
where r is the radius of the plate. In this case, the radius is doubled: r'=2r. Therefore, the new area will be
While the separation between the plate was unchanged (d); so, the new capacitance will be
So, the capacitance has increased by a factor 4; therefore, the new voltage is
which means
Answer:
the efficiency of hydralic is 79.88%
Explanation:
convert mm to m
1mm = (1/1000)m
diameter of pipe upsteam
d₁= 90mm= 0.09m
diameter of pipe downsteam
d₂= 30mm = 0.03m
finding velocity of upsteam
recall Q=A₁V₁
V₁=Q/A₁
V₁=3.14m/s
velocity of downsteam
V₂= Q/A₂
V₂= 28.29m/s
mass flow rate
m= ρQ
ρ is the density of water
m = 1000× 0.02
m= 20kg/s
the efficiency of hydralic is 79.88%
