Answer:
Rewirable or Kit – Kat Type Fuses are a type of Low Voltage (LV) Fuses. They are most commonly used in house wiring, small industries and other small current applications. Rewirable Fuses consists of two main parts: a Fuse Base, which contains the in and out terminal, and a Fuse Carrier, which holds the Fuse Element.
B . it should be convert energy.
Entropy is an extensive property of a thermodynamic system. It quantifies the number Ω of microscopic configurations (known as microstates) that are consistent with the macroscopic quantities that characterize the system (such as its volume, pressure and temperature).[1] Under the assumption that each microstate is equally probable, the entropy
S
S is the natural logarithm of the number of microstates, multiplied by the Boltzmann constant
The members of these groups make up the majority of voters in many districts thus this be considered a problem.
<u>Option: D</u>
<u>Explanation:</u>
Interest groups play a key role in US politics. Such organizations are made up of wealthy and powerful members who often seek to impose some form of leverage in politicians to promote their goals and agendas. Across the years via many campaigns, they have understood how to speak and manipulate elected leaders and apply leverage to get the kind of legislation that is in their favor. Here the majority of voters in several districts are standing due to group members, as we recognize the interest group belongs to a body in which it uses different methods of lobbying to influence others.
Answer:
A) 26V
Explanation:
(a) the potential difference between the plates
Initial capacitance can be calculated using below expresion
C1= A ε0/ d1
Where d1= distance between = 2.70 mm= 2.70× 10^-3 m
ε0= permittivity of space= 8.85× 10^-12 Fm^-1
A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2
If we substitute the values we
C1= A ε0/ d1
=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3
C1=2.589 ×10^-12 F= 2.59 pF
Initial charge can be determined using below expresion
q1= C1 × V1
V1=2.589 ×10^-12 F
V1= voltage=7.90 V
If we substitute we have
q1= 2.589 ×10^-12 × 7.90
q1= 20.45×10^-12C
20.45 pC
Final capacitance can be calculated as
C2= A ε0/ d2
d2=8.80 mm= /8.80× 10^-3
7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3
C1=0.794 ×10^-12 F= 0.794 pF
Final charge= initial charge
q2=q1 (since the battery is disconnected)
q2=q1= 20.45 pC
Final potential difference
V2= q/C2
= 20.45/0.794
= 26V
