The capacitance of two square parallel plates will be 1.777 farads.
<h3>What is the capacitance of two parallel plates?</h3>
The body's capacity to hold electric charge is constrained by capacitance. There is a capacitance for each capacitor. Two metallic sheets of area A, spaced by d, make up the standard parallel-plate capacitor.
The parallel plate capacitor formula is given by:
C = ε Ad
Two square parallel plates 6.7 cm on a side are separated by 1.8 mm of paraffin. The dielectric constant of paraffin is 2.2. Then the capacitance of two square parallel plates will be given as,
C = ε Ad
C = 2.2 x 0.67 x 0.67 x 1.8
C = 1.777 farad
Two parallel square plates will have a capacitance of 1.777 farads.
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Answer:
Trenches are the deepest part of the oceans, narrow and long depressions on the sea floor. One good example is the Mariana Trench, the deepest part of the ocean
Explanation:
Answer: option (D)
Explanation:
The potential energy of each of the students is given below as
P.E(student A) = mgh, where m = mass of student A, g is acceleration due to gravity and h = height of the high dive structure.
The mass of student B is twice as much as that of A, hence his mass is 2m and his potential energy is given below as
P.E ( student B) =2mgh = 2(mgh)
Recall that the relationship between potential energy and work done is that
Work done = - (change in potential)
For student A, work done = - mgh
For student B, work done = - 2mgh
From the equations above it can be seen that student B will do twice the work in getting to the high dive structure than student A hence validating option D.
One should never anchor in a narrow channel, until unless required very importantly. One should stay to the starboard side, and use a prolonged blast. The announcement must be done to alarm the nearby vessels, about your approach. The vessel should be kept at the outer limit of the starboard side.
Answer:
the force acting on the team mate is 1.19 kN.
Explanation:
given,
mass = 196 lbm
while tackling, the deceleration is from velocity 6.7 m/s to 0 m/s
time taken for deceleration = 0.5 sec
F = mass × acceleration
acceleration = 
= -13.4 m/s²
1 lbs = 0.453 kg
196 lbs = 196 × 0.453 = 88.79 kg
F = 88.79 × 13.4
F = 1189.786 N = 1.19 kN
hence, the force acting on the team mate is 1.19 kN.
