The main difference between a series and a parallel circuit is that

1 answer:

3 0

C) is correct

series circuit - in the same path : current flow on one path so they are equal on each component and equal to the source's. voltage on each components may be different.

parallel circuit - between same nodes : voltage of the components are equal and equal to the source's. current on each components may be different.

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What is newton's second law of motion?

lord [1]

Answer:

Newton's second law of motion can be formally stated as follows: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

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Two pendulum bobs have equal masses and lengths (8.100 m). bob a is initially held horizontally while bob b hangs vertically at

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Since both hv same mass and elsstic collision, so their velocity will exchange. Bob A will stop and bob B will move with speed of A just before the collision.

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Which is the best definition to describe a transfer of energy?

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Answer:

Explanation:

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2 years ago

a cone is constructed by cutting a sector from a circular sheet of metal with radius . the cut sheet is then folded up and welde

Svet_ta [14]

The expression for the radius and height of the cone can be obtained from

the property of a function at the maximum point.

$The \ radius, \ of \ the \ base \ of \ the \ cone \ is \ \sqrt{ \dfrac{3}{4}} \times radius \ of \ circular \ sheet \ metal$

The height of the cone is half the length of the radius of the circular sheet metal.

Reasons:

The part used to form the cone = A sector of a circle

The length of the arc of the sector = The perimeter of the circle formed by the base of the cone.

$Volume \ of \ a \ cone = \dfrac{1}{3} \cdot \pi \cdot r^3 \cdot h$

$Volume \ of \ a \ cone, \, V = \dfrac{1}{3} \cdot \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}$

θ/360·2·π·s = 2·π·r

Where;

s = The radius of he circular sheet metal

h = s² - r²

$\dfrac{dV}{dr} = \dfrac{d}{dr} \left(\dfrac{1}{3} \cdot \pi \cdot r^3 \cdot \sqrt{(s^2- r^2)}\right) = \dfrac{\pi \cdot (3 \cdot r^2 \cdot s^2 - 4 \cdot r^4)}{\sqrt{(s^2- r^2)}} = 0$