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3 years ago

Which of these statements articulate one of Kirchhoff's laws? (Select all that apply)

Physics

1 answer:

aleksley [76]3 years ago

8 0

Answer: E) and F)

Explanation:

E) In any closed loop, the sum of all voltage drops must be zero, as it is stated in the Kirchhoff Voltage Law (KVL).

This is a direct consequence of the energy conservation principle, as if it could be possible to return to the starting point in a loop with a different voltage, this energy could be used as a infinite energy source, which is against laws of nature, specially Thermodynamic's 2nd law.

F) At any junction, the sum of the currents into it, must be equal to the sum of the currents that leave the junction, as it is stated in the Kirchhoff"s Current Law (KCL).

This is a direct consequence of the charge conservation principle, as the charge can't neither be accumulated nor be destroyed at any point.

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3 years ago

Now let’s apply the work–energy theorem to a more complex, multistep problem. In a pile driver, a steel hammerhead with mass 200

andrew11 [14]

Answer:

a) v = 7.67

b) n = 81562 N

Explanation:

Given:-

- The mass of hammer-head, m = 200 kg

- The height at from which hammer head drops, s12 = 3.00 m

- The amount of distance the I-beam is hammered, s23 = 7.40 cm

- The resistive force by contact of hammer-head and I-beam, F = 60.0 N

Find:-

(a) the speed of the hammerhead just as it hits the I-beam and

(b) the average force the hammerhead exerts on the I-beam.

Solution:-

- We will consider the hammer head as our system and apply the conservation of energy principle because during the journey of hammer-head up till just before it hits the I-beam there are no external forces acting on the system:

ΔK.E = ΔP.E

K_2 - K_1 = P_1- P_2

Where, K_2: Kinetic energy of hammer head as it hits the I-beam

K_1: Initial kinetic energy of hammer head ( = 0 ) ... rest

P_2: Gravitational potential energy of hammer head as it hits the I-beam. (Datum = 0)

P_1: Initial gravitational potential energy of hammer head

- The expression simplifies to:

K_2 = P_1

Where, 0.5*m*v2^2 = m*g*s12

v2 = √(2*g*s12) = √(2*9.81*3)

v2 = 7.67 m/s

- For the complete journey we see that there are fictitious force due to contact between hammer-head and I-beam the system is no longer conserved. All the kinetic energy is used to drive the I-beam down by distance s23. We will apply work energy principle on the system:

Wnet = ( P_3 - P_1 ) + W_friction

Wnet = m*g*s13 + F*s23

n*s23 = m*g*s13 + F*s23

Where, n: average force the hammerhead exerts on the I-beam.

s13 = s12 + s23

Hence,

n = m*g*( s12/s23 + 1) + F

n = 200*9.81*(3/0.074 + 1) + 60

n = 81562 N

6 0

3 years ago

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