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3 years ago

14

What is a good analogy for potential energy in a circuit?

1 answer:

AlexFokin [52]3 years ago

8 0

Answer: find the answer in the explanation

Explanation: The potential energy in a circuit between two points is the workdone in taking units positive charge from one point to the other in the circuit. Its S.I unit is Joule.

Because voltage of a battery or cell in a circuit is the potential energy difference across its terminal

The potential energy difference between two points is one volt, if the workdone in taking one Coulomb of positive charge from one point to the other is one Joule.

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A train travels 76 kilometers in 2 hours and then 54 kilometers in 5 hours .What is the average speed ?

gizmo_the_mogwai [7]

V = d ÷ t --> bc d=vt
V = (76+54)÷(2+5) = 130÷7 = 18.57km/hr

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3 years ago

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3 years ago

Can someone help me ASAP

Phantasy [73]

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2 years ago

Which sequence shows electromagnetic waves arranged in a decreasing order of their wavelengths?

Hunter-Best [27]

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2 years ago

The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca

d1i1m1o1n [39]

Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

$U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}$

the elastic potencial energy is given by:

$U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J$

The work is defined as:

$W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J$

this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

$U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J$

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

$8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s$

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3 years ago