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Mademuasel [1]
3 years ago
6

Visually, how can you distinguish between an ac and dc power supply?

Physics
2 answers:
anyanavicka [17]3 years ago
8 0

Alternating current (AC), is the form of electrical energy that consumers use whenever they plug televisions, fans, bulb, AC, and any other appliances into a socket.

Direct Current (DC)

There are many uses of direct current (DC). It can be used to charge battery and large power supplies for motors and electronic system.

<h2>Further Explanation</h2>

Alternating Current is the form whereby electric power is provided for residences and businesses. Alternating current may be produced using a device known as alternator.  

With Alternating current (AC), electrons don’t really flow in AC; they vibrate from positive to negative and back from negative to positive. They really vibrate back and forth. In AC, there is no continuous vibration unlike in DC.

Direct current (DC)

Direct current power in large quantities can be used in production of aluminum. It can also be used in other electrochemical processes.

With Direct current (DC), there is a movement of electron in one direction, that is, from negative (-) to positive (+). It is a constant current that flows continuously until it is switch off or the source of its power runs out.  An example of a DC power is a battery.

They may flow throw a conductor such as wire; they can also flow through insulators, semiconductors or vacuum as in ion beams or electron.  

LEARN MORE:

  • AC power and DC power brainly.com/question/13033819

KEYWORDS:

  • direct current
  • alternating current
  • battery
  • appliances
  • power supply
Temka [501]3 years ago
5 0

Answer:

By the use of slow motion camera.

Explanation:

Visually, it is very hard to differentiate between an ac and dc power supply. But Since, we that In Ac supply polarity changes 100 times in a second ( because frequency of ac supply is 50 Hz generally). Whereas, Dc gives a steady power supply. So, in slow motion camera we can easily capture the flickering light tubes which won't happen in case of dc supply.

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2 years ago
A 50.0-g object connected to a spring with a force constant of 35.0 n/m oscillates with an amplitude of 4.00 cm on a frictionles
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A) The total energy of the system is sum of kinetic energy and elastic potential energy:
E=K+U= \frac{1}{2}mv^2 +  \frac{1}{2}kx^2
where
m is the mass
v is the speed
k is the spring constant
x is the elongation/compression of the spring

The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
x=A=4.00 cm = 0.04 m
then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to
E=U= \frac{1}{2}kA^2 =  \frac{1}{2}(35.0 N/m)(0.04 m)^2=0.028 J

b) When the position of the object is 
x=1.00 cm = 0.01 m
the potential energy of the system is
U= \frac{1}{2}kx^2 =  \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J
and so the kinetic energy is
K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J
since the mass is m=50.0 g=0.05 kg, and the kinetic energy is given by
K= \frac{1}{2}mv^2
we can re-arrange the formula to find the speed of the object:
v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.026 J}{0.05 kg} }=1.02 m/s

c) The potential energy when the object is at 
x=3.00 cm=0.03 m
is
U= \frac{1}{2}kx^2 =  \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J
Therefore the kinetic energy is
K=E-U=0.028 J-0.016 J = 0.012 J

d) We already found the potential energy at point c, and it is given by
U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J
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