Question about Physio-Control 803732 Lifepak 9 Defibrillator Shock Advisory Adapter

It works normal after some time. as we test it on 200 joules chrging level increase after each discharge.when charging level rech to 200 joules it works normal

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m stands for mili like 1/1000 and W stands for watts which is the measurement of joule/second (joules per second). So 5mW means the laser's power is 5 miliwatts or it is putting out 0.005 joules of energy per second. The higher the number the more joules of energy are being emitted.

Sep 11, 2013 | Office Equipment & Supplies

Batteries in the UPS typically last 3 to 5 years and it depends upon several factors. Including the number of times the unit must go on battery power and environmental conditions. There are usually several batteries in the UPS and while the battery voltage of each battery may show 13 volts when tested with a multi-meter. This is a float voltage and a true indication of the battery voltage and its condition requires each battery to be tested under a load. If one battery is faulty in a set of batteries it will causes the whole battery system to fail and indicate a battery fault.

Also a faulty charging circuit will not charge the batteries and will cause a fault condition warning.

on Dec 06, 2009 | Opti - UPS ES800C 800VA 480W 1050 JOULES...

If the rechargeable battery is more than 3 - 4 years old then the battery could be worn out. Rechargeable batteries have a finite number of charge and discharge cycles and will lose their charge capacity over time, i.e. won't charge to 100% and gradually the charge reduces until the battery won't charge up at all. OR The battery shows a 100% charge but when the adapter is disconnected the battery drops off to zero capacity in a very short time. If the battery drops to an unacceptable charge level then the battery needs to be replaced. Rechargeable batteries will fail if stored in a discharged state for long periods.

Oct 24, 2012 | Pentax Optio Cameras

Sorry to say, but your units are incoherent.

The mJ/cm^2 is a unit of energy flux (energy per unit surface area) whereas mW is a unit of power ( energy per unit time). There is a connecting dimension missing. It might be helpful to figure out what you need to do to know that 1mJ = 1mWx1s= 1mW.s

The mJ/cm^2 is a unit of energy flux (energy per unit surface area) whereas mW is a unit of power ( energy per unit time). There is a connecting dimension missing. It might be helpful to figure out what you need to do to know that 1mJ = 1mWx1s= 1mW.s

Feb 18, 2011 | Office Equipment & Supplies

A
unit of energy used for convenience in atomic systems. Specifically, it
is the change in energy of an electron, or of any particle having a
charge numerically equal to that of an electron, when it is moved
through a difference of potential of 1 mks volt. Its value (in mks units) is obtained from the equation *W* = *qV*, where *W* is energy in joules, *q* the charge in coulombs, and *V* the potential difference in volts. For a potential difference of 1 volt and the electronic charge of 1.602 × 10?19 coulomb, the electronvolt is 1.602 × 10?19 joule. *See also* Electron; Ionization potential.

Read more: http://www.answers.com/topic/electronvolt#ixzz1AHEXkh1j

Conversion factors:

The kinetic energy gained by an electron by falling through a potential difference of 1 volt. 1 eV is equivalent to 1.6 × 10-12 ergs. 1,000 eV is referred to as 1 kilo electron volt, or keV, and 1,000,000 eV are referred to as 1 mega electron volt, or MeV.

Read more: http://www.answers.com/topic/electronvolt#ixzz1AHInNmzI

From this equation W = qV = (5 x 1.6 × 10?19 Coulomb) x 8900 v = 44.500 x 1.6 × 10?19 Joule

= 44.500 eV (because 1 eV = 1.602176487(40)×10?19 J)

= 44.500 x 10?6 MeV ( because 1 MeV = 1.000.000 eV)

K.E. = 0.0445 MeV

Well, I hope this right, I'm not so sure my self :P.

Hope this help and be kind to rate please, thanks.

(^_^)

Read more: http://www.answers.com/topic/electronvolt#ixzz1AHEXkh1j

Conversion factors:

- 1 eV = 1.602176487(40)×10?19 J (the conversion factor is numerically equal to the elementary charge expressed in coulombs).

The kinetic energy gained by an electron by falling through a potential difference of 1 volt. 1 eV is equivalent to 1.6 × 10-12 ergs. 1,000 eV is referred to as 1 kilo electron volt, or keV, and 1,000,000 eV are referred to as 1 mega electron volt, or MeV.

Read more: http://www.answers.com/topic/electronvolt#ixzz1AHInNmzI

From this equation W = qV = (5 x 1.6 × 10?19 Coulomb) x 8900 v = 44.500 x 1.6 × 10?19 Joule

= 44.500 eV (because 1 eV = 1.602176487(40)×10?19 J)

= 44.500 x 10?6 MeV ( because 1 MeV = 1.000.000 eV)

K.E. = 0.0445 MeV

Well, I hope this right, I'm not so sure my self :P.

Hope this help and be kind to rate please, thanks.

(^_^)

Jan 06, 2011 | Cameras

Ifthe laptop battery is more than 3 - 4 years old then the battery could be wornout. Rechargeable batteries have a finite number of charge and discharge cyclesand will lose their charge capacity over time, i.e. won't charge to 100% andgradually the charge reduces until the battery won't charge up atall.
ORThe battery shows a 100% charge but when the adapter is disconnected thebattery drops off to zero capacity in a very short time.
Ifthe battery drops to an unacceptable charge level then the battery needs to bereplaced.

Dec 14, 2010 | Acer Aspire One PC Notebook

Theoritcally, the energy (in Joules) will be exactly the same as all you are doing is effectively doubling the capacitance.

In practical terms, the energy will be very slightly less after connecting them as the charge on the charged capacitor has to be equally distrbuted across itself and the uncharged one, by the flow of electrons via the electrical connection between them. This connection will have a finite resistance (allbeit very low if the connection is not very long), therefore some energy will be dispersed as heat as a result of electrons flowing through this resistance (Amps squared times the resistance times the time it takes for the charge to equally distribute - Joules heating effect & Ohms law).

In practical terms, the energy will be very slightly less after connecting them as the charge on the charged capacitor has to be equally distrbuted across itself and the uncharged one, by the flow of electrons via the electrical connection between them. This connection will have a finite resistance (allbeit very low if the connection is not very long), therefore some energy will be dispersed as heat as a result of electrons flowing through this resistance (Amps squared times the resistance times the time it takes for the charge to equally distribute - Joules heating effect & Ohms law).

Apr 27, 2010 | Computers & Internet

OK, this Microwave has got a failed magnetron, and/or a High power Diode has gone faulty. You need to get it repaired by a professional, Always ask for a "Quote" because often, replacement parts, and labor, can often exceed or equal the cost of a new one, these days. Self repair of Microwaves is very **dangerous**, as there is nearly always lethal Voltages and Currents, even with an OFF, and Disconnected Microwaves, you see, inside, there is a GIANT Capacitor, this capacitor, feeds off the Magnetron,it can hold, for some considerable amount of time, enough Joules of energy, to *kill anyone that touches it.* **Before any work is done, the first thing that must be done is to, by using and earthed "Shorting Bar" and touch it to each terminal of the capacitor to discharge it,** There will be a BIG Spark, and a Bang, as the energy discharges. Once that is done, then they are safe to work,on, but if ever turned on again, say to test, then, it must again, it be shorted out. (This is where some service people come unstuck, they forget, after discharging once, that, after they had tuned it on, say to test, and then off again, that the capacitor, had immediately "Charged Up", just waiting....) Just as an extra, more service people are killed each year from Microwaves than any other single thing.

Mar 31, 2010 | Panasonic Microwave Ovens

Send the flash to the manufacture or their authorized service rep for proper service. These capacitors are charged to high voltage and are discharged with very high current.

There are serious safety issues with these capacitors. The flash manufactures normaly will not give out service information or supply support for self service. There are serious liability and safety issues involved.

For personal interest, below is a very good article on basic capcitor theory and history. If you have a background in physics and math, this article can be well appreciated.

http://en.wikipedia.org/wiki/Capacitor

Jerry G.

There are serious safety issues with these capacitors. The flash manufactures normaly will not give out service information or supply support for self service. There are serious liability and safety issues involved.

For personal interest, below is a very good article on basic capcitor theory and history. If you have a background in physics and math, this article can be well appreciated.

http://en.wikipedia.org/wiki/Capacitor

Jerry G.

Jul 08, 2008 | Metz Mecablitz 60CT4 TTL Flash

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