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Dmitry_Shevchenko [17]
3 years ago
12

_____ technology uses radio frequency identification (RFID) tags to identify and monitor the movement of each individual product

, from a factory floor to the retail checkout counter.
Physics
1 answer:
Lena [83]3 years ago
4 0

Answer:

Electronic product code is the correct answer.

Explanation:

  • Electronic product code is an identifier used to provides the identity of different physical material.
  • Electronic product code has a number that is used to identify information about the product origin, destination and production date.
  • Electronic Product Code run on the principle of radio frequency tag.
  • Electronic Product Code provides several benefits to consumers such as it used to track the inventory data and help to provide the right products to the client.
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Suppose a scientist conducts a series of experiments and the results are so amazing she wants to share them with other experts i
nataly862011 [7]

Answer:

The results have not been through the rigorous process of peer review

Explanation:

When a scientist conducts a study and obtains results, those results ought to be submitted to a reputable journal where the results would go through the rigorous protocol of peer review.

During this process, the reliability of the data presented is ascertained before the results are published for other scientists to see.

If the results are hurriedly published on the internet, many researchers who come in contact with the work may be fed with inaccurate information.

7 0
2 years ago
An ice skater spins at 2.5 rev/s when his arms are extended. He draws his arms in and spins at 10.0 rev/s. By what factor does h
Rainbow [258]

Answer:

The moment of inertia decreased by a factor of 4

Explanation:

Given;

initial angular velocity of the ice skater, ω₁ = 2.5 rev/s

final angular velocity of the  ice skater, ω₂ = 10.0 rev/s

During this process we assume that angular momentum is conserved;

I₁ω₁ = I₂ω₂

Where;

I₁ is the initial moment of inertia

I₂ is the final moment of inertia

I_2 = \frac{I_1 \omega_1}{\omega_2} = \frac{I_1*2.5}{10} \\\\I_2 = 0.25I_1 = \frac{1}{4}I_1

Therefore, the moment of inertia decreased by a factor of 4

4 0
2 years ago
What is the best explanation for why a magnet is different from a regular piece of metal?
arlik [135]

Answer:

in a magnet there is a magnetic field that draws ever mental to it

5 0
2 years ago
Why does it rain a lot?
Vaselesa [24]

Answer:

Clouds are made up of tiny water droplets. ... As more and more droplets join together they become too heavy and fall from the cloud as rain. Warm air can hold more moisture than cool air. When the warmer air is cooled and the moisture condenses, it often rains more heavily.

Explanation:

hope it helps

3 0
3 years ago
Two parallel plate capacitors 1 and 2 are identical except that capacitor 1 has charge +q on one plate and charge −q on the othe
Grace [21]

Answer:

a) the capacitance is the same for both capacitors.

b) The potential difference between the plates for the capacitor with charge +2q, is double of the one for the capacitor  with charge +q.

c) The electric field magnitude between the plates for the capacitor with charge +2q, is double of the one for the capacitor with charge +q

d) The energy stored between the plates for the capacitor with charge +2q, is 4 times the value for the one with charge +q  

Explanation:

a) The capacitance of a capacitor, by definition, is as follows:

C = \frac{q}{V}

Appying Gauss' Law to one of plates, it can be showed, that the capacitance (for a parallel plates capacitor) can be  expressed as follows:

C = ε*A / d

As it can be seen, it does not depend on the charge. so we conclude that the capacitance must be the same for both capacitors, due to they are identical except for the value of the charge on the plates.

b) By definition, as we said above, the capacitance is equal to the proportion between the charge of one of the plates, and the potential difference between them.

If this proportion must remain the same, and one of the capacitors has the double of  the charge than the other, the potential difference must be the double also.

c) Applying Gauss' law, to the surface of one of  the plates, and assuming a constant surface charge density σ, it can be  showed that the  electric field can be calculated as follows:

E*A = Q/ε₀ as σ=Q/A

⇒ E = σ/ε₀

As σ is directly proportional to the charge (being the area A the same), we conclude that the electric field for the capacitor with charge +2q must be the double than the one for the capacitior with charge +q.

c) The electric potential energy, stored between plates of a capacitor, can be written as follows:

Ue = \frac{1}{2} *\frac{q^{2}}{C}

If the capacitance remains the same, we can conclude that the electric potential energy for the capacitor with charge +2q, as the charge is raised to the 2nd power, must be 4 times the one for the capacitor with charge +q.

4 0
3 years ago
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