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mojhsa [17]
3 years ago
12

Transformers will not work on ______ electrical systems.

Physics
2 answers:
Andru [333]3 years ago
5 0

Transformers only work with AC.

If you put DC into a transformer, all that comes out is smoke.

(a)

Sphinxa [80]3 years ago
3 0

Transformers will not work on direct current in electrical systems.

Answer: Option A

<u>Explanation: </u>

Transformers work only on the alternate current of electrical system. And they are not suitable to work upon with the direct current. As the direct current has the property to transfer high voltage directly and is unable to direct the current with high and low voltages, causes the transformer to burn out the coil leading to burning of the transformer circuit and causing smoked transformers.

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A substance that accelerates the rate of a chemical reaction is called a(an)
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<span>A substance that accelerates the rate of chemical reaction is called a catalyst. It serves as an alternative pathway for the reaction product. The increase of rate of chemical reaction is because catalyst has low activiation energy than the original pathway. The low activation energy will increase the amount of molecules that can benefit in the energy created by the catalyst.</span>
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3 years ago
a uniform rod is hung at onen end and is partially submerged in water. If the density of the rod is 5/9 than of wter, find the f
34kurt

Answer:

    \frac{h_{liquid} }{ h_{body} } = 5/9

Explanation:

This is an exercise that we can solve using Archimedes' principle which states that the thrust is equal to the weight of the desalted liquid.

         B = ρ_liquid  g V_liquid

let's write the translational equilibrium condition

         B - W = 0

let's use the definition of density

        ρ_body = m / V_body

        m = ρ_body  V_body

        W = ρ_body  V_body  g

we substitute

          ρ_liquid  g  V_liquid = ρ_body  g  V_body

          \frac{\rho_{body}   }{\rho_{liquid} } } =  \frac{V_{liquid}   }{V_{body} } }

In the problem they indicate that the ratio of densities is 5/9, we write the volume of the bar

          V = A h_bogy

Thus

          \frac{V_{liquid} }{V_{1body} } = \frac{ h_{liquid} }{h_{body} }

we substitute

           5/9 = \frac{h_{liquid} }{ h_{body} }

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Vinil7 [7]

Answer:

hope it helps

plz mark as brainliest

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Explanation:

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It would be oraganic matter I think.
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