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vlada-n [284]
2 years ago
7

B

Physics
1 answer:
ololo11 [35]2 years ago
3 0

Answer:

Thermal capacity of the object = 720 j/°C

Explanation:

Given:

Specific heat capacity of metal = 360J/(kg°C)

Mass of given object = 2 kg

Find:

Thermal capacity of the object

Computation:

Thermal capacity = Specific heat x Mass

Thermal capacity of the object = Specific heat capacity of metal x Mass of given object

Thermal capacity of the object = 360 x 2

Thermal capacity of the object = 720 j/°C

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Answer:

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Explanation:

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2 years ago
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Which is the best example of the use of imagery in a sentence?
Talja [164]

Answer:

Common Examples of Imagery

Taste: The familiar tang of his grandmother's cranberry sauce reminded him of his youth. Sound: The concert was so loud that her ears rang for days afterward. Sight: The sunset was the most gorgeous they'd ever seen; the clouds were edged with pink and gold.

I hope it's helpful!

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3 years ago
Consider a spherical volume of space that is large enough to be considered homogeneous. Also consider a particle on the surface
LenKa [72]

Answer:

Option A applies.

A.  Greater than its escape speed from the mass within the volume

Explanation:

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3 years ago
You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that pla
lesantik [10]

Answer:

When they are connected in series

     The  50 W bulb glow more than the 100 W bulb

Explanation:

From the question we are told that

     The power rating  of the first bulb is P_1  = 100 \ W

      The power rating of the second bulb is  P_2  =  50 \ W

     

Generally the power rating of the first bulb is mathematically represented as

      P_1  =  V^2 R

Where  V is the normal household voltage which is constant for both bulbs

  So  

        R_1  =  \frac{V^2}{P_1 }

substituting values

        R_1  =  \frac{V^2}{100}

Thus the resistance of the second bulb would be evaluated as

       R_2  =  \frac{V^2}{50}

From the above calculation we see that

        R_2  >  R_1

This power rating of the first bulb can also be represented mathematically as  

        P_  1  =  I^2_1  R_1

This power rating of the first bulb can also be represented mathematically as    

       P_  2  =  I^2_2 R_2

Now given that they are connected in series which implies that the same current flow through them so

       I_1^2 =  I_2^2

This means  that

       P \ \alpha  \  R

So  when they are connected in series

     P_2  >  P_1

This means that the 50 W bulb glows more than the 100 \ W bulb

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