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seraphim [82]
3 years ago
5

A wave with a greater amplitude will transfer . . . . \

Physics
1 answer:
Orlov [11]3 years ago
8 0

Answer:

More energy

Explanation:

The amount of energy carried by a wave is related to the amplitude of the wave itself. In particular, the amount of energy carried by the wave is proportional to the square of the amplitude of the wave:

E \propto A^2

where

E is the energy

A is the amplitude

This means, for instance, that if the amplitude of a wave is doubled, the energy it carries increases by a factor 4.

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Human industrialization, because without greenhouse gases the planet would be less warm

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A 4 kg box on the ground has a total of three horizontal forces on it as shown. The acceleration of the box is...
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C

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What is the specific heat of the masses in this experiment? Infer the substance the masses are made of and explain your inferenc
Liono4ka [1.6K]

The metal whose specific heat capacity is close to the obtained value is aluminum.

<h3>What is specific heat capacity</h3>

The specific heat capacity of an object is the heat required to raise a unit mass of the substance by 1 kelvin.

Q= mc\Delta \theta

where;

  • c is the specific heat capacity
  • Δθ is change in temperature

Let the mass of the water = 50 g

mass of the metal for this first trial = 50 g

The heat gained by the water is calculated as follows

Q = 50 \times 4.184 \times 8.4\\\\Q = 1757.28 \ J

Specific heat capacity of the metal for the first trial is calculated as follows;

Heat gained by water = Heat lost by metal

C = \frac{Q}{m\Delta T} = \frac{1757.28}{50\times 8.4} = 4.184 \ J/g^oC

Specific heat capacity of the metal for the second trial;

mass of metal = 200 - mass of water = 150 g

C_2 = \frac{1757.28}{150 \times 15.2} = 0.77 \ J/g^oC

Specific heat capacity of the metal for the third trial;

C_3 = \frac{1757.28}{250 \times 20.8} = 0.34\ J/g^oC

Specific heat capacity of the metal for the fourth trial;

C_4 = \frac{1757.28}{350 \times 25.4} = 0.19\ J/g^oC

Specific heat capacity of the metal for the fifth trial;

C_5 = \frac{1757.28}{450 \times 29.6} = 0.13\ J/g^oC

Average specific heat capacity

C = \frac{4.184 + 0.77 + 0.34+ 0.19 + 0.13 }{5} = 1.12 \ J/g^oC = 1120 J/kg^oC

The metal whose specific heat capacity is close to the above value is aluminum.

Learn more about specific heat capacity here: brainly.com/question/16559442

8 0
2 years ago
1.
Jet001 [13]

Answer:

1. high frequency and high energy.

2. short wavelengths and high frequencies.

3. less energy and long wavelengths.

4. high frequencies and high energy

5. X-rays

6. infrared waves

7. radio waves.

8. Gamma Rays

9. Microwaves

10. infrared waves

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Are particularly useful in explaining deformation and strengthening in metallic materials.
vampirchik [111]

Answer: value

Explanation:

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