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uranmaximum [27]
3 years ago
11

Need Some Help Please :)

Physics
2 answers:
joja [24]3 years ago
6 0
1. The amount of energy carried by the wave is related to the Amplitude of the wave.
2. A mechanical wave requires an initial energy input, Once this initial energy is added the wave travels through the medium until all it's energy is transferred.
yanalaym [24]3 years ago
6 0

Answer:

  1. The higher the wave, the higher the frequency
  2. Energies are transferred in mechanical waves through a medium

Explanation:

<u>How frequency relates to the energy of the wave</u>

Literally, frequency of a wave refers to the number of waves that move past a certain point during a given amount of time, (although it is often measured per second )

So, this means that when more waves moves past the given points, the frequency of the wave will get increased and if other wise, the frequency if the wave will get reduced, this is so because wave frequency is related to wave energy.

Since all that waves really are is traveling energy, the more energy in a wave, the higher its frequency. The lower the frequency is, the less energy in the wave.

<u>How energy is transferred through mechanical wave</u>

Mechanical waves are such that they're not capable of transmitting energy via vacuum. Hence, they need a medium to transmit these energies from one point to another.

All type of mechanical waves need a form of medium in order to transmit.

Take for instance:

  • A slinky wave requires the coils of the slinky;
  • A water wave requires water

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39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th
tatuchka [14]

<u>Answer:</u> The final temperature of the solution is 80.14^oC

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, \text{heat}_{absorbed}=\text{heat}_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 39 g

m_2 = mass of coffee = 166 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 83^oC

c_1 = specific heat of aluminium = 0.904J/g^oC

c_2 = specific heat of coffee= 4.1801J/g^oC

Putting all the values in equation 1, we get:

39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]

T_{final}=80.14^oC

Hence, the final temperature of the solution is 80.14^oC

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