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3 years ago

Why do you think the combined wave is more powerful than either the transverse or longitudinal wave with the same amplitude

Physics

1 answer:

liq [111]3 years ago

6 0

Answer:

Explanation:

The combined wave only end up been more powerful than the Longitudinal wave. This means, the transverse wave is more powerful than the combined wave. In transverse wave, the oscillation is perpendicular to the direction of the wave, while in longitudinal wave, the motion of the movement of the object is parallel to the movement of the wave. And in combined wave, the movement of the medium is in a circular manner,

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A current exists whenever electric charges move. If ΔQ is the net charge that passes through a surface during a time period Δt,

jeka57 [31]

Answer:

It represents the change in charge Q from time t = a to t = b

Explanation:

As given in the question the current is defined as the derivative of charge.

I(t) = dQ(t)/dt ..... (i)

But if we take the inegral of the equation (i) for the time interval from t=a to

t =b we get

Q =∫_a^b▒〖I(t) 〗 dt

which shows the change in charge Q from time t = a to t = b. Form here we can say that, change in charge is defiend as the integral of current for specific interval of time.

5 0

3 years ago

Consider the following True/False statements:

Ainat [17]

Answer:

6) False

7) True

8) False

9) False

10) False

11) True

12) True

13) True

14) True

Explanation:

The spacing between two energy levels in an atom shows the energy difference between them. Clearly, B has a greater value of ∆E compared to A. This implies that the wavelength emitted by B is greater than A while B will emit fewer, more energetic photons.

When atoms jump from lower to higher energy levels, photons are absorbed. The kinetic energy of the incident photon determines the frequency, wavelength and colour of light emitted by the atom.

The energy level to which an atom is excited is determined by the kinetic energy of the incident electron. As the voltage increases, the kinetic energy of the electron increases, the further the atom is from the source of free electrons, the greater the required kinetic energy of free electron. When electrons are excited to higher energy levels, they must return to ground state.

4 0

3 years ago

A body weighing 108N moves with speed of 5m/s in a horizontal

poizon [28]

Answer:

i) 5 $m$ / $s^{2}$ ii) 54 $N$ iii) 54 $N$

Explanation:

i) $a = \frac{v^{2}}{r}$ ⇒ $a$ = 5² ÷ 5 = 5 $m/s^{2}$

ii) $m = \frac{W}{g}$ ⇒ $m$ = 108 ÷ 10 = 10.8 $kg$ , $F = ma$ ⇒ $F =$ 10.8 × 5 = 54 $N$

iii) F1 = F2 = 54 $N$

4 0

3 years ago

What is the basic unit of matter? atom cell bacterium molecule Description

Zielflug [23.3K]

The atom is the most basic unit of matter

6 0

3 years ago

Read 2 more answers

ome metal oxides can be decomposed to the metal and oxygen under reasonable conditions. 2 Ag2O(s) → 4 Ag(s) + O2(g) Thermodynami

yuradex [85]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

Explanation :

The given balanced chemical reaction is,

$2Ag_2O(s)\rightarrow 4Ag(s)+O_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{product}$

$\Delta H^o=[n_{Ag}\times \Delta H_f^0_{(Ag)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta H_f^0_{(Ag_2O)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[4mole\times (0kJ/mol)+1mole\times (0kJ/mol)}]-[2mole\times (-31.1kJ/mol)]$

$\Delta H^o=62.2kJ=62200J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{product}$

$\Delta S^o=[n_{Ag}\times \Delta S_f^0_{(Ag)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]-[n_{Ag_2O}\times \Delta S_f^0_{(Ag_2O)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[4mole\times (42.55J/K.mole)+1mole\times (205.07J/K.mole)}]-[2mole\times (121.3J/K.mole)]$

$\Delta S^o=132.67J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 298 K.

$\Delta G^o=(62200J)-(298K\times 132.67J/K)$

$\Delta G^o=22664.34J=22.66kJ$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $62.2kJ,132.67J/K\text{ and }22.66kJ$ respectively.

6 0

3 years ago