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SVEN [57.7K]
3 years ago
6

What type of wave travels through matter only??

Physics
2 answers:
Olenka [21]3 years ago
7 0

A mechanical wave can only travel through matter.

Lorico [155]3 years ago
3 0
Mechanical waves travel through water because waves transfer from one place to another
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Suppose a ball had a potential energy of 5 J when you dropped it.What would be it’s kinetic energy just as it hit the ground?
timama [110]

Answer:

Due to Conservation of Energy just as the ball hits the ground it's potential energy is assumed zero

Therefore disregarding air resistance all energy is converted into potential energy.

So KE = PE  

(5 J)

4 0
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Which of the following statements best describes earths mineral resources
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It is D, you can not replace minerals. This makes them valuable.
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On the way to school, the bus speeds up from 20 m/s to 36 m/s in 4 seconds. What distance
Tresset [83]
Answer B. 112 m



Step-by-Step Explanation

initial velocity u = 20 m /s
final velocity v = 36 m /s
time taken t = 4 s
acceleration = (v - U) / t
= (36 - 20) / 4
a=4m/s2
from the formula
7-u2=2as , sis distance covered
putting the values
362-202=2×4×s
1296 - 400 = 8 x S
S= 112 m
4 0
2 years ago
the force exerted by gravity,electricity,and magnetism act at a distance with no physical connection. for this reason, they are
Soloha48 [4]
Its called the electric force
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3 years ago
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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
4 years ago
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