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Rzqust [24]
3 years ago
11

A motorcycle is capable of accelerating at 4.7 m/s^2, starting from rest, how far can it travel in 3 seconds?

Physics
2 answers:
tankabanditka [31]3 years ago
7 0
21.15m should be the answer ^_^
maw [93]3 years ago
5 0

Answer: 21.15m

Explanation:

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The component of waves must have same frequency and phase.


when the component of waves vibrate at the same rate and attain maximum point at the same time, reinforcement of the waves amplitude occur to cause a constructive interference.However, when the two waves are out of phase where one is at minimum when the other is at maximum a destructive interference happens.
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You place your pencil on your desk it stays there
zepelin [54]

Answer:

yes

Explanation:

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Solutions with ions that react with acids or bases to lessen their effects are fubrefs
Softa [21]

Answer:

Buffers

Explanation:

A buffer solution is a solution containing weak acids and their salts or weak bases and their salts.

A buffer solution is an equilibrium system that resists changes in pH or pOH when a small amount of an acid or base is added hence it is a solution of fairly constant pH value.

4 0
3 years ago
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
The body weighing 8kg moves in a straight line uniformly accelerated with an acceleration of 3m / s? on a horizontal surface, un
Usimov [2.4K]

Answer:

pls explain this to me

Explanation:

I will not answer it

3 0
3 years ago
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