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3 years ago

7

Fill in the correct answers. A big wheel with twice the circumference of a small wheel will rotate with __ the force and _ t

he speed of the small wheel.
A) half, twice
B) Three times, half
C) twice, half
D) half, three times

2 answers:

hichkok12 [17]3 years ago

4 0

I think it’s c

hope this helped

Lina20 [59]3 years ago

3 0

Answer:

C.

Explanation:

""""""" Twice,Half """"""

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a bus starting from rest moves with a uniform acceleration of 0.1 metre per second square for 2 minutes find the speed acquired

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S ?
U 0m/s
V ?
A 0.1m/s^2
T 2min (120 sec)

S=ut+0.5at^2
S=0(120 sec)+0.5(0.1m/s^2)(120 sec)^2
S=720m

Distance double 720m*2=1440m

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V^2=(0)^2+2(0.1 m/s^2)(1440m)
V^2=288
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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

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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.

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maria [59]

Answer:

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Explanation:

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On Ella's fifth birthday, her parents told her she had to stop whining or no one would want to play with her. She tried very har

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3 years ago

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A candy-filled piñata is hung from a tree for Elia's birthday. During an unsuccessful attempt to break the 4.4-kg piñata, Tonja

Klio2033 [76]

Answer: v = 0.6 m/s

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Momentum is calculated as Q = m.v

For the piñata problem:

$Q_{i}=Q_{f}$

$m_{p}v_{p}_{i}+m_{s}v_{s}_{i}=m_{p}v_{p}_{f}+m_{s}v_{s}_{f}$

Before the collision, the piñata is not moving, so $v_{p}_{i}=0$ .

After the collision, the stick stops, so $v_{s}_{f}=0$ .

Rearraging, we have:

$m_{s}v_{s}_{i}=m_{p}v_{p}_{f}$

$v_{p}_{f}=\frac{m_{s}v_{s}_{i}}{m_{p}}$

Substituting:

$v_{p}_{f}=\frac{(0.54)(4.8)}{(4.4)}$

$v_{p}_{f}=$ 0.6

Immediately after being cracked by the stick, the piñata has a swing speed of 0.6 m/s.

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3 years ago