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Lady bird [3.3K]
3 years ago
11

As a 3.0 kg bucket is being lowered into a 10 m deepwell, starting from top, the tension in the rope is 9.8 N. theacceleration o

f the bucket will be:____________A) 6.5 m/s2 downwardB) 9.8 m/s2 downwardC) zeroD) 3.3 m/s2 upwardE) 5.6 m/s2 upward
Physics
1 answer:
777dan777 [17]3 years ago
8 0

Answer:

A) 6.5 m/s²

Explanation:

Mass of the bucket, m = 3.0 kg

depth of the well, d = 10 m

tension on the rope, T = 9.8 N

The net downward force on the bucket is given as;

T = mg - ma

where;

a is downward acceleration of the bucket

9.8 = (3 x 9.8) - 3a

9.8 = 29.4 - 3a

3a = 29.4 - 9.8

3a = 19.6

a = 19.6 / 3

a = 6.53 m/s² downwards

Therefore, the acceleration of the bucket is 6.53 m/s² downwards

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Answer:

Explanation:

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Given

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≥  . 5254 x ⁻²⁵

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12.56 x 10⁻⁹ ≥ λ  

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3 years ago
Hello everyone. This is a question about Dimensional Analysis and I came across this question but I am unable to wrap my head ar
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Answer:

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Explanation:

I assume you mean this:

A = B² + 2B⁴/C²

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B² = [L]²/[T]²

B = [L]/[T]

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3 years ago
Why do scientists believe that dark matter exists even though it cannot be seen?
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Explain whether demand is likely to be elastic or inelastic for Big Macs. A. Elastic comma since many other fast food items coul
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Explanation:

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Since Big Macs are (i) a luxury good, and (ii) have close substitutes (other burgers available at McDonalds and other fast food stores), we will say their elasticity is greater than 1.  

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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
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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}

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where,

\Delta H^o = enthalpy of reaction = ?

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\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)]

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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

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

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As we know that,

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6 0
3 years ago
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