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mash [69]
3 years ago
8

What are factors that would impede positive parenting and family relationships?

Physics
2 answers:
Oduvanchick [21]3 years ago
5 0

Answer:

Factors that would impede positive parenting and family relationships are discussed below.

Explanation:

Various factors would impede positive parenting and family relationships such as

  • Jealousy is the major cause of impede family relationship and lead to unhealthy family relationship.
  • Acting controlling and extra possessiveness lead to unhealthy family relationship and develops doubt for each other in a family and also the same in the case of positive parenting.
  • The day to day lifestyle in which no give time to each other in the family leads to unhealthy family relationships and exactly the same in parenting also.

fomenos3 years ago
4 0
•respect •love •comunication •understanding that's it hope this answer ur question
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How is a family defined
natulia [17]

Answer:Family

Explanation: Family is defined as people you love, people that you can count on. People that make you happy. Family isnt only people who are related to you by blood, we are all a family. Certain people are your family.

3 0
3 years ago
In one cycle a heat engine absorbs 450 J from a high-temperature reservoir and expels 290 J to a low-temperature reservoir. If t
Vitek1552 [10]

Answer:

So the ratio will be \frac{T_L}{T_H}=-0.171

Explanation:

We have given heat engine absorbs 450 joule from high temperature reservoir

So Q=450j

As the heat engine expels 290 j

So work done W = 290 J

We know that efficiency \eta =\frac{W}{Q}=\frac{290}{450}=0.6444

It is given that efficiency of the engine only 55 % of Carnot engine

So efficiency of Carnot engine =\frac{0.6444}{0.55}=1.171

Efficiency of Carnot engine is \eta =1-\frac{T_L}{T_H}

1.171 =1-\frac{T_L}{T_H}

\frac{T_L}{T_H}=-0.171

3 0
3 years ago
Read 2 more answers
Help Please I Don't Know.
Tom [10]
1)
HCl: hydrogen, chloride
3CO2: carbon, oxygen
2Na2SO4:sodium, sulphur, oxygen.

2)
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8 0
3 years ago
(a) Consider the initial-value problem dA/dt = kA, A(0) = A0 as the model for the decay of a radioactive substance. Show that, i
murzikaleks [220]

Answer:

a) t = -\frac{ln(2)}{k}

b) See the proof below

A(t) = A_o 2^{-\frac{t}{T}}

c) t = 3T \frac{ln(2)}{ln(2)}= 3T

Explanation:

Part a

For this case we have the following differential equation:

\frac{dA}{dt}= kA

With the initial condition A(0) = A_o

We can rewrite the differential equation like this:

\frac{dA}{A} =k dt

And if we integrate both sides we got:

ln |A|= kt + c_1

Where c_1 is a constant. If we apply exponential for both sides we got:

A = e^{kt} e^c = C e^{kt}

Using the initial condition A(0) = A_o we got:

A_o = C

So then our solution for the differential equation is given by:

A(t) = A_o e^{kt}

For the half life we know that we need to find the value of t for where we have A(t) = \frac{1}{2} A_o if we use this condition we have:

\frac{1}{2} A_o = A_o e^{kt}

\frac{1}{2} = e^{kt}

Applying natural log we have this:

ln (\frac{1}{2}) = kt

And then the value of t would be:

t = \frac{ln (1/2)}{k}

And using the fact that ln(1/2) = -ln(2) we have this:

t = -\frac{ln(2)}{k}

Part b

For this case we need to show that the solution on part a can be written as:

A(t) = A_o 2^{-t/T}

For this case we have the following model:

A(t) = A_o e^{kt}

If we replace the value of k obtained from part a we got:

k = -\frac{ln(2)}{T}

A(t) = A_o e^{-\frac{ln(2)}{T} t}

And we can rewrite this expression like this:

A(t) = A_o e^{ln(2) (-\frac{t}{T})}

And we can cancel the exponential with the natural log and we have this:

A(t) = A_o 2^{-\frac{t}{T}}

Part c

For this case we want to find the value of t when we have remaining \frac{A_o}{8}

So we can use the following equation:

\frac{A_o}{8}= A_o 2^{-\frac{t}{T}}

Simplifying we got:

\frac{1}{8} = 2^{-\frac{t}{T}}

We can apply natural log on both sides and we got:

ln(\frac{1}{8}) = -\frac{t}{T} ln(2)

And if we solve for t we got:

t = T \frac{ln(8)}{ln(2)}

We can rewrite this expression like this:

t = T \frac{ln(2^3)}{ln(2)}

Using properties of natural logs we got:

t = 3T \frac{ln(2)}{ln(2)}= 3T

8 0
3 years ago
Select the correct answer.
andrew-mc [135]
I think the answer would be C
7 0
3 years ago
Read 2 more answers
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