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3 years ago

14

angel has two chores after school. she rakes leaves for 3/4 hour and spends 1/2 hour washing the car. how long does angel spend

on her chores in all

2 answers:

Semenov [28]3 years ago

8 0

I think it is 3/4+2/4 (because 1/2 is equal) =5/4 or 1 1/4 or1hr and 15min

Hope I helped

Mark me as brainiest please

Alona [7]3 years ago

5 0

I takes her 1 1/4 of an hour

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Please help me Conrad has 6 more marbles than Rory. If r represents the numbers of marbles that Rory has, which expression repre

Blababa [14]

Answer: r + 6

Step-by-step explanation:

Shown in the question, r represents the numbers of marbles that Rory has. Conrad has 6 more marbles than Rory, which means than is r + 6.

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If p is the hypothesis of a conditional statement and q is the conclusion, which is represented by ~p → ~q?

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Conditional statement is a statement with a hypotesis and a conclusion:

If $\text{ \underline{ hypothesis } } p$ , then $\text{ \underline { conclusion } } q$ or mathematically $p\rightarrow q$ .

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If you negate (that means stick a "not" in front of) both the hypothesis and conclusion, you get the inverse: $\neg p\rightarrow \neg q$ .

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4 years ago

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It is in the 10th floor

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3 years ago

A closed cylindrical vessel contains a fluid at a 5MPa pressure. The cylinder, which has an outside diameter of 2500mm and a wal

Julli [10]

Answer:

1) Increase in the diameter equals 3.5 mm

2) Increase in the length equals $0.0003724L_{i}$ where $L_{i}$ is the initial length of the vessel.

Step-by-step explanation:

The diametric strain in the vessel is given by

$\epsilon_{D} =\epsilon_{diam}-\nu \epsilon _{axial}$

We have

$\epsilon _{diam}=\frac{\sigma _{hoop}}{E}\\\\\sigma _{hoop}=\frac{\Delta P\times D}{2t}\\\\\therefore \epsilon _{diam}=\frac{\Delta P\times D}{2t\times E}$

Applying values we get

$\therefore \epsilon _{diam}=\frac{5\times 10^{6}\times 2.5}{2\times 20\times 10^{-3}\times 193\times 10^{9}}\\\\\therefore \epsilon _{diam}=\frac{5}{3088}$

Similarly axial strain is given by

$\epsilon _{diam}=\frac{\sigma _{axial}}{E}$

$\sigma _{axial}=\frac{\Delta P\times D}{4t}\\\\\therefore \epsilon _{axial}=\frac{\Delta P\times D}{4t\times E}$

Applying values we get

$\therefore \epsilon _{axial}=\frac{5\times 10^{6}\times 2.5}{4\times 20\times 10^{-3}\times 193\times 10^{9}}\\\\\therefore \epsilon _{diam}=\frac{2.5}{3088}$

Hence The effect of axial strain along the diameter is given by

$-\nu \epsilon _{axial}$

Applying values we get

$-\nu \epsilon _{axial}=-0.27\times \frac{2.5}{3088}=-0.0002185$

hence

$\epsilon _{D} =\frac{5}{3088}-0.0002185\\\\\epsilon =0.00140$

Now by definition of strain we have

$\epsilon _{D} =\frac{D_{f}-D_{i}}{D_{i}}\\\\\therefore D_{f}=D_{i}+\epsilon D_{i}\\\\D_{f}=2.5+0.0014\times 2.5\\\\\therefore D_{f}=2503.5mm$

Increase in the diameter is thus 3.5 mm

Using the same procedure for axial strain we have

$\epsilon_{axial} =\epsilon_{axial}-\nu \epsilon _{diam}$

Applying values we get

$\epsilon_{axial} =\frac{2.5}{3088}-0.27\times \frac{5}{3088}$

$\epsilon_{axial} =0.0003724$

Now by definition of strain we have

$\epsilon _{axial} =\frac{L_{f}-L_{i}}{L_{i}}\\\\\therefore \Delta L=0.0003724L_{i}$

where $L_{i}$ is the initial length of the cylinder.

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4 years ago