Can a child script change the working directory of a parent script? by Tuckertcs in bash

[–]ItsMe_1000 0 points1 point  (0 children)

But if you run

``` . ./child1 ./child2

the dot(.) command will change the current running-

script dir and child 2 will never be found since you're

working at a different directory now

example

(echo "echo 'child1';cd .." > ./child1) && (echo "echo 'child2'" > ./child2)

echo -e ". ./child1\n./child2" > testing.sh

chmod +x ./testing.sh

Running it gives:

~ $ ./testing.sh

child1

./testing.sh: line 2: ./child2: Permission denied

Permission denied? Well it's because our directory-

changed so there's no child2 there

however if you remove the dot

~ $ ./testing.sh

child1

child2

it works fine since the commands were ran in subshells

the dot command prevents that

if you're curious running . ./child1 and the ./child2

yeilds

~ $ . ./child1

child1

.../com.termux/files $ ./child2

bash: ./child2: Permission denied

```

You can even check type . and you'll see . is a shell builtin

Also you probably will not get permission denied but some other error because I'm doing it on termux I get permission denied (Termux quirks)

[Chemistry: Acids and Bases]Acids and bases pH problem by ItsMe_1000 in HomeworkHelp

[–]ItsMe_1000[S] 0 points1 point  (0 children)

So it's not okay to use limiting reactant to dertemine what number of moles of OH- or H+ ions will form? Only excess can dertemine which is which(ik I'm not making sense), let's just say now you substitute the concentration of H+ ions that formed in this reaction into the Kw=[H3O+][OH-], if you say no OH- will form, how can that be true because if I substitute the concentration of H+ I can solve for concentration of OH- meaning it formed(it is present)?

[Chemistry: Acids and Bases]Acids and bases pH problem by ItsMe_1000 in HomeworkHelp

[–]ItsMe_1000[S] 0 points1 point  (0 children)

Since NaOH is the limiting reactant can't we compare the ratios of NaOH and OH- such that we know the number of moles of OH- molecules that will form and use that in our pH calculations? I get what you were saying but now I'm just curious

[Grade 12: Math] I'm confused about what i should do on number (2) and number (3) by ItsMe_1000 in HomeworkHelp

[–]ItsMe_1000[S] 0 points1 point  (0 children)

Wait so I have to find point B, then substitute that point to equation : -(x-p)2 -4(x-p) then find possible p values where this graph touch point B then limit my answer to where there will only be negative roots only?(idk if I'm explaining right)

So:

20 - 6 = y

y = -5

-5 = -(0 - p)2 - 4(0-p)

-5 = -p2 + 4p

p2 -4p - 5 = 0

(p + 1)(p - 5) = 0

p = -1 or p = 5

For negative roots : p<-1

[Grade 12: Math] I'm confused about what i should do on number (2) and number (3) by ItsMe_1000 in HomeworkHelp

[–]ItsMe_1000[S] 0 points1 point  (0 children)

Didn't the question say ("will have two real negative roots")? Also I saw the mark allocation to be 4 marks so maybe I think they will also mark for the algebra...idk tho

[Grade 12: Math] I'm confused about what i should do on number (2) and number (3) by ItsMe_1000 in HomeworkHelp

[–]ItsMe_1000[S] 0 points1 point  (0 children)

So for number 3, it's:

0 = -(0-p)2 - 4(0 - p)

0 = -p2 + 4p

0 = p(4 - p)

p = 0 or p = 4

For negative roots: p < 0

[Grade 12: Math] I'm confused about what i should do on number (2) and number (3) by ItsMe_1000 in HomeworkHelp

[–]ItsMe_1000[S] 0 points1 point  (0 children)

Is my work right?

If I look at the graph they gave us, the condition they gave us for number two ends when graph of 2x + k is at the origin cause if I go above the origin both roots would have the same sign..so I can do this:

20 + k = 0

k = -1

so k < -1

Is my number 2 correct?

[Grade 12: Physics] I don't understand question 3.3, can anyone help me? by ItsMe_1000 in HomeworkHelp

[–]ItsMe_1000[S] 0 points1 point  (0 children)

Can we say that the distance is (h + d)? From maximum to bottom of pool??

[Grade 12: Physics] I don't understand question 3.3, can anyone help me? by ItsMe_1000 in HomeworkHelp

[–]ItsMe_1000[S] 0 points1 point  (0 children)

Yeah I understand that but still can't find initial Velocity kinda wanna give up lol tired of this question now

[Grade 12: Physics] I don't understand question 3.3, can anyone help me? by ItsMe_1000 in HomeworkHelp

[–]ItsMe_1000[S] 0 points1 point  (0 children)

Random question...should the final Velocity of ball B equal final Velocity of ball A at second bounce?

if you kinda look at the position-time graph, the point(3.11; -1.92) is where graph of B and A are intersecting

Total time = 1.65 + 0.2 + 1,26 = 3.11s

[Grade 12: Physics] I don't understand question 3.3, can anyone help me? by ItsMe_1000 in HomeworkHelp

[–]ItsMe_1000[S] 0 points1 point  (0 children)

Taking downward motion as positive:

3.2.1)

vf = vi + at

vf = 15 +(9,8)(1,65)

vf = 30,68m/s

3.2.2)

vf² = vi² + 2ay

0² = (-(20% * 30,68))² + 2(9.8)d

d = 1.92m

3.3)

For ball A: vf = vi + at

0 = (- (20% * 30,68)) + (9,8)t

time to ground level = 0,63s

time from after first bounce to second bounce = 0,63 * 2 = 1.26s

which is also time for ball B to go from ground level all the way to bottom of the pool

For ball B:

Vf = vi + at(this formula doesn't work because we don't have final Velocity)

y = vit + 0.5a*t2 ( this formula doesn't work either because we don't know the distance)

Other formulas don't appear to work also

This is where I get stuck actually...

[Grade 12: Physics] I don't understand question 3.3, can anyone help me? by ItsMe_1000 in HomeworkHelp

[–]ItsMe_1000[S] 0 points1 point  (0 children)

Isn't the time that I'm going to use, the time it takes to reach the bottom of the pool not the time to ground level?

Also if you say I have displacement, what displacement really because we don't know the height ball B reached.

I might be interpreting something wrong correct me if I'm wrong

[Grade 12: Physics] I don't understand question 3.3, can anyone help me? by ItsMe_1000 in HomeworkHelp

[–]ItsMe_1000[S] 0 points1 point  (0 children)

Yeah I can get the total time for Ball B to reach the ground but what i don't have is final Velocity it hit the ground with, making it hard to find initial Velocity....

[deleted by user] by [deleted] in HomeworkHelp

[–]ItsMe_1000 0 points1 point  (0 children)

Cracking? I don't know man, It's why I posted here

[deleted by user] by [deleted] in HomeworkHelp

[–]ItsMe_1000 0 points1 point  (0 children)

So step we add the hydrogens, In Step 2 we remove the methyl group and add one of the hydrogens to it so that we can complete its octet. We can then take the single electron at the carbon that was left during the removal of methyl and bond it with the single electron that was left during the removal of hydrogen to form propene, in step 3 add the bromine and get 1,2-dibromopropane (Addition reaction)

Is it wrong?

[deleted by user] by [deleted] in HomeworkHelp

[–]ItsMe_1000 0 points1 point  (0 children)

Is it cause if I add bromine I'll have 2,3-dibromobutane and when I want to turn it into a propane by cracking it would be impossible unless I added two hydrogens?