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m_a_m_a [10]
3 years ago
13

Lets try to get to 100 sub before charismas day Jordan Gracia 32 sub and 5 videos

Engineering
2 answers:
Natasha_Volkova [10]3 years ago
7 0
Keep on posting this and maybe even more people will subscribe to you :)
and I hope you do achieve it
dedylja [7]3 years ago
4 0

Answer:

make this the brainliest and ill do it

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How could the location of tests affect the performance of a catapult ?
Luba_88 [7]

Answer:

It could affect how far the projectile travels

Explanation:

Facing Uphill: Moves less far

Downhill: Moves further

3 0
3 years ago
Read 2 more answers
Câu 1: Tìm từ trong 2 câu thơ sau liên quan đến các từ: hình chiếu, tia chiếu, mặt phẳng chiếu.
xeze [42]

Explanation:

are bhai brainly aap english me questions kar ne ko hai

ye kon si bhasha hai ??????

4 0
2 years ago
A hypothetical A-B alloy of composition 57 wt% B-43 wt% A at some temperature is found to consist of mass fractions of 0.5 for b
Dennis_Churaev [7]

Answer:

composition of alpha phase is 27% B

Explanation:

given data

mass fractions  = 0.5 for both

composition = 57 wt% B-43 wt% A

composition = 87 wt% B-13 wt% A

solution

as by total composition Co = 57 and by beta phase composition  Cβ = 87  

we use here lever rule that is

Wα = Wβ   ...............1

Wα = Wβ = 0.5

now we take here left side of equation

we will get

\frac{C_\beta - Co}{C_\beta - Ca}   = 0.5

\frac{87 - 57}{87 - Ca} = 0.5  

solve it we get

Ca = 27

so composition of alpha phase is 27% B

8 0
3 years ago
Using the Rayleigh criterion, calculate the minimum feature size that can be resolved in a system with a 0.18 NA lens when g-lin
Vladimir79 [104]

Answer:

a)

# for a g line, R = 1.211 μm

# for an I-line, R = 1.013 μm

b)

# for a g line, R = 0.726 μm

# for an I-line, R = 0.243 μm

c)

# for a g line, R = 0.605 μm

# for an I-line, R = 0.608 μm

Explanation:

We know that;

Rayleigh Resolution R = 0.5 × λ/NA

for a g line, λ = 436 nm

for an I-line λ = 365 nm

a)

Now when NA = 0.18

# for a g line, λ = 436 nm

R = 0.5 × 436/0.18 =  1.211 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.18 =  1.013 μm

b)

when NA = 0.30

# for a g line, λ = 436 nm

R = 0.5 × 436/0.30 =  0.726 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 =  0.243 μm

c)

when NA = 0.36

# for a g line, λ = 436 nm

R = 0.5 × 436/0.36 =  0.605 μm

# for an I-line λ = 365 nm

R = 0.5 × 365/0.30 =  0.608 μm

6 0
3 years ago
جائت فكرة ربط الحواسيب لغرض نقل البيانات و مشاركتها و بعدها بفتره قصيره جائت إمكانية مشاركة الموارد بين الحواسيب صح ام خطأ​
melamori03 [73]

Answer:

بدلاً من ذلك يُشار إليه باسم مشاركة أو مشاركة شبكة ، الدليل المشترك هو دليل أو مجلد يمكن الوصول إليه من قبل العديد من المستخدمين على الشبكة. هذه هي الطريقة الأكثر شيوعًا للوصول إلى المعلومات ومشاركتها على شبكة محلية

Explanation:

5 0
3 years ago
Read 2 more answers
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