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stellarik [79]
3 years ago
8

particles of mass 2m and m are fixed at the extreme of a light string of lenght l.find the centre of mass and the moment of iner

tia of the system of particles from the lighter particle.
Mathematics
1 answer:
Sever21 [200]3 years ago
3 0
Centre of mass will be at a distance of (L / 3) from mass 2m.

moment of inertia = m(2L/3)^2 + (2m)(L/3)^2
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tatuchka [14]
0.21 is greater than 0.12 because it in the 20s
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1. Find the value of
spin [16.1K]

9514 1404 393

Answer:

  1. -√5
  2. 3/5
  3. -4/5

Step-by-step explanation:

The relevant relations are ...

  sec = ±√(tan² +1)

  cos = 1/sec

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Sine and Cosecant are positive in quadrants I and II. Cosine and Secant are positive in quadrants I and IV.

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1. sec(θ) = -√((-2)² +1) = -√5

2. cos(θ) = 1/sec(θ) = 1/(5/3) = 3/5

3. csc(θ) = -1/√(1 -(-3/5)²) = -√(16/25) = -4/5

8 0
2 years ago
1. x + 2x - 1 = 38<br> Inverse operation
Lena [83]

Answer:

x=13

Step-by-step explanation:

First, you add the x's together: 2x+x=3x

Now you have:3x-1=38

Add 1 to both sides: 3x=39

1 cancel on the left side of the equation

Divide 3 on both sides: x=39/3

3 cancels on the left side and 39 divided by 3 is 13:

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Hope this helps!

6 0
3 years ago
Solve,justify and check: 3(7+5n)=6n+6(1+4n)
vlabodo [156]

Answer:

1=n

Step-by-step explanation:

Step 1- Distribute into the parenthesis.

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Step 2- Multiply

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Step 3- Add common variables to simplify.

21+15n= (24n+6n)+6

21+15n= 30n+6

Step 4- Subtract the smallest variable to both sides.

21+15n= 30n+6

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21= 15n+6

Step 5- Subtract 6 to both sides.

21= 15n+6

-6           -6

15= 15n

Step 6- Divide both sides by 15.

<u>15</u>= <u>15n</u>

15   15

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4 0
3 years ago
Read 2 more answers
7,14/5, 8/25 find the 7th term
tiny-mole [99]
I believe that would be 7
4 0
3 years ago
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