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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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. If you saved $2.00 on january 1, $4.00 on february 1, $8.00 on march 1, $16.00 on april 1, and so on. how much money would you
SIZIF [17.4K]
=2+4+8+16+32+64+128+256+512+1024+2048+4096
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4 0
3 years ago
Read 2 more answers
I need both of these answered
olga_2 [115]
Q18)
7/0.75 = 28/3
28/3(45) = 420 -> 420/60 = 7
Therefore the answer is B. 7 minutes.
Q20)
60.90/5 = 12.18
12.18(7) = 85.26
85.26+14.99 = 100.25
Therefore the answer is D. $100.25.
6 0
2 years ago
A square has sides measuring 3 feet. If the sides of a square are doubled, will the area also double? Explain
Kryger [21]
Hello!

Let's find the area and see!

A = s^2

A = 3^2

A = 6 ft^2

Now, if you double 3, you get 6. 3 × 2 = 6 Find the area:

A = 6^6

A = 36 ft^2

ANSWER:

No, the area will not be doubled. The area was 9 ft^2 when the sides were 3 feet long. If you double 9, you get 18, since 9 × 2 = 18.

36 is not the double of 9, so no, the area is not double.
8 0
3 years ago
Matthew invested $3,000 into two accounts. One account paid 3% interest and the other paid 8% interest. He earned 4% interest on
boyakko [2]

<u>Answer:</u>

<em>Mathew invested</em><em> $600 and $2400</em><em> in each account.</em>

<u>Solution:</u>

From question, the total amount invested by Mathew is $3000. Let p = $3000.

Mathew has invested the total amount $3000 in two accounts. Let us consider the amount invested in first account as ‘P’

So, the amount invested in second account = 3000 – P

Step 1:

Given that Mathew has paid 3% interest in first account .Let us calculate the simple interest (I_1) earned in first account for one year,

\text {simple interest}=\frac{\text {pnr}}{100}

Where  

p = amount invested in first account

n = number of years  

r = rate of interest

hence, by using above equation we get (I_1) as,  

I_{1}=\frac{P \times 1 \times 3}{100} ----- eqn 1

Step 2:

Mathew has paid 8% interest in second account. Let us calculate the simple interest (I_2) earned in second account,

I_{2} = \frac{(3000-P) \times 1 \times 8}{100} \text { ------ eqn } 2

Step 3:

Mathew has earned 4% interest on total investment of $3000. Let us calculate the total simple interest (I)

I = \frac{3000 \times 1 \times 4}{100} ----- eqn 3

Step 4:

Total simple interest = simple interest on first account + simple interest on second account.

Hence we get,

I = I_1+ I_2 ---- eqn 4

By substituting eqn 1 , 2, 3 in eqn 4

\frac{3000 \times 1 \times 4}{100} = \frac{P \times 1 \times 3}{100} + \frac{(3000-P) \times 1 \times 8}{100}

\frac{12000}{100} = \frac{3 P}{100} + \frac{(24000-8 P)}{100}

12000=3P + 24000 - 8P

5P = 12000

P = 2400

Thus, the value of the variable ‘P’ is 2400  

Hence, the amount invested in first account = p = 2400

The amount invested in second account = 3000 – p = 3000 – 2400 = 600  

Hence, Mathew invested $600 and $2400 in each account.

3 0
3 years ago
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Which of the following statements is true?
boyakko [2]

Answer:

B) Aluminum and steel are good conductors of electricity.

Step-by-step explanation:

5 0
2 years ago
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