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Alex
3 years ago
15

A 2.0 kg cart moving right at 5.0 on a frictionless track collides with a 3.0 kg cart initially at rest. The 2.0 kg

Mathematics
1 answer:
Wewaii [24]3 years ago
4 0

final speed of the 3.0 kg cart is 2.67 m/s .

<u>Step-by-step explanation:</u>

Here we have , A 2.0 kg cart moving right at 5.0 on a friction less track collides with a 3.0 kg cart initially at rest. The 2.0 kg  cart has a final speed of 1.0 to the left. We will use law of conservation of momentum: For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.

m_1(v_1)+m_2(v_2) =m_1(v_3)+m_2(v_4)

⇒ m_1(v_1)+m_2(v_2) =m_1(v_3)+m_2(v_4)

⇒ 2(5)+3(0) =2(1)+3(v_4)

⇒ 10 =2+3(v_4)

⇒ 3(v_4) = 8

⇒ (v_4) = \frac{8}{3}

⇒ (v_4) = 2.67

Therefore, final speed of the 3.0 kg cart is 2.67 m/s .

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Fantom [35]

Part 1: You can simplify a_n to

\dfrac{3n+1}{3n}-\dfrac{3n}{3n+1} = \dfrac1{3n}+\dfrac1{3n+1}

Presumably, the sequence starts at <em>n</em> = 1. It's easy to see that the sequence is strictly decreasing, since larger values of <em>n</em> make either fraction smaller.

(a) So, the sequence is bounded above by its first value,

|a_n| \le a_1 = \dfrac13+\dfrac14 = \boxed{\dfrac7{12}}

(b) And because both fractions in a_n converge to 0, while remaining positive for any natural number <em>n</em>, the sequence is bounded below by 0,

|a_n| \ge \boxed{0}

(c) Finally, a_n is bounded above and below, so it is a bounded sequence.

Part 2: Yes, a_n is monotonic and strictly decreasing.

Part 3:

(a) I assume the choices are between convergent and divergent. Any monotonic and bounded sequence is convergent.

(b) Since a_n is decreasing and bounded below by 0, its limit as <em>n</em> goes to infinity is 0.

Part 4:

(a) We have

\displaystyle \lim_{n\to\infty} \frac{10n^2+1}{n^2+n} = \lim_{n\to\infty}10+\frac1{n^2}}{1+\frac1n} = 10

and the (-1)ⁿ makes this limit alternate between -10 and 10. So the sequence is bounded but clearly not monotonic, and hence divergent.

(b) Taking the limit gives

\displaystyle\lim_{n\to\infty}\frac{10n^3+1}{n^2+n} = \lim_{n\to\infty}\frac{10+\frac1{n^3}}{\frac1n+\frac1{n^2}} = \infty

so the sequence is unbounded and divergent. It should also be easy to see or establish that the sequence is strictly increasing and thus monotonic.

For the next three, I'm guessing the options here are something to the effect of "does", "may", or "does not".

(c) may : the sequence in (a) demonstrates that a bounded sequence need not converge

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(e) does : this is true and is known as the monotone convergence theorem.

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3 years ago
Teresas restraunt bill comes to $29.99 before tax. If the sales tax is 6.25% and she tips the waiter 20%. What is the total cost
alisha [4.7K]

Answer:

$37.86

Step-by-step explanation:

Multiply 29.99 by 0.0625 to get the sales tax

29.99x0.0625=1.87

Multiply 29.99 by 0.20 to get the tip

29.99x0.20=5.998

Round 5.998 to 6

Add 1.87 to 29.99

29.99+1.87=31.86

Add 6 to 31.86

31.86+6=37.86


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Step-by-step explanation:

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