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kramer
2 years ago
4

Ley de Hooke

Physics
1 answer:
sp2606 [1]2 years ago
4 0

Answer:

1)  F = 58.45 N ,  2) m = 4,490 kg ,  3)  x = -0.10 m , 4)   k = 1750 N / m

5)  k = 19.6 N / m

Explanation:

This problem is formed by small exercises regarding Hooke's law and Young's modulus

Hooke's law

1) The elastic force is

         F = k x

         F = 835 0.07

         F = 58.45 N

2) from the condition of static equilibrium the spring force is equal to the hanging weight

          W = F

          m g = k x

          m = k x / g

          m = 1100 0.04 / 9.8

          m = 4,490 kg

3) F = -k x

          x = - F / k

         x = - 10/100

         x = -0.10 m

4) F = k x

         k = F / x

         k = 700 / 0.40

         k = 1750 N / m

5) k = 19.6 / 1

         k = 19.6 N / m

Young's modulus

the expression for the unit elongation a one dimension is

             F / A = Y Dx / L

1) they ask us for the Young module

let's look for the area the wire

          A = pi r2 = pi d2 / 4

          A = pi 0.0022 2/4

          A = 3.80 10-6 m²

         Y = (F / A) / (Dx / L)

we substitute

         Y = (390 / 3.90 10-6) / (0.10 / 120)

         Y = 1 10-4 / 8.33 10-4

         Y = 0.12 N / m²

2) F / A = Y Dx / L

the young modulus of steel is 20 10 10 N / m²

           F / A = 20 1010 0.003 / 2

           F / A = 3 108 N / m²

3) marine force occurs when elongation is equal to initial length

               F / A = Y 1

               F = Y A

               F = 20 1010 3 10-4

               F = 6 10 7 N

4) Dx = L F / (A Y)

           W = F = m g

    Dx = 1.2 (400 9.8) / 20 1010 3 10-4)

            Dx = 7.84 10-5 m

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Three couples and two single individuals have been invited to an investment seminar and have agreed to attend. Suppose the proba
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Answer:

a) Probability mass function of x

x P(X=x)

0 0.0602

1 0.0908

2 0.1700

3 0.2050

4 0.1800

5 0.1550

6 0.0843

7 0.0390

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0 0.0602

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Explanation:

Probability of arriving late = 0.43

Probability of coming late = 0.57

Let's start with the probability P(X=0) that exactly 0 people arrive late, the probability P(X=1) that exactly 1 person arrives late, the probability P(X=2) that exactly 2 people arrive late, and so on up to the probability P(X=8) that 8 people arrive late.

Interpretation(s) of P(X=0)

The two singles must arrive on time, and the three couples also must. It follows that P(X=0) = (0.57)⁵ = 0.0602

Interpretation(s) of P(X=1)

Exactly 1 person, a single, must arrive late, and all the rest must arrive on time. The late single can be chosen in 2 ways. The probabiliy that (s)he arrives late is 0.43.

The probability that the other single and the three couples arrive on time is (0.57)⁴

It follows that

P(X=1) = (2)(0.43)(0.57)⁴ = 0.0908

Interpretation(s) of P(X=2)

Two late can happen in two different ways. Either (i) the two singles are late, and the couples are on time or (ii) the singles are on time but one couple is late.

(i) The probability that the two singles are late, but the couples are not is (0.43)²(0.57)³

(ii) The probability that the two singles are on time is (0.57)²

Given that the singles are on time, the late couple can be chosen in 3 ways. The probability that it is late is 0.43 and the probability the other two couples are on time is (0.57)².

So the probability of (ii) is (0.57)²(3)(0.43)(0.57)² which looks better as (3)(0.43)(0.57)⁴ It follows that

P(X=2) = (0.43)²(0.57)³ + (3)(0.43)(0.57)⁴ = 0.0342 + 0.136 = 0.1700

Interpretations of P(X=3).

Here a single must arrive late, and also a couple. The late single can be chosen in 2 ways. The probability the person is late but the other single is not is (0.43)(0.57).

The late couple can be chosen in 3 ways. The probability one couple is late and the other two couples are not is (0.43)(0.57)². Putting things together, we find that

P(X=3) = (2)(3)(0.43)²(0.57)³ = 0.2050

Interpretation(s) P(X=4)

Since we either (i) have the two singles and one couple late, or (ii) two couples late. So the calculation will break up into two cases.

(i) Two singles and one couple late

Two singles' probability of being late = (0.43)² and One couple being late can be done in 3 ways, so its probability = 3(0.43)(0.57)²

(ii) Two couples late, one couple and two singles early

This can be done in only 3 ways, and its probability is 2(0.57)³(0.43)²

P(X=4) = (3)(0.43)³(0.57)² + (3)(0.57)³(0.43)² = 0.0775 + 0.103 = 0.1800

Interpretations of P(X=5)

For 5 people to be late, it has to be two couples and 1 single person.

For couples, The two late couples can be picked in 3 ways. Probability is 3(0.43)²(0.57)

The late single person can be picked in two ways too, 2(0.43)(0.57)

P(X=5) = 2(3)(0.43)³(0.57)² = 0.1550

Interpretations of P(X=6)

For 6 people to be late, we have either (i) the three couples are late or (ii) two couples and the two singles.

(i) Three couples late with two singles on time = (0.43)³(0.57)²

(ii) Two couples and two singles late

Two couples can be selected in 3 ways, so probability = 3(0.43)²(0.57)(0.43)²

P(X=6) = (0.43)³(0.57)² + 3(0.43)⁴(0.57) = 0.0258 + 0.0585 = 0.0843

Interpretation(s) of P(X=7)

For 7 people to be late, it has to be all three couples and only one single (which can be picked in two ways)

P(X=7) = 2(0.57)(0.43)⁴ = 0.0390

Interpretations of P(X=8)

Everybody had to be late

P(X=8) = (0.43)⁵ = 0.0147

6 0
2 years ago
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