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avanturin [10]
2 years ago
11

-A coconut falls out of a tree 12.0 m above the ground and hits a bystander 3.00 m tall on the top of the head. It bounces back

up 1.50 m before falling to the ground. If the mass of the coconut is 2.00 kg, calculate the Eg of the coconut relative to the ground at each of the following sites: Assume no loss due to friction or air resistance.
a.Eg while it is still in the tree
b.Eg when it hits the bystander on the head
c.Eg when it bounces up to its maximum height
d. At what point will it have the most Ek?
Physics
1 answer:
Llana [10]2 years ago
3 0

Explanation:

Eg = mgh

a. Eg = (2.00 kg) (9.8 m/s²) (12.0 m)

Eg = 235 J

b. Eg = (2.00 kg) (9.8 m/s²) (3.00 m)

Eg = 58.8 J

c. Eg = (2.00 kg) (9.8 m/s²) (4.50 m)

Eg = 88.2 J

d. When the coconut hits the bystander:

Ek = 235 J − 58.8 J = 176 J

When the coconut hits the ground:

Ek = 88.2 J − 0 J = 88.2 J

Ek is the greatest when the coconut hits the bystander.

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Milk containing 3.7% fat and 12.8% total solids is to be evaporated to produce a product containing 7.9% fat. What is the yield
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Answer:

the yield of product is YP=46.835 % and the concentration of solids is

Cs = 27.33%

Explanation:

Assuming that all the solids and fats remains in the milk after the evaporation, then the mass of product mP will be

Mass of fat in 100 kg of milk = 100 kg* 0.037 = mP* 0.079

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then the yield YP of the product is

YP= mP / 100 kg =  46.835 kg / 100 kg = 46.835 %

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Cs = 100 kg* 0.128 / 46.835 kg  = 0.2733 = 27.33%

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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

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