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Artyom0805 [142]
2 years ago
14

Car enthusiasts often lower their cars closer to the ground as a matter of style. James wants to lower his car by replacing all

four of his original coil springs with new ones offered as an option by the car manufacturer. Note that changing the coil springs on a car can unsafely affect its handling. Consult your national or state laws before altering the suspension on any car. The new springs will be identical to the original springs, except the force constant will be 5355.00 N/m smaller. When James removes the original springs, he discovers that the length of each spring expands from 8.55 cm (its length when installed) to 12.00 cm (its length with no load placed on it). If the mass of the car body is 1455.00 kg, by how much will the body be lowered with the new springs installed, compared to its original height?
Physics
1 answer:
Alexandra [31]2 years ago
5 0

Answer:

\Delta h=0.0364\ m=3.64\ cm

Explanation:

Given:

  • change in stiffness constant of the spring on replacing the original springs, \delta k=5355\ N.m^{-1}
  • mass of the car, m=1455\ kg
  • initial length of the original car-spring before compression, l_i=12\ cm=0.12\ m
  • final length of the original car-spring after compression, l_f=8.55\ cm=0.0855\ m

So, weight of the car:

w=m.g

w=1455\times 9.81

w=14273.55\ N

<u>Now the spring constant of original spring:</u>

w=4k_o.(l_i-l_f) (since 4 springs are in parallel)

14273.55=4k_o\times (0.12-0.0855)

k_o=103431.522\ N.m^{-1}

<u>So the stiffness constant of the new springs:</u>

k_n=k_o-\delta k

k_n=103431.522-5355

k_n=98076.522\ N.m^{-1}

<u>Now the height lowered:</u>

w=k_n.4\Delta h (since 4 springs are in parallel)

14273.55=4\times98076.522\times \Delta h

\Delta h=0.0364\ m=3.64\ cm

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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
lions [1.4K]

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

8 0.0147

b) Cumulative Distribution function of X

x F(x)

0 0.0602

1 0.1510

2 0.3210

3 0.5260

4 0.7060

5 0.8610

6 0.9453

7 0.9843

8 1.0000

The cumulative distribution function gives 1.0000 as it should.

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
The value of specific heat for copper is 390 J/kg⋅C∘, for aluminun is 900 J/kg⋅C∘, and for water is 4186 J/kg⋅C∘.
abruzzese [7]

Answer:

The equilibrium temperature is

21.97°c

Explanation:

This problem bothers on the heat capacity of materials

Given data

specific heat capacities

copper is Cc =390 J/kg⋅C∘,

aluminun Ca = 900 J/kg⋅C∘,

water Cw = 4186 J/kg⋅C∘.

Mass of substances

Copper Mc = 235g

Aluminum Ma = 135g

Water Mw = 825g

Temperatures

Copper θc = 255°c

Water and aluminum calorimeter θ1= 16°c

Equilibrium temperature θf =?

Applying the principle of conservation of heat energy, heat loss by copper equal heat gained by aluminum calorimeter and water

McCc(θc-θf) =(MaCa+MwCw)(θf-θ1)

Substituting our data into the expression we have

235*390(255-θf)=

(135*900+825*4186)(θf-16)

91650(255-θf)=(3574950)(θf-16)

23.37*10^6-91650*θf=3.57*10^6θf- +57.2*10^6

Collecting like terms and rearranging

23.37*10^6+57.2*10^6=3.57*10^6θf+91650θf

8.2*10^6=3.66*10^6θf

θf=80.5*10^6/3.6*10^6

θf =21.97°c

5 0
1 year ago
Recall that in the equilibrium position, the upward force of the spring balances the force of gravity on the weight. Use this co
natima [27]

Recall that in the equilibrium position, the upward force of the spring balances the force of gravity on the weight is given below.

Explanation:

Measure unstretched length of spring, L.  E.g. L = 0.60m.

Set mass to a convenient value (e.g. m = 0.5kg).

Hang mass.

Measure new spring length, L'. E.g. L' = 0.70m.

Calculate extension: e = L' - L = 0.70 – 0.60 = 0.10m

Use mg = ke (in equilibrium weight = tension)

k = mg/e

Don't know what value you are using for example.  Suppose it is 10N/kg (same thing as 10m/s²).

k = 0.5*10/0.10 = 50 N/m

Repeat for a few different masses.  (L always stays the same.)

Take the average of your k values.

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