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Finding the work done in stretching or compressing a spring. Hooke's Law for Springs.
According to Hooke's law, the force required to compress or stretch a spring from an equilibrium position is given by F(x)=kx, for some constant k. The value of k (measured in force units per unit length) depends on the physical characteristics of the spring. The constant k is called the spring constant and is always positive.

In this problem we assume that the force applied doesn't distort the metal in the spring.

A 2 m spring requires 11 J to stretch to 2.4 m. Find the force function, F(x), for the spring described.

Answers

Answer 1

Answer:

Answer:

Check the explanation

Step-by-step explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

Answer 2

Answer:

Final answer:

To find the force function of a spring using Hooke's Law, you first identify the spring constant 'k' using the given work done and extension. In this case, we found 'k' to be 137.5 N/m. Hence, the force function F(x) for the spring comes out to be 137.5x N.

Explanation:

The problem revolves around Hooke's Law, which is used to determine the force needed to stretch or compress a spring by a certain distance away from its equilibrium position. This law can be mathematically represented as F(x)=kx, where 'F(x)' represents the force applied, 'k' is the spring constant, and 'x' is the distance.

In this question, the work done (W) to stretch the spring is given as 11 J, and the extension (Δx) is 0.4 m (from 2 m to 2.4 m). The work done on a spring is calculated by the equation W = 1/2 * k * (Δx)^2. From this, you can solve for 'k' value. Once you have 'k', you can find the force function F(x) for the spring.

1. Calculate 'k' using the work done equation:

11 J = 1/2 * k * (0.4 m)^2 ➔ k = 137.5 N/m

2. Substitute 'k' in F(x):

F(x) = 137.5 N/m * x

Hence, the force function F(x) = 137.5x N is required to extend the spring by 'x' metres.

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As sample size increases, which of the following is true for a t-distribution-Distribution will get taller and SD will increase
-distribution sill get taller and SD will decrease
-distribution will get shorter and SD will decrease
Distribution will get shorter and SD will increase

Answers

Answer:

Distribution will get taller and SD will decrease.

Step-by-step explanation:

Sample Size and Standard Deviation:

In a t-distribution, sample size and standard deviation are inversely related.

A larger sample size results in decreased standard deviation and a smaller sample size will result in increased standard deviation.

Sample Size and Shape of t-distribution:

As we increase the sample size, the corresponding degree of freedom increases which causes the t-distribution to like normal distribution. With a considerably larger sample size, the t-distribution and normal distribution are almost identical.

Degree of freedom = n - 1

Where n is the sample size.

The shape of the t-distribution becomes more taller and less spread out as the sample size is increased

Refer to the attached graphs, where the shape of a t-distribution is shown with respect to degrees of freedom and also t-distribution is compared with normal distribution.

We can clearly notice that as the degree of freedom increases, the shape of the t-distribution becomes taller and narrower which means more data at the center rather than at the tails.

Also notice that as the degree of freedom increases, the shape of the t-distribution approaches normal distribution.

Final answer:

In a t-distribution, as the sample size increases, the distribution becomes 'shorter', and the standard deviation decreases following the law of large numbers. The increased sample size reduces variability and introduces less deviation from the mean.

Explanation:

As the sample size increases for a t-distribution, the distribution tends to approach a normal distribution shape, which means the distribution will get 'shorter'. Additionally, the standard deviation (SD) would generally decrease as the sample size increases. This is due to the fact that when sample size increases, a smaller variability is introduced, hence less deviation from the mean.

To illustrate, imagine rolling a dice. If you roll it a few times, you may end up with quite a bit of variation. If you roll it a hundred times, however, the numbers should average out closer to the expected value (3.5 for a six-sided dice), and the standard deviation (a measure of variability) would decrease.

In conclusion, when the sample size increases, a t-distribution will get 'shorter' and SD will decrease. This concept is often referred as the law of large numbers.

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Find the perimeter of trapezoid WXYZ with vertices W(2, 3), X(4, 6), Y(7, 6), and Z(7,3). Leave your answer in simplest radical form.

Answers

Answer:

$Perimeter = 11 + √(13)$

Step-by-step explanation:

To find the perimeter of WXYZ we need to find the length of all four sides: WX, XY, YZ and WZ.

To find the length of each side, we can use the formula for the distance of two points:

$distance = √((x_1 - x_2)^2 + (y_1 - y_2)^2)$

So we have that:

$WX = √((2 - 4)^2 + (3 - 6)^2) = √(13)$

$XY = √((4 - 7)^2 + (6 - 6)^2) = 3$

$YZ = √((7 - 7)^2 + (6 - 3)^2) = 3$

$WZ = √((2 - 7)^2 + (3 - 3)^2) = 5$

The perimeter is:

$Perimeter = WX + XY + YZ + WZ$

$Perimeter = √(13) + 3 + 3 + 5 =11 + √(13)$

NEED HELP ON THIS ASAP PLS

Answers

Answer:

65.4

for this triangle we use the SOHCAHTOA rules because it is a right angled triangle.

so the rule used here is $Tan x= (opposite)/(adjacent)$

$Tan (17)= (20)/(x)$

we cross multiply both sides

$(tan(17))/(1) =(20)/(x)$

$tan(17)x=20$

$x=(20)/(tan(17)) =65.41705237$

the question said to the nearest tenth, so 65.4 is the final answer

In this card game, a player is dealt 10 cards. A player wins $1,000,000 if he is dealt 8 hearts. Use combinatorics to set up your problem a. Calculate the probability a player is dealt exactly 8 hearts b. Calculate the probability a player is dealt exactly 8 hearts if the first tweo cards he was dealt were diamonds. 2

Answers

Assume a standard deck of 52 cards with 4 suits of 13 cards each.

a. There are $\dbinom{13}8\dbinom{39}2$ ways of being dealt a hand consisting of 8 hearts and 2 non-hearts, so the probability of being dealt such a hand is

$\frac{\dbinom{13}8\dbinom{39}2}{\dbinom{52}{10}}\approx0.0000602823$

b. This time, the non-hearts specifically belong to the suit of diamonds, for which there are $\dbinom{13}2$ ways of getting drawn, so the probability is