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DaniilM [7]
2 years ago
11

If a force F⃗ acts on an object as that object moves through a displacement s⃗ , the work done by that force equals the scalar p

roduct of F⃗ and s⃗ : W=F⃗ ⋅s⃗ . A certain object moves through displacement s⃗ =(4.00m)i^+(5.00m)j^. As it moves it is acted on by force F⃗ , which has x-component Fx = -12.0 N (1 N = 1 newton is the SI unit of force). The work done by this force is 26.0 N⋅m = 26.0 J (1 J = 1 joule = 1 newton-meter is the SI unit of work). Find the y-component of F⃗ .
Mathematics
2 answers:
Norma-Jean [14]2 years ago
7 0

Answer:

14.8 N

Step-by-step explanation:

Oliga [24]2 years ago
4 0

Answer:

Fy=14.8N

Step-by-step explanation:

the dot product between \vec{F} \\ and \vec{S}

is given by:

W=\vec{F}\cdot\vec{S}= S_xF_x+SyFy

this gives the equation :

26= (4.00)(-12.0)+(5.00)Fy

26= -48+(5.00)Fy

adding 48 to both sides of the equation:

74= (5.00)Fy

dividing by 5, we get

Fy=74/5.00=14.8

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Quincy feeds his dog 2.75 cups food each day each cup of dog food costs $1.25. He multiplies the numbers together to determine h
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2times1 because 2.75 take the 75 off and 1.25 take the 25 off so you multiply 2 and 1 so 2times1 equals 2

6 0
2 years ago
Tina has 60 grams of popcorn. She wants to give all of her popcorn to her 4 friends. She says there are 2 ways that she can give
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Answer:

you can give 15 grams to each friend or you can give 30 grams to 2 friend and share with each other.

Step-by-step explanation:

6 0
2 years ago
Power series of y''+x^2y'-xy=0
Ray Of Light [21]
Assuming we're looking for a power series solution centered around x=0, take

y=\displaystyle\sum_{n\ge0}a_nx^n
y'=\displaystyle\sum_{n\ge1}na_nx^{n-1}
y''=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}

Substituting into the ODE yields

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}+\sum_{n\ge1}na_nx^{n+1}-\sum_{n\ge0}a_nx^{n+1}=0

The first series starts with a constant term; the second series starts at x^2; the last starts at x^1. So, extract the first two terms from the first series, and the first term from the last series so that each new series starts with a x^2 term. We have

\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}=2a_2+6a_3x+\sum_{n\ge4}n(n-1)a_nx^{n-2}

\displaystyle\sum_{n\ge0}a_nx^{n+1}=a_0x+\sum_{n\ge1}a_nx^{n+1}

Re-index the first sum to have it start at n=1 (to match the the other two sums):

\displaystyle\sum_{n\ge4}n(n-1)a_nx^{n-2}=\sum_{n\ge1}(n+3)(n+2)a_{n+3}x^{n+1}

So now the ODE is

\displaystyle\left(2a_2+6a_3x+\sum_{n\ge1}(n+3)(n+2)a_{n+3}x^{n+1}\right)+\sum_{n\ge1}na_nx^{n+1}-\left(a_0x+\sum_{n\ge1}a_nx^{n+1}\right)=0

Consolidate into one series starting n=1:

\displaystyle2a_2+(6a_3-a_0)x+\sum_{n\ge1}\bigg[(n+3)(n+2)a_{n+3}+(n-1)a_n\bigg]x^{n+1}=0

Suppose we're given initial conditions y(0)=a_0 and y'(0)=a_1 (which follow from setting x=0 in the power series representations for y and y', respectively). From the above equation it follows that

\begin{cases}2a_2=0\\6a_3-a_0=0\\(n+3)(n+2)a_{n+3}+(n-1)a_n=0&\text{for }n\ge2\end{cases}

Let's first consider what happens when n=3k-2, i.e. n\in\{1,4,7,10,\ldots\}. The recurrence relation tells us that

a_4=-\dfrac{1-1}{(1+3)(1+2)}a_1=0\implies a_7=0\implies a_{10}=0

and so on, so that a_{3k-2}=0 except for when k=1.

Now let's consider n=3k-1, or n\in\{2,5,8,11,\ldots\}. We know that a_2=0, and from the recurrence it follows that a_{3k-1}=0 for all k.

Finally, take n=3k, or n\in\{0,3,6,9,\ldots\}. We have a solution for a_3 in terms of a_0, so the next few terms (k=2,3,4) according to the recurrence would be

a_6=-\dfrac2{6\cdot5}a_3=-\dfrac2{6\cdot5\cdot3\cdot2}a_0=-\dfrac{a_0}{6\cdot3\cdot5}
a_9=-\dfrac5{9\cdot8}a_6=\dfrac{a_0}{9\cdot6\cdot3\cdot8}
a_{12}=-\dfrac8{12\cdot11}a_9=-\dfrac{a_0}{12\cdot9\cdot6\cdot3\cdot11}

and so on. The reordering of the product in the denominator is intentionally done to make the pattern clearer. We can surmise the general pattern for n=3k as

a_{3k}=\dfrac{(-1)^{k+1}a_0}{(3k\cdot(3k-3)\cdot(3k-2)\cdot\cdots\cdot6\cdot3\cdot(3k-1)}
a_{3k}=\dfrac{(-1)^{k+1}a_0}{3^k(k\cdot(k-1)\cdot\cdots\cdot2\cdot1)\cdot(3k-1)}
a_{3k}=\dfrac{(-1)^{k+1}a_0}{3^kk!(3k-1)}

So the series solution to the ODE is given by

y=\displaystyle\sum_{n\ge0}a_nx^n
y=a_1x+\displaystyle\sum_{k\ge0}\frac{(-1)^{k+1}a_0}{3^kk!(3k-1)}

Attached is a plot of a numerical solution (blue) to the ODE with initial conditions sampled at a_0=y(0)=1 and a_1=y'(0)=2 overlaid with the series solution (orange) with n=3 and n=6. (Note the rapid convergence.)

7 0
2 years ago
What is sin 45 expressed as a square root?
V125BC [204]
Sin(45) is equal to \frac{ \sqrt{2} }{2} in degrees.
8 0
2 years ago
Read 2 more answers
On average, Betsy read 1 page of her book every 1.5 minutes. Her book has 116 pages. Raymond starts a 94- page book on Saturday
natka813 [3]

Answer:

14 minutes

Step-by-step explanation:

given,

given,

Betsy time taken to read one page = 1.5 minutes

Number of pages in the Betsy book = 116 pages

number of pages in Raymond Book = 94 pages

time taken by the Raymond  to read = 11.38 - 8.30

                                                             = 3.08 hrs

time taken by Raymond = 188 minutes

time taken by Betsy to read book = 1.5 x 116

                                                         = 174 minutes

Raymond take more time than Betsy by  = 188 - 174

                                                                   = 14 minutes

Raymond take 14 minute more than Betsy

4 0
2 years ago
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