All categories

2 years ago

How much work is done if an object travels 55m due to a force of 20N?

Mathematics

1 answer:

bulgar [2K]2 years ago

7 0

Force=20N
Displacement=55m

$\\ \rm\longmapsto Work\:Done=Force\times displacement$

$\\ \rm\longmapsto Work\:Done=20(55)$

$\\ \rm\longmapsto Work\:Done=1100J$

You might be interested in

#1 Identify the characteristics of the quadratic graph

shusha [124]

Vertex: (1,-3)
Y intercept: -3
Axis of symmetry x=1
Domain all real numbers
I only got these the rest I don’t know because I just started learning it but these are correct.

8 0

2 years ago

The equation x2 − 6x − 27 = 0 when solved is:

Amiraneli [1.4K]

Answer:

-3 , 9

Step-by-step explanation:

Sum = - 6

Product = -27

Factors = 3, -9

x² - 6x-27 = 0

x² + 3x - 9x - 9*3 = 0

x(x + 3) - 9(x + 3) = 0

(x + 3) (x - 9) = 0

x +3 = 0 ; x - 9 = 0

x = - 3 ; x = 9

Solution: x = -3 , 9

8 0

3 years ago

How many kilograms of tin will be required to make 250 kg of an alloy, if the alloy must contain 81% copper, 14% tin and 5% lead

uysha [10]

Answer:

Step-by-step explanation:

all are obeying a constant ratio

copper 81
tin 14⇒?
lead 5
total 100 ⇒250
?=250*14/100=35

5 0

2 years ago

Read 2 more answers

y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence

sukhopar [10]

Let

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots$

Differentiating twice gives

$\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots$

$\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n$

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

$\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0$

$\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0$

Then the coefficients in the power series solution are governed by the recurrence relation,

$\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}$

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

$k=0 \implies n=0 \implies a_0 = a_0$

$k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}$

$k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}$

$k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}$

It should be easy enough to see that

$a_{n=2k} = \dfrac{a_0}{(2k)!}$

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

$k = 0 \implies n=1 \implies a_1 = a_1$

$k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}$

$k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}$

$k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}$

so that

$a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}$

So, the overall series solution is

$\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)$

$\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}$

4 0

2 years ago

What websites are similar to brainly?

Maslowich

Websites similar to Brainly: Quizlet and Socratic

Can I pls have brainliest :(

7 0

2 years ago