Ask question

Ask question

All categories

4 years ago

11

The weight that a horizontal beam can support varies inversely as the length of the beam. Suppose that a 2-m beam can support 31

0 kg. How many kilograms can a 10-m beam support?

2 answers:

iren [92.7K]4 years ago

8 0

W = kL where W = weight supported, L = length of the beam and k is constant of variation.

Substituting the given values:-

310 = 2k
k = 155

so equation of variation is W = 155L

For length 10 m the weight it can support is

W = 155*10 = 1550 kg answer

maria [59]4 years ago

4 0

Answer:

62 kg.

Step-by-step explanation:

Let w be the weight of beam and l be the length of beam.

We have been given that the weight that a horizontal beam can support varies inversely as the length of the beam.

We know that inversely proportion quantities are in form: $y=\frac{k}{x}$ , where k is constant of proportionality.

$w=\frac{k}{l}$

First of all, we will find the value of k using the $w=310$ and $l=2$ in inversely proportion equation as:

$310=\frac{k}{2}$

$310*2=\frac{k}{2}*2$

$620=k$

Now we will substitute $620=k$ and $l=10$ , in our given equation as:

$w=\frac{620}{10}$

$w=62$

Therefore, a 10-m beam can support 62 kilograms.

You might be interested in

A scientist has discovered an organism that replicates at a rate of five per hour and lives for one week. Each replicated organi

Assoli18 [71]

At hour one, there is 1 organism. At hour two, there is five more organism. At hour three. there will be 6 x 5 = 30 more organisms, etc

This is a geometric sequence with a = 1 (i.e. number of organisms at hour one), r = 5 (number of reprications per hour).

The total number of organisms at hour seven is the sum of the first seven terms of the sequence given by a(r^n - 1)/(r - 1) = 1(5^7 - 1)/(5 - 1) = (78125 - 1)/4 = 78124/4 = 19,531

4 0

3 years ago

Sarah sees an antelope and a giraffe on a nature show. The antelope is 3 feet tall. The giraffe is 18 feet tall. How many times

Rashid [163]

Answer:

6

6×3=18

This is the answer

4 0

3 years ago

Read 2 more answers

Find the area of the triangle<br> A)<br> B)<br> 09<br> 11 ft2<br> D)<br> 224²

Romashka-Z-Leto [24]

Answer:

A

Step-by-step explanation:

Area of a triangle = $\frac{bh}{2}$

where b = base length and h = height

the triangle shown has a base length of 2 and 1/2 and a height 4. and 2/5

So area =

$\frac{2\frac{1}{2} *4\frac{2}{5} }{2} \\\\2\frac{1}{2} *4\frac{2}{5} =11\\\\\frac{11}{2} =5.5$

5.5 also = $5\frac{1}{2}$

Hence, the correct answer is A

7 0

3 years ago

4. Lisa walked 48 blocks in 3 hours.<br> blocks per hour

dlinn [17]

Answer:

The number of blocks in one hour is the same that the unit rate, so 16 blocks per hour

Step-by-step explanation:

we know that

To find out the unit rate, divide the total blocks by the total time

so

$\frac{48}{3}\ \frac{blocks}{hours}=16\ \frac{blocks}{hour}$

3 0

3 years ago

Solve the given initial-value problem. x' = 1 2 0 1 − 1 2 x, x(0) = 2 7

Ilia_Sergeevich [38]

I'll go out on a limb and guess the system is

$\mathbf x'=\begin{bmatrix}\frac12&0\\1&-\frac12\end{bmatrix}\mathbf x$

with initial condition $\mathbf x(0)=\begin{bmatrix}2&7\end{bmatrix}^\top$ . The coefficient matrix has eigenvalues $\lambda$ such that

$\begin{vmatrix}\frac12-\lambda&0\\1&-\frac12-\lambda\end{vmatrix}=\lambda^2-\dfrac14=0\implies\lambda=\pm\dfrac12$

The corresponding eigenvectors $\eta$ are such that

$\lambda=\dfrac12\implies\begin{bmatrix}\frac12-\frac12&0\\1&-\frac12-\frac12\end{bmatrix}\eta=\begin{bmatrix}0&0\\1&-1\end{bmatrix}\eta=\begin{bmatrix}0\\0\end{bmatrix}$
$\implies\eta=\begin{bmatrix}1\\1\end{bmatrix}$

$\lambda=-\dfrac12\implies\begin{bmatrix}\frac12+\frac12&0\\1&-\frac12+\frac12\end{bmatrix}\eta=\begin{bmatrix}1&0\\1&0\end{bmatrix}\eta=\begin{bmatrix}0\\0\end{bmatrix}$
$\implies\eta=\begin{bmatrix}0\\1\end{bmatrix}$

So the characteristic solution to the ODE system is

$\mathbf x(t)=C_1\begin{bmatrix}1\\1\end{bmatrix}e^{t/2}+C_2\begin{bmatrix}0\\1\end{bmatrix}e^{-t/2}$

When $t=0$ , we have

$\begin{bmatrix}2\\7\end{bmatrix}=C_1\begin{bmatrix}1\\1\end{bmatrix}+C_2\begin{bmatrix}0\\1\end{bmatrix}=\begin{bmatrix}C_1\\C_1+C_2\end{bmatrix}$

from which it follows that $C_1=2$ and $C_2=5$ , making the particular solution to the IVP

$\mathbf x(t)=2\begin{bmatrix}1\\1\end{bmatrix}e^{t/2}+5\begin{bmatrix}0\\1\end{bmatrix}e^{-t/2}$

$\mathbf x(t)=\begin{bmatrix}2e^{t/2}\\2e^{t/2}+5e^{-t/2}\end{bmatrix}$

5 0

4 years ago