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3 years ago

the ratio of the length of an airplane wing to its width is 6 to1 . if the length of a wing is 12.5 meters ,how wide must it be

Mathematics

1 answer:

erastova [34]3 years ago

3 0

If you set up a proportion, this will make sense.

$\frac{6}{1}= \frac{12.5}{x}$

Now, Cross Multiply

6x = 12.5

x = 25/12

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In a circus performance, a monkey is strapped to a sled and both are given an initial speed of 3.0 m/s up a 22.0° inclined track

Aloiza [94]

Answer:

Approximately $0.31\; \rm m$ , assuming that $g = 9.81\; \rm N \cdot kg^{-1}$ .

Step-by-step explanation:

Initial kinetic energy of the sled and its passenger:

$\begin{aligned}\text{KE} &= \frac{1}{2}\, m \cdot v^{2} \\ &= \frac{1}{2} \times 14\; \rm kg \times (3.0\; \rm m\cdot s^{-1})^{2} \\ &= 63\; \rm J\end{aligned}$ .

Weight of the slide:

$\begin{aligned}W &= m \cdot g \\ &= 14\; \rm kg \times 9.81\; \rm N \cdot kg^{-1} \\ &\approx 137\; \rm N\end{aligned}$ .

Normal force between the sled and the slope:

$\begin{aligned}F_{\rm N} &= W\cdot \cos(22^{\circ}) \\ &\approx 137\; \rm N \times \cos(22^{\circ}) \\ &\approx 127\; \rm N\end{aligned}$ .

Calculate the kinetic friction between the sled and the slope:

$\begin{aligned} f &= \mu_{k} \cdot F_{\rm N} \\ &\approx 0.20\times 127\; \rm N \\ &\approx 25.5\; \rm N\end{aligned}$ .

Assume that the sled and its passenger has reached a height of $h$ meters relative to the base of the slope.

Gain in gravitational potential energy:

$\begin{aligned}\text{GPE} &= m \cdot g \cdot (h\; {\rm m}) \\ &\approx 14\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times h\; {\rm m} \\ & \approx (137\, h)\; {\rm J} \end{aligned}$ .

Distance travelled along the slope:

$\begin{aligned}x &= \frac{h}{\sin(22^{\circ})} \\ &\approx \frac{h\; \rm m}{0.375}\end{aligned}$ .

The energy lost to friction (same as the opposite of the amount of work that friction did on this sled) would be:

$\begin{aligned} & - (-x)\, f \\ = \; & x \cdot f \\ \approx \; & \frac{h\; {\rm m}}{0.375}\times 25.5\; {\rm N} \\ \approx\; & (68.1\, h)\; {\rm J}\end{aligned}$ .

In other words, the sled and its passenger would have lost (approximately) $((137 + 68.1)\, h)\; {\rm J}$ of energy when it is at a height of $h\; {\rm m}$ .

The initial amount of energy that the sled and its passenger possessed was $\text{KE} = 63\; {\rm J}$ . All that much energy would have been converted when the sled is at its maximum height. Therefore, when $h\; {\rm m}$ is the maximum height of the sled, the following equation would hold.

$((137 + 68.1)\, h)\; {\rm J} = 63\; {\rm J}$ .

Solve for $h$ :

$(137 + 68.1)\, h = 63$ .

$\begin{aligned} h &= \frac{63}{137 + 68.1} \approx 0.31\; \rm m\end{aligned}$ .

Therefore, the maximum height that this sled would reach would be approximately $0.31\; \rm m$ .

7 0

2 years ago

Complete the proof.

trasher [3.6K]

You circle the points on both sides and then you calculate them and get what you need...

5 0

2 years ago

Sherman has 3 cats and 2 dogs. he wants to buy a toy for each of his pets. Sherman has 22$ to spend on pet toys. How much can he

WARRIOR [948]

Sherman wants to spend on each pet, so we are assuming he wants to spend the same amount. He has 5 pets in all (3 cats, 2 dogs)

Divide 22 with 5

22/5 = 4.4

He can spend $4.40, or in fraction form:

4 4/10 (Mixed fraction)

hope this helps

6 0

3 years ago

Hi can you guys help me with this please

Sav [38]

Answer:

Step-by-step explanation:

i think

4 0

2 years ago

Read 2 more answers

An ordinary (fair) die is a cube with the numbers 1 through 6 on the sides (represented by painted spots). Imagine that such a d

Anika [276]

Answers:

P(A) = 7/12

P(B) = 1/4

====================================================

Explanation:

Instead of having one die, let's say we have two dice. I'll make one red and the other blue.

I'll be using the dice chart shown below. The red and blue values add up to the black numbers inside the table. For instance, we have 1+1 = 2 in the upper left corner. There are 6*6 = 36 sums total.

Using that table, we can see the following:

There are 6 copies of "7"
There are 5 copies of "8"
There are 4 copies of "9"
There are 3 copies of "10"
There are 2 copies of "11"
There is 1 copy of "12"

In total, we have 6+5+4+3+2+1 = 21 instances where the two dice add to something larger than 6.

This is out of 36 ways to roll two dice.

Therefore P(A) = 21/36 = (3*7)/(3*12) = 7/12

-----------------------------------

If a number is divisible by 4, then it is a multiple of 4.

The multiples of 4 found in the table are: 4, 8, 12

We have

3 copies of "4"
5 copies of "8"
1 copy of "12"

This gives 1+5+3 = 9 values that are a multiple of 4

P(B) = 9/36 = (1*9)/(4*9) = 1/4

8 0

2 years ago