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3 years ago

13

Show how he can make pieces of copper wire whose lenghts in feet are one of each of the whole numbers from 1 to 20 inclusive, wh

ile minimising the amount of wastage.

1 answer:

katovenus [111]3 years ago

3 0

Hudyydjdjgthdfhtfyhthfyjfjfyyfjfyjfyj

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Karen made 175 quarts of punch for her son and 11.3 quarts for

Alenkinab [10]

Answer:

186.3 quarts of punch

Step-by-step explanation:

175+11.3=186.3

it asks how much did she make in total so you would need to add the 2 together to get the total

6 0

3 years ago

The differnce between a number divide by 3 and 4 is 20

Talja [164]

Answer:

(n/3)/4 =20

Step-by-step explanation:

i think this is what your looking for if not im sorry

have a blessed day :)

3 0

2 years ago

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

X-3= 0 and 4x+2y-1= 0

Assoli18 [71]

X-3=0
x=3

4x+2y-1=0
4(3)+2y-1=0
12+2y-1=0
11+2y=0
2y=-11
y=-5.5

3 0

2 years ago

Read 2 more answers

The sum of the squares of two consecutive negative integers is 61. Find the smaller of the two integers

Marysya12 [62]

$x^2+(x+1)^2=61\\ x^2+x^2+2x+1-61=0\\ 2x^2+2x-60=0\ \ /:2\\ x^2+x-30=0\\ \Delta=1^2-4\cdot(-40)=1+120=121\ \ \Rightarrow\ \ \sqrt{\Delta} =11\\ \\ x_1= \frac{-1-11}{2} = \frac{-12}{2} =-6,\ \ \ \ x_2= \frac{-1+11}{2} = \frac{10}{2}=5\\ \\Ans.:x=-6$

3 0

3 years ago