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3 years ago

Need help with #8 plssssssssssssss

Mathematics

1 answer:

Aleksandr-060686 [28]3 years ago

3 0

I'm sorry, I don't see the question. Could you please retake the photo in better quality so I could answer it? Thanks!

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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

Estimate the Circumference of this circle.<br> 10m

Rina8888 [55]

Answer:

62.8

Step-by-step explanation:

10*2=20

20*3.14= 62.8

HOPE IT HELPS

7 0

2 years ago

Solve the following linear equation for k.<br> 2 (k-5) + 3k = k + 6

kiruha [24]

Answer:

2(k-5)+3k=k+6

-3k. -3k

2(k-5)=-2k+6

/2. /2

k-5=-k+3

+k. +k

2k-5=3

+5. +5

2k=8

/2. /2

k=4

7 0

2 years ago

Lines a and b are parallel Line cis perpendicular to both line a and line b. Which

prohojiy [21]

Answer:

see below

Step-by-step explanation:

The slopes of parallel lines are the same. The slopes of perpendicular lines are negative reciprocals of each other, hence their product is -1.

___

For the most part, the concept of adding slopes of lines does not relate to parallel or perpendicular lines in any way.

6 0

2 years ago

Read 2 more answers

What equation represents a line that when if graphed, passes through the points (0, -1) and

Keith_Richards [23]

Y= 2x-1
Explanation: First, you find slope. If the first point it x-axis point is 0 and the second x-axis point is 1 then you already know the answer is over 1, so you have to figure out how many up you are going on the y-axis from -1 to 1. Which is 2, so your slope is 2/1 or just 2. Next, you find your y-intercept, which is where the line crosses the y-axis, and the easy way to find this is finding what y is equal to when x is equal to 0, in the problem you are told (0, -1) as a point, so they gave you your y-intercept right there, which is -1. Finally you right the equation in slip intercept form (y= mx+b) which is y= 2x-1

5 0

3 years ago