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3 years ago

Explain your answer !! Have a nice day Will give braisnlt

Mathematics

1 answer:

lora16 [44]3 years ago

3 0

Answer:

The equations are:

$4H + 3P = 13.75$

$2H + 5P = 11.25$

$Hotdog = 2.50$

$Pretzel = 1.25$

Step-by-step explanation:

Let:

$H = hot\ dog$

$P =Pretzel$

For Tina:

$4H + 3P = 13.75$

For Doug:

$2H + 5P = 11.25$

Solving for the price of both items:

We have:

$4H + 3P = 13.75$ --- (1)

$2H + 5P = 11.25$ --- (2)

Multiply (2) by 2

2 * ( $2H + 5P = 11.25$ )

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

$4H + 10P = 22.50$ --- (3)

Subtract (3) from (1)

$4H + 3P = 13.75$

$4H + 10P = 22.50$

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

$4H - 4H + 3P - 10P = 13.75 - 22.50$

$3P - 10P = 13.75 - 22.50$

$-7P = -8.75$

Solve for P

$P = \frac{-8.75}{-7}$

$P = 1.25$

Substitute 1.25 for P in (2)

$2H + 5P = 11.25$

$2H + 5 * 1.25 = 11.25$

$2H + 6.25 = 11.25$

Collect Like Terms

$2H = 11.25-6.25$

$2H = 5$

Solve for H

$H = \frac{5}{2}$

$H = 2.50$

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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.