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3 years ago

Can somone explain to me how to do this and help me thansk!!!

Mathematics

1 answer:

Gnesinka [82]3 years ago

6 0

=> x² = (2.4)² + (4.8)²

=> x² = 5.76 + 23.04

=> x² = 28.8

=> x = 5.366563146

=> x = 5.4

Ans: 2nd option (5.4)

Hope this helps!

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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

3 years ago

Several friends bought flowers to make table centerpieces. write the ratios of purple flowers to white flowers for each friend.

Zina [86]

The ratio for Rowan = 4:1.

The ratio for Marcia = 10:1.

The ratio for Jillian = 5:1.

The ratio for Lulua = 6:1.

The smallest ratio among these for purple to white flowers is for Rowan, that is, 4:1.

Ratios are fractions in the simplest form representing the relationship between two quantities.

The ratio of purple to white flowers for each friend can be shown as:-

Rowan:

The number of purple flowers = 8

The number of white flowers = 2

Thus, the fraction of purple to white flowers = 8/2 = 4/1.

Thus, the ratio for Rowan = 4:1.

Marcia:

The number of purple flowers = 10

The number of white flowers = 1

Thus, the fraction of purple to white flowers = 10/1.

Thus, the ratio for Marcia = 10:1.

Jillian:

The number of purple flowers = 10

The number of white flowers = 2

Thus, the fraction of purple to white flowers = 10/2 = 5/1.

Thus, the ratio for Jillian = 5:1.

Lulua:

The number of purple flowers = 18

The number of white flowers = 3

Thus, the fraction of purple to white flowers = 18/3 = 6/1.

Thus, the ratio for Lulua = 6:1.

The smallest ratio among these for purple to white flowers is for Rowan, that is, 4:1.

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