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2 years ago

The height of

Mathematics

1 answer:

Amiraneli [1.4K]2 years ago

4 0

Hey there! I'm happy to help!

This glass is most likely a cylinder. To find the volume of a cylinder you take the area of the circle base and multiply it by the height.

To find the area of a circle, you square the radius and multiply it by pi (we will use 3.14 for pi).

Our radius is 3cm.

We square it.

3²=9

Multiply by 3.14.

9×3.14=28.26

Now that we have our circle base area, we multiply it by the height (7 cm).

28.26×7= 197.82

This is how many cubic centimeters the glass can hold if it is completely full. However, we want half full, so we can just divide by 2.

197.82/2=98.91

We can also multiply our circle base by 3.5, which will give us half of the full height.

So, the height of the water is 3.5 cm and the volume of the water is 98.1 cm³.

Have a wonderful day and keep on learning! :D

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Which expression is equivalent to (x Superscript 27 Baseline y) Superscript one-third?

alexgriva [62]

Answer:

x Superscript 9 Baseline (RootIndex 3 StartRoot y EndRoot)

OR x^9/(∛y)

Step-by-step explanation:

Given the indicinal equation

(x^27/y)^1/3

To find the corresponding expression, we will simplify the equation as shown:

(x^27/y)^⅓

= (x^27)^⅓/y⅓

= {x^(3×9)}^⅓/y⅓

= x^9/y⅓

= x^9/(∛y)

The right answer is x Superscript 9 Baseline (RootIndex 3 StartRoot y EndRoot)

6 0

3 years ago

Read 2 more answers

Solve for x -2x + 4 = 6x - 12

aleksklad [387]

Answer:

x=2

Step-by-step explanation:

-2x+4=6x-12 Add 2x to both sides.

+2x +2x

4=8x-12 Add 12 to both sides.

+12 +12

16=8x Divide by 8 on both sides.

/8 /8

2=x

Hope this helps you out and have a great day (～^▽^)～

5 0

3 years ago

Read 2 more answers

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.