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3 years ago

Help me please ......

Mathematics

1 answer:

Kruka [31]3 years ago

8 0

Since the block is a cube, the dimentions from corner to corner equal all the same, as long as the pieces are equal, which in this case, they are. :) So basically what you need to do is find the perimeter of one side of the cube, and that's the answer.

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Ryan wanted to know the height of a cell-phone tower neighboring his property he walked 80 feet from the base of the tower and m

spin [16.1K]

Since Ryan is 5 feet tall, we will assume that the height of the cell-phone tower is x + 5.
Since the given is 54 degrees.
Tan 54degrees = x/80(feet)
x=80(tan54)
x=110
To finalize the answer, we will add up the height of the cell-phone tower to the answer.
110 + 5 = 115 feet
So the answer to the question is that the height of the cell-phone tower is 115 ft.

3 0

3 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

C). The local bus service has 2 lines of buses that start together at 8

jok3333 [9.3K]

Answer:

3 times

Step-by-step explanation:

Firstly we calculate the number of minutes between 8 and 11 am

The number of hours is 3 hours

The number of minutes is 180 minutes since 1 hour is 60 minutes

Now out of 180, to know the number of times that they both leave, we need to get the multiples of both between 0 and 180

The multiples are;

60, 120, 180

This means that they leave together 3 times

8 0

2 years ago

50 POINTS!!Different project management steps are initiating, planning, executing, monitoring and controlling, and closing a pro

garri49 [273]

Answer: Planning because you need to start somewhere if you dont then you dont have any of the other steps

Good luck lovie :)

8 0

3 years ago

An above-ground swimming pool in the shape of a cylinder has a diameter of 18 feet and a height of 4.5 feet. If the pool is fill

laila [671]

The volume of a cylinder is

   (pi) (radius²) (height) .

Radius = 1/2 diameter.
Radius of this pool = (1/2) (18 ft) = 9 ft

The pool is a cylinder with height of 4.5 feet.
The water in it is also a cylinder, but only 4 ft high.

Volume of the water =

   (pi) x (radius²) x (height)

   = (pi) x (9 ft)² x (4 ft)

   = (pi) x (81 ft²) x (4 ft)

   = (pi) x (324 ft³) = 1,017.9 ft³ .

7 0

3 years ago

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