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3 years ago

6

The local high school is installing new bleachers at the stadium and must also add handrails to meet code. The students know the

bleachers are 8 m tall, and they measure the depth of the bleachers at 7 m. How long must the handrails be to go along the bleachers from bottom to top? Use a component table to solve.

1 answer:

Advocard [28]3 years ago

5 0

Answer:

The handrails must be approximately 10.63 meters long

Explanation:

The given parameters are;

The height of the bleachers, h = 8 m

The depth of the bleachers, d = 7 m

The length of the hand rails to go along the bleachers from bottom to top is given by Pythagoras' Theorem as follows;

The length of the hand rail = √(d² + h²)

∴ The length of the hand rail = √(7² + 8²) = √113 ≈ 10.63

In order for the handrails to go along the bleachers from top to bottom, they must be approximately 10.63 meters long.

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0.12m/s

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As the wavelength of A wave in a uniform medium increases, its speed will what?

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The speed of a wave in a uniform medium doesn't depend on its wavelength.

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What is the ideal banking angle for a gentle turn of 1.20-km radius on a highway with a 105 km/h speed limit (about 65 mi/h), as

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4.14°

Explanation:

given:

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1) <em>convert your given </em>

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b) v = 105 km/h to m/s = 29.2 m/s

2) <em>plug into your ideal banking angle equation</em>

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A spaceship is traveling through deep space towards a space station and needs to make a course correction to go around a nebula.

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Answer:

Magnitude = 3.64 × $10^6$

စ = 43.9°

Explanation:

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solution

we will consider here

Px = 1.7 × $10^6$

Py = 0

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tan စ = Hy ÷ Hx ......................3

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3 years ago

Tubby and his twin brother Libby have a combined mass of 200 kg and are zooming along in a 100 kg amusement park bumper car at 1

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$p_{i}=p_{f}$ (1)

Being:

$p_{i}=m_{1}V_{i} + m_{2}U_{i}$

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Where:

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$U_{i}=0 m/s$ is the velocity of Flubby with the car before the collision

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$U_{f}$ is the velocity of Flubby with the car after the collision

So, we have the following:

$m_{1}V_{i} + m_{2}U_{i}=m_{1}V_{f} + m_{2}U_{f}$ (2)

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$U_{f}=\frac{300 kg(10 m/s-4.12 m/s)}{125 kg}$ (4)

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$U_{f}=14.1 m/s$

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3 years ago