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2 years ago

9. How does the length of the hypotenuse in a right triangle relate to the lengths of the legs? (2 points)

2 answers:

6 0

<span>The pythagorean theorem addresses the length of the hypotenuse in relation to the length of the legs. The square root of the length of the hypotenuse is equal to the sum of one leg squared plus the other leg squared. In other words, A squared plus B squared equals C squared where A and B are the lengths of the legs of the triangle and C is the length of the hypotenuse.</span>

seropon [69]2 years ago

4 0

Answer:

hypotenuse² = base² +height²

Explanation:

Let us suppose a right angled triangle having b as base, h as hypotenuse and p as perpendicular distance.

The relation between the length of legs, the base and the hypotenuse is called Pythagoras theorem. It is given by :

$h^2=b^2+p^2$

hypotenuse² = base² +height²

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A ball of plasticine is released from rest at height of 2.2 m above the ground. After touching the ground, the plasticine ball c

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The magnitude of the acceleration of the ball while coming to rest is 477.43 m/s²

The direction of the acceleration of the ball is downwards

The given parameters

initial velocity of the ball, u = 0

height above the ground, h = 2.2 m

time of motion of the ball, t = 96 ms = 0.096 s

The magnitude of the acceleration of the ball while coming to rest is calculated as;

let the downwards direction of the acceleration be positive

$h = ut + 0.5 at^2\\\\h = 0 + 0.5at^2\\\\h = 0.5 at^2\\\\a = \frac{h}{0.5t^2} \\\\a = \frac{2.2}{0.5 \times 0.096^2} \\\\a = 477.43 \ m/s^2$

The direction of the acceleration of the ball is downwards

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2 years ago

When monochromatic light shines perpendicularly on a soap film (n = 1.33) with air on each side, the second smallest nonzero fil

Anika [276]

Let us start from considering monochromatic light as an incidence on the film of a thickness t whose material has an index of refraction n determined by their respective properties.

From this point of view part of the light will be reflated and the other will be transmitted to the thin film. That additional distance traveled by the ray that was reflected from the bottom will be twice the thickness of the thin film at the point where the light strikes. Therefore, this relation of phase differences and additional distance can be expressed mathematically as

$2t + \frac{1}{2} \lambda_{film} = (m+\frac{1}{2})\lambda_{film}$

We are given the second smallest nonzero thickness at which destructive interference occurs.

This corresponds to, m = 2, therefore

$2t = 2\lambda_{film}$

$t = \lambda_{film}$

The index of refraction of soap is given, then

$\lambda_{film} = \frac{\lambda_{vacuum}}{n}$

Combining the results of all steps we get

$t = \frac{\lambda_{vacuum}}{n}$

Rearranging, we find

$\lambda_{vacuum} = tn$

$\lambda_{vacuum} = (278)(1.33)$

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2 years ago

Explain how smoking damages the respiratory system and name two conditions that can develop when a person smokes. ribe the locat

kipiarov [429]

Answer

Explanation:

Smoking can cause lung disease by damaging your airways and the small air sacs (alveoli) found in your lungs. Lung diseases caused by smoking include COPD, which includes emphysema and chronic bronchitis. Cigarette smoking causes most cases of lung cancer.

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2 years ago

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other than serving as a conduction pathway what is a major function of the pons why is the medulla the most vital part of the br

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1. What is the major function of the pons?
-Other than serving as a conduction pathway, the pons also contains a nucleus that is essential for breathing.
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3 years ago

Calculate the air pressure in the pressurized tank, if h1 = 0. 18 m, h2 = 0. 2m and h3 = 0. 25m. The density of the mercury, wat

fomenos

The air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625 N/m².

<h3 /><h3>What is pressure?</h3>

The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure.

It is denoted by P. The pressure relative to the ambient pressure is known as gauge pressure.

Pressure is found as the product of the density,acceleraton due to gravity and the height.

P₁=ρ₁gh₁

P₁=13,600 kg/m³×9.81 (m/s²)×0.18 m

P₁=24014.88 N/m²

P₂=ρ₂gh₂

P₂= 1000 kg/m³×9.81 (m/s²)×00.2 m

P₂=196.2 N/m²

P₃=ρ₃gh₃

P₃=850 kg/m³×9.81 (m/s²)×0.25

P₃=2084.625 N/m²

Hence,the air pressure in the pressurized tank will be 24014.88 N/m²,196.2 N/m²,2084.625 N/m².

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