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Anna71 [15]
2 years ago
8

In an experiment in which molten naphthalene is allowed to cool, the cooling curve shown below was obtained, the temperature 80∘

C is known as
A. cooling temperature

B. boiling point

C. melting point

D. vaporization point

Physics
1 answer:
notka56 [123]2 years ago
7 0

Answer:

pa heart follow back follow pa brainliest narin po if my answer is correct for you THANK

Explanation:

In an experiment in which molten naphthalene is allowed to cool, the cooling cur...

Question

In an experiment in which molten naphthalene is allowed to cool, the cooling curve in fig 5 was obtained, the temperature

80

∘

<h2>C.</h2>

is known as

Options

A) cooling temperature

B) boiling point

C) melting point √√√√√√√√√√※※※ ↑↑ANSWER

D) vaporization point

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A submarine emerges 1/9 of its volume when it partially floats on the sea surface. For make it completely submerge it is necessa
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4,524,660 N

Explanation:

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Are plants carnivores<br>Ture or False​
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2 years ago
A cannon, located 60.0 m from the base of a vertical 25.0-m-tall cliff, shoots a 15-kg shell at 43.0o above the horizontal towar
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Answer:

a)   v₀ = 32.64 m / s , b)  x = 59.68 m

Explanation:

a) This is a projectile launching exercise, we the distance and height of the cliff

         x = v₀ₓ t

         y = v_{oy} t - ½ g t²

We look for the components of speed with trigonometry

         sin 43 = v_{oy} / v₀

         cos 43 = v₀ₓ / v₀

         v_{oy} = v₀ sin 43

         v₀ₓ = v₀ cos 43

Let's look for time in the first equation and substitute in the second

         t = x / v₀ cos 43

         y = v₀ sin 43 (x / v₀ cos 43) - ½ g (x / v₀ cos 43)²

          y = x tan 43 - ½ g x² / v₀² cos² 43

          1 / v₀² = (x tan 43 - y) 2 cos² 43 / g x²

           v₀² = g x² / [(x tan 43 –y) 2 cos² 43]

Let's calculate

          v₀² = 9.8 60 2 / [(60 tan 43 - 25) 2 cos 43]

          v₀ = √ (35280 / 33.11)

          v₀ = 32.64 m / s

.b) we use the vertical distance equation with the speed found

         y = v_{oy} t - ½ g t²

         .y = v₀ sin43 t - ½ g t²

        25 = 32.64 sin 43 t - ½ 9.8 t²

        4.9 t² - 22.26 t + 25 = 0

         t² - 4.54 t + 5.10 = 0

We solve the second degree equation

         t = (4.54 ±√(4.54 2 - 4 5.1)) / 2

         t = (4.54 ± 0.46) / 2

         t₁ = 2.50 s

         t₂ = 2.04 s

The shortest time is when the cliff passes and the longest when it reaches the floor, with this time we look for the horizontal distance traveled

         x = v₀ₓ t

         x = v₀ cos 43 t

         x = 32.64 cos 43  2.50

         x = 59.68 m

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Real-world examples of power
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