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makkiz [27]
3 years ago
5

A stone was dropped off a cliff and hit the ground with a speed of 96 ft/s. What is the height of the cliff? (Use 32 ft/s2 for t

he acceleration due to gravity.)
Physics
1 answer:
nadezda [96]3 years ago
4 0
The initial velocity of the stone is 0 ft/s. Given the initial velocity (Vi), final velocity (Vf), and acceleration due to gravity (g), the distance may be calculated through the equation,
                                     d = ((Vf)² - (Vi)²) / 2g
Substituting the known values,
                                     d = ((96 ft/s)² - 0))/ (2x32.2)
The value of d is 143.10 ft. 
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1 year ago
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3 years ago
The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actu
Juliette [100K]

\orange{\underline{\huge{\bold{\textit{\green{\bf{QUESTION}}}}}}}

The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actual depth of the liquid in the tank?

{\bold{\blue{GIVEN}}}

REFRACTIVE INDEX = 1.3

APPARENT DEPTH = 7.7 cm

{ \bold{\green{To  \: Find}}}

REAL DEPTH OF THE OBJECT.

{\red{FORMULA   \:  \: USED }}

Reflective  \:  Index  =  \frac{Real \:  Depth}{Apparent  \: Depth }

\huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}

Refractive Index = 1.3

Apparent Depth = 7.7 cm

Putting the values in the formula:-

Reflective  \:  Index  =  \frac{Real \:  Depth}{Apparent  \: Depth }  \\  \\ 1.3 =  \frac{Real \:  Depth}{7.7 \: cm}  \\ \\  1.3 \times 7.7 = Real \:  Depth \\  \\ 10.01 \:  \: cm = Real \:  Depth

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2 years ago
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