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lisov135 [29]
3 years ago
11

Suppose Sammy Sosa hits a home run which travels 361. ft (110. m). Leaving the bat at 50 degrees above the horizontal, how high

did it travel?

Physics
1 answer:
Mashcka [7]3 years ago
5 0

Answer:

The horizontal distance of Sosa is 276.526 ft or 84.28 meter.

Explanation:

As shown in the figure, let point O is the starting point of Sosa. She travels 361 ft at an angle 50 degree with the horizontal.

sin 50 = \frac{OM}{OP}

0.7660 = h / 361

h = 276.526 ft

h = 84.28 meter

The horizontal distance of Sosa is 276.526 ft or 84.28 meter.

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When using a different calorimeter, and mixing 50 ml of hot water at 65 degrees c with 60 ml of water in the calorimete
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The specific heat capacity of the calorimeter used in mixing the water is determined as 21.87 J/g⁰C.

<h3>Conseervation of energy</h3>

The heat capacity of the calirometer is determined by applying the principle of conservation of energy.

Heat lost by the hot water = Heat gained by the calirometer

Q _w = Q_c\\\\M_w C_w\Delta \theta _w = M_c C_c\Delta \theta _c

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  • M is mass

mass = density x volume = ρV

Density of water = 1 g/ml

Mass of hot water = 1 x (50) = 50 g

Mass of water in calorimeter = 1 x (60) = 60 g

<h3>Equilibrium temperature</h3>

\Delta T_c = 5.5\\\\T - 25 = 5.5\\\\T = 30.5 \ ^0C

<h3>Specific heat capacity of the calirometer</h3>

50 \times 4.184 \times (65 - 30.5) = 60 \times C_c \times (30.5 - 25)\\\\7217.4 = 330C_c\\\\C_c = \frac{7217.4}{330} \\\\C_c = 21.87 \ J/g^0C

Thus, the specific heat capacity of the calorimeter used in mixing the water is determined as 21.87 J/g⁰C.

The complete question is below

When using a different calorimeter, and mixing 50 ml of hot water at

65 degrees C with 60 ml of water in the calorimeter at 25 degrees C, the temperature of the calorimeter increased by 5.5 degrees C.

a. Calculate the heat capacity of this calorimeter?

Learn more about heat capacity here: brainly.com/question/16559442

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