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3 years ago

Madison was driving at 40 mph and went 80 miles. How long did it take madison?

2 answers:

monitta3 years ago

3 0

$v=40\ mph\\\\s=80\ miles\\\\t=?\\--------------\\v=\frac{s}{t}\to vt=s\to t=\frac{s}{v}\\--------------\\t=\frac{80\ miles}{40\ mph}=2\ h\leftarrow Answer$

elena-s [515]3 years ago

3 0

2 hours! This is because she's driving 40 mph and has to drive 80 miles so 80/40= 2

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3 years ago

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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<h3>Conseervation of energy</h3>

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

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

where;

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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