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oksian1 [2.3K]
3 years ago
12

A car drives around a horizontal, circular track at constant speed. Consider the following three forces that act on the car: (1)

The upward normal force exerted on the car by the road, (2) the downward gravitational force on the car, (3) and the frictional force that is directed toward the center of the circular path.
Which of these forces does zero work on the car as the car moves along the circular path?  1, 2, and 3   3   1 and 2 
 1   2   1 and 3 
 2 and 3 
Physics
1 answer:
musickatia [10]3 years ago
6 0

Answer: 1, 2 and 3.

Explanation:

By definition, work, is the process through which a force, applied on an object, produces a displacement of this object, and can be expressed as the dot product of the force vector and the displacement vector.

It can be also understood as the projection of the force in the direction of the displacement, so, if both vectors are perpendicular, as the projection of one vector along the other is nul, if a force and the displacement are perpendicular, no work is done.

In our case, the 3 forces mentioned, are perpendicular respect the displacement.

For normal force and gravity, as they act vertically, and the car moves along an horizontal trajectory, both are perpendicular, so no work is done.

For the friction force, as it is the only horizontal force present, it is just the centripetal force that keeps the car moving in a circular path.

It always aims to the center of the circle, along the radius at any point, so it is always perpendicular to the displacement also.

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Time for the change = 4.0sec

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3 years ago
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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
Paha777 [63]

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

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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