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gregori [183]
2 years ago
11

The Hot-Wheel car was accelerating across the table at 0.25 meters per second squared (m/s^2) from an applied force of 10 N. Wha

t is the mass of the HotWheel car? *
Chemistry
1 answer:
SOVA2 [1]2 years ago
4 0

Answer:

Sorry if its wrong im very busy rn

Explanation:

Consider the motion of a Hot Wheels car beginning from rest at an elevated position. The Hot Wheels car rolls down a hill and begins its motion across a level surface. Along the level surface, the Hot Wheels car collides with a box and skids to a stop over a given distance. How could work and energy be utilized to analyze the motion of the Hot Wheels car? Would the total mechanical energy of the Hot Wheels car be altered in the process of rolling down the incline or in the process of skidding to a stop? Or would the total mechanical energy of the Hot Wheels car merely be conserved during the entire motion?

Of course the answers to these questions begin by determining whether or not external forces are doing work upon the car. If external forces do work upon the car, the total mechanical energy of the car is not conserved; the initial amount of mechanical energy is not the same as the final amount of mechanical energy. On the other hand, if external forces do not do work upon the car, then the total mechanical energy is conserved; that is, mechanical energy is merely transformed from the form of potential energy to the form of kinetic energy while the total amount of the two forms remains unchanged.

While the Hot Wheels car moves along the incline, external forces do not do work upon it. This assumes that dissipative forces such as air resistance have a negligible affect on the car's motion. This is a reasonable assumption for the low speeds of the car and its streamline characteristics. Since external forces do not do work on the car, the total mechanical energy of the car is conserved while moving along the incline. As the work-energy bar charts in the animation below depict, energy is transformed from potential energy (the stored energy of position) to kinetic energy (the energy of motion). The car gains speed as it loses height. The bar chart also depicts the fact that the total amount of mechanical energy is always the same; when the two forms are added together, the sum is unchanging.

When the Hot Wheels car collides with the box and skids to a stop, external forces do a significant amount of work upon the car. The force of friction acts in the direction opposite the car's motion and thus does negative work upon the car. This negative works contributes to a loss in mechanical energy of the car. In fact, if 0.40 Joules of mechanical energy are lost, then -0.40 Joules of work are done upon the car. As this work is done, the mechanical energy of the car (in the form of kinetic energy) is transformed into non-mechanical forms of energy such as sound and heat.

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How many moles of oxygen are necessary to react completely with four moles of propane (CH)?
steposvetlana [31]

Answer:

20 mole of oxygen

Explanation:

1 mole of proprane reacts with 5 moles of oxygen so 4 time 5 equals 20

7 0
3 years ago
You have 8 moles of a gas at 250 K in a 6 L container. What is the pressure of the gas? Show your work
Bogdan [553]

Hello:

In this case, we will use the Clapeyron equation:

P = ?

n = 8 moles

T = 250 K

R = 0.082 atm.L/mol.K

V = 6 L

Therefore:

P * V = n *  R * T

P * 6 = 8 * 0.082* 250

P* 6 = 164

P = 164 / 6

P = 27.33 atm


Hope that helps!

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Answer:

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Explanation:

8 0
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Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water
madreJ [45]

Explanation:

The given data is as follows.

         Temperature of dry bulb of air = 25^{o}C

          Dew point = 10^{o}C = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

        Partial pressure of water vapor (P_{a}) = vapor pressure of water at a given temperature (P^{s}_{a})

Using Antoine's equation we get the following.

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}

            ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}

                               = 0.17079

                   P^{s}_{a} = 1.18624 kPa

As total pressure (P_{t}) = atmospheric pressure = 760 mm Hg

                                   = 101..325 kPa

The absolute humidity of inlet air = \frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                  \frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}

                 = 0.00735 kg H_{2}O/ kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg H_{2}O/ kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

                 0.07 kg - 0.00735 kg

              = 0.06265 kg H_{2}O for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

                0.06265 kg \times 100

                  = 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

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antiseptic1488 [7]

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