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gregori [183]
3 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]3 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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A tank of oxygen has a volume of 1650 L. The temperature of the gas inside is 35?C. If there are 9750 moles of oxygen in the tan
Paul [167]

Answer:

2192.64 PSI.

Explanation:

  • From the general law of ideal gases:

<em>PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the container in L (V = 1650 L).

n is the no. of moles of the gas in mol (n = 9750 mol).

R is the general gas constant (R = 0.082 L.atm/mol.K).

T is the temperature of the gas in (T = 35°C + 273 = 308 K).

∴ P = nRT/V = (9750 mol)(0.082 L.atm/mol.K)(308 K)/(1650 L) = 149.2 atm.

  • <u><em>To convert from atm to PSI:</em></u>

1 atm = 14.696 PSI.

<em>∴ P = 149.2 atm x (14.696 PSI/1.0 atm) = 2192.64 PSI.</em>

4 0
3 years ago
In the titration of 50. 0 mL of 0. 400 M HCOOH with 0. 150 M LiOH, how many mL of LiOH are required to reach the equivalence poi
mart [117]

The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL

<h3>Balanced equation </h3>

HCOOH + LiOH —> HCOOLi + H₂O

From the balanced equation above,

The mole ratio of the acid, HCOOH (nA) = 1

The mole ratio of the base, LiOH (nB) = 1

<h3>How to determine the volume of LiOH </h3>
  • Molarity of acid, HCOOH (Ma) = 0.4 M
  • Volume of acid, HCOOH (Va) = 50 mL
  • Molarity of base, LiOH (Mb) = 0.15 M
  • Volume of base, LiOH (Vb) =?

MaVa / MbVb = nA / nB

(0.4 × 50) / (0.15 × Vb) = 1

20 / (0.15 × Vb) = 1

Cross multiply

0.15 × Vb = 20

Divide both side by 0.15

Vb = 20 / 0.15

Vb = 133.3 mL

Thus, the volume of the LiOH solution needed is 133.3 mL

Learn more about titration:

brainly.com/question/14356286

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How might you explain the difference between the pH values of the 0.01 M HCl (pH = 1.1) and the 0.01 M HC2H3O2 (pH = 3.6)?
JulsSmile [24]
The difference is due to the degree of dissociation of the substances. HCl dissociates completely into ions when added to water, while this is not the case for HC₂H₃O₂; therefore, HCl has a lower pH.
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The answer is A. 6.2 calories
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What amount of energy is required to change a spherical drop of water with a diameter of 1.80 mm to three smaller spherical drop
Gekata [30.6K]
This is a straightforward question related to the surface energy of the droplet. 

<span>You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. </span>

<span>With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. </span>

<span>The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ </span>

<span>The five smaller droplets need to have the same volume as the original. Therefore </span>

<span>5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. </span>

<span>Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. </span>

<span>Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². </span>

<span>The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ </span>
<span>From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. </span>

<span>Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets. </span>
7 0
3 years ago
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