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zvonat [6]
3 years ago
11

At point A the roller coaster car has 480 Joules of gravitational potential energy. At point C it only has 290 Joules of gravita

tional potential energy. How much kinetic energy does the car have at point C?
Select one:

480 Joules

0 Joules

190 Joules

290 Joules
Physics
1 answer:
ruslelena [56]3 years ago
6 0

This problem is providing the gravitational potential energy at two points, A and C, 480 J and 290 J respectively, so that the kinetic energy at point C is required. According to the law of conservation of mass; which turns out to be 190 J considering to the following:

<h3>Law of conservation of energy:</h3>

When objects are in both vertical and horizontal motion or rest, they are able to carry specific amounts of energy, which are potential and kinetic, respectively; the former depending on the mass and gravity and the latter on the mass and velocity.

Thus, when a roller coaster is quiet at a specific point with no vertical movement, we say it has its maximum energy, and as it starts moving, the addition of the kinetic and potential energies must equal to the maximum one because it is not supposed to be lost, but conserved.

In such a way, when the roller coaster moves to a lower position, it decreases it potential energy and increases the kinetic one, according to:

E_{max}=U+K

Whereas U stands for the potential energy and K for the kinetic one. Hence, since the maximum energy is 480 J and the potential energy at point C is 290 J, the kinetic energy must be:

E_{max}=U+K\\\\K=480J-290J\\\\K=190 J

Learn more about the law of conservation of energy: brainly.com/question/21288807

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Answer:

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p_2 = Neutrino has a momentum = 7.3\times 10^{-24}\ kg m/s

M = total mass = 2.34\times 10^{-26}\ kg

In the x axis as the momentum is conserved

Mv_x=m_1v_1\\\Rightarrow v_x=\dfrac{m_1v_1}{M}\\\Rightarrow v_x=\dfrac{9.11\times 10^{−31}\times 5.7\times 10^7}{2.34\times 10^{-26}}\\\Rightarrow v_x=2219.10256\ m/s

In the y axis

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The resultant velocity is

R=\sqrt{v_x^2+v_y^2}\\\Rightarrow R=\sqrt{2219.10256^2+311.96581^2}\\\Rightarrow R=2240.92365\ m/s

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4 years ago
Inside a 30.2 cm internal diameter stainless steel pan on a gas stove water is being boiled at 1 atm pressure. If the water leve
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Answer:

Q = 20.22 x 10³ W = 20.22 KW

Explanation:

First we need to find the volume of water dropped.

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r = radius of pan = 30.2 cm/2 = 15.1 cm = 0.151 m

h = height drop = 1.45 cm = 0.0145 m

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V = π(0.151 m)²(0.0145 m)

V = 1.038 x 10⁻³ m³

Now, we find the mass of the water that is vaporized.

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where,

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Therefore,

m = (1000 kg/m³)(1.038 x 10⁻³ m³)

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Now, we calculate the heat required to vaporize this amount of water.

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Now, for the rate of heat transfer:

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