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2 years ago

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17. Calculate the amount of gravitational potential energy at the top of one 4 points hill. The mass of the coaster is 500 kg. T

he height from the hill to the valley is 59.6 m. Use 9.81 m/s^2 as the acceleration due to gravity. Record your calculations in joules. (Hint: Gravitational PE = mgh)
A. 292.338 m/s^2

B. 3037 J

C. 82.3 kg

D. 292,338 J

1 answer:

MissTica2 years ago

4 0

Answer: D. 292,338 J

This is the correct answer :)

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Each of the colors of light in the visible light spectrum has...

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The wavelength of a wave on a string is 1.2 meters. If the speed of the wave is 60 meters/second, what is its frequency?

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2. How have human activities and natural processes influenced the gradual or sudden change in global temperature? (You need a mi

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Answer:

Human activities and natural processes have influenced the change in the global temperature by the following processes

1) Green house gas such as carbon dioxide, methane, ozone, nitrous oxide and fluorinated gases produced by the combustion of fossil fuels the use of industrial chemicals, the production of coal, and natural gas

2) Deforestation which reduces the natural process of conversion of carbon dioxide to oxygen, thereby, increasing the greenhouse gases in the atmosphere

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Explanation:

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3 years ago

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

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Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

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3 years ago

On a calm day with no wind, you can run a 1500-m race at a velocity of 4.0 m/s. If you run the same race on a day when you have

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Answer:

The time taken to finish the race is 750 s.

Explanation:

The velocity of the person on the day of wind is slowed down by 2.0 m/s. So the person's velocity on the day of wind is 4-2=2 m/s.

The relation between time, speed and distance is t=v/d

Given d=1500 m and calculated v= 2 m/s.

t=1500/2

t=750 s.

Learn more about distance formula.

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1 year ago

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