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NISA [10]
3 years ago
15

How are energy and distance up the ramp related

Physics
1 answer:
AfilCa [17]3 years ago
8 0
I don't exactly understand the question but I'll try my best to answer it and help you. You need energy in order to go up the ramp and cover the distance. I hope this helps! :)
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The nuclear power used for electricity is produced by
svet-max [94.6K]

Answer:

<h3>b.fission. </h3>

Explanation:

<h3>Please mark my answer as a brainliest.Please follow me ❤❤❤</h3>
5 0
2 years ago
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An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a
S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, KE_{t} = \frac{3}{2}k_{b}T

k_{b} = 1.38\times 10^{- 23} J/k = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, \lambda_{e}:

\lambda_{e} = \frac{h}{p_{e}}

\lambda_{e} = \frac{h}{m_{e}{v_{e}}              (1)

where

h = Planck's constant = 6.626\times 10^{- 34}m^{2}kg/s

p_{e} = momentum of an electron

v_{e} = velocity of an electron

m_{e} = 9.1\times 10_{- 31} kg = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T

}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}

}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}

v_{e} = 2.86\times 10^{5} m/s                    (2)

Using eqn (2) in (1):

\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm

Now, to calculate the de-Broglie wavelength of proton, \lambda_{e}:

\lambda_{p} = \frac{h}{p_{p}}

\lambda_{p} = \frac{h}{m_{p}{v_{p}}                             (3)

where

m_{p} = 1.6726\times 10_{- 27} kg = mass of proton

v_{p} = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T

}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}

}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}

v_{p} = 6.674\times 10^{3} m/s                               (4)                    

Using eqn (4) in (3):

\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm

7 0
3 years ago
A boy shoves his stuffed toy zebra down a frictionless chute. It starts at a height of 1.69 m above the bottom of the chute with
Veseljchak [2.6K]

Answer:

x = 6.94 m

Explanation:

For this exercise we can find the speed at the bottom of the ramp using energy conservation

Starting point. Higher

            Em₀ = K + U = ½ m v₀² + m g h

Final point. Lower

            Em_{f} = K = ½ m v²

            Em₀ = Em_{f}

            ½ m v₀² + m g h = ½ m v²

            v² = v₀² + 2 g h

             

Let's calculate

             v = √(1.23² + 2 9.8 1.69)

             v = 5.89 m / s

In the horizontal part we can use the relationship between work and the variation of kinetic energy

            W = ΔK

            -fr x = 0- ½ m v²  

               

Newton's second law

              N- W = 0

     

The equation for the friction is

               fr = μ N

               fr = μ m g

We replace

             μ m g x = ½ m v²

             x = v² / 2μ g

Let's calculate

            x = 5.89² / (2 0.255 9.8)

            x = 6.94 m

6 0
3 years ago
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Mass, volume and density are all properties of
snow_tiger [21]

Answer:

Properties of matter

Explanation:

All properties of matter are either extensive or intensive and either physical or chemical. Extensive properties, such as mass and volume, depend on the amount of matter that is being measured. Intensive properties, such as density and color, do not depend on the amount of matter.

7 0
3 years ago
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Two bodies of masses 1000kg and 2000kg are separated 1km which is the gravitational force between them
denpristay [2]

Answer:

1.33×10⁻¹⁰ N

Explanation:

F = GMm / r²

where G is the gravitational constant,

M and m are the masses of the objects,

and r is the distance between them.

F = (6.67×10⁻¹¹ N/m²/kg²) (1000 kg) (2000 kg) / (1000 m)²

F = 1.33×10⁻¹⁰ N

3 0
3 years ago
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