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Viefleur [7K]
3 years ago
14

What is the speed of an electron that has been accelerated from rest through a potential difference of 1020 V?

Physics
1 answer:
sashaice [31]3 years ago
8 0
<span>a. KE in electron volts is 1020 eV. 
b. KE in Joules is e(1020) = (1.6022E-19)(1020) = 1.634E-16 
c. KE = (1/2)mv^2, so v = sqrt[2*KE/m] = 18.94E6 m/s

note: m is the mass of an electron = 9.109e-31 kg

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>
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Which of the following are coefficients you could use in a balanced equation?
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a)   x₁ = 290.50 feet ,  x₂ = 169.74 feet , b)  v_max= 41 mph

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For this exercise we will work in two parts, the first with Newton's second law to find the acceleration of vehicles

X Axis          fr = m a

Y Axis          N-W = 0

                    N = W = mg

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We replace

                 μ mg = ma

                 a = μ g

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μ              a (feet / s2)

0.599       19.168

0.536       17,152

0.480       15.360

0.350        11.200

These are the acceleration values, for the maximum distance we use the minimum acceleration (a₁ = 11,200 feet / s²) and for the minimum braking distance we use the maximum acceleration (x₂ = 19,168 feet / s²)

                 v² = v₀² - 2 a x

When the speed stops it is zero

                 x₁ = v₀² / 2 a₁

                         

Let's reduce speed

            v₀ = 55mph (5280 foot / 1 mile) (1h / 3600s) = 80,667 feet / s²

Let's calculate the maximum braking distance

            x₁ = 80.667² / (2 11.2)

            x₁ = 290.50 feet

The minimum braking distance

            x₂ = 80.667² / (2 19.168)

            x₂ = 169.74 feet

b) maximum speed to stop at distance x = 155 feet

            0 = v₀² - 2 a x

            v₀ = √2 a x

We calculate the speed for the two accelerations

             v₀₁ = √ (2 11.2 155)

             v₀₁ = 58.92 feet / s

       

             v₀₂ = √ (2 19.168 155)

             v₀₂ = 77.08 feet / s

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