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melisa1
12 days ago
14

The final velocity, V, of an object under constant acceleration can be found using the formula uppercase V squared = v squared +

2 a s, where v is the initial velocity (in meters per second), a is acceleration (in meters per second), and s is the distance (in meters). What is the formula solved for a?
uppercase V squared minus v squared minus 2 x = a
uppercase V squared minus v squared + 2 s = a
StartFraction uppercase V squared minus v squared Over 2 s EndFraction = a
StartFraction uppercase V squared plus v squared Over 2 s EndFraction = a
Mathematics
2 answers:
zzz [9K]12 days ago
8 0
The equation resolved for a is StartFraction uppercase V squared minus v squared Over 2 s EndFraction = a.
babunello [8.4K]12 days ago
6 0
C.
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Max rides his scooter toward Kim and then passes her at a constant speed. His distance in feet, d, from Kim t seconds after he s
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a resorvoir can be filled by an inlet pipe in 24 hours and emptied by an outlet pipe 28 hours. the foreman starts to fill the re
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The ground-state wave function for a particle confined to a one-dimensional box of length L is Ψ=(2/L)^1/2 Sin(πx/L). Suppose th
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Respuesta:

(a) 4.98x10⁻⁵

(b) 7.89x10⁻⁶

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(e) 2.9x10⁻²

Explicación paso a paso:

La probabilidad (P) de encontrar la partícula está dada por:

P=\int_{x_{1}}^{x_{2}}(\Psi\cdot \Psi) dx = \int_{x_{1}}^{x_{2}} ((2/L)^{1/2} Sin(\pi x/L))^{2}dx  

P = \int_{x_{1}}^{x_{2}} (2/L) Sin^{2}(\pi x/L)dx     (1)

La solución de la integral de la ecuación (1) es:

P=\frac{2}{L} [\frac{X}{2} - \frac{Sin(2\pi x/L)}{4\pi /L}]|_{x_{1}}^{x_{2}}  

(a) La probabilidad de encontrar la partícula entre x = 4.95 nm y 5.05 nm es:

P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{4.95}^{5.05} = 4.98 \cdot 10^{-5}    

(b) La probabilidad de encontrar la partícula entre x = 1.95 nm y 2.05 nm es:

P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{1.95}^{2.05} = 7.89 \cdot 10^{-6}  

(c) La probabilidad de encontrar la partícula entre x = 9.90 nm y 10.00 nm es:

P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{9.90}^{10.00} = 1.89 \cdot 10^{-4}    

(d) La probabilidad de encontrar la partícula en la mitad derecha de la caja, es decir, entre x = 0 nm y 50 nm es:

P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{50.00} = 0.5

(e) La probabilidad de encontrar la partícula en el tercio central de la caja, es decir, entre x = 0 nm y 100/6 nm es:

P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{16.7} = 2.9 \cdot 10^{-2}

Espero que te ayude.

3 0
3 days ago
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