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Otrada [13]
3 years ago
14

In the equation for centripetal force, which expression represents the centripetal acceleration of the object? mv2 StartFraction

m over r EndFraction StartFraction v Superscript 2 Baseline over r EndFraction StartFraction m times v Superscript 2 Baseline over r EndFraction
Physics
2 answers:
Vinvika [58]3 years ago
8 0

Answer:

c

Explanation:

Sphinxa [80]3 years ago
3 0

Answer: \frac{V^{2}}{r}

Explanation:

According to Newton's 2nd Law of motion the force F is proportional to the mass Fm and acceleration a:

F=m.a (1)

On the other hand, the equation for the Centripetal force is:

F=\frac{mV^{2}}{r} (2)

Where:

V is the velocity

r is the radius of the circular motion

Making (1) and (2) equal:

m.a=\frac{mV^{2}}{r} (3)

Hence:

a=\frac{V^{2}}{r} This is the expression for the centripetal acceleration

It should be noted, this acceleration is directed toward the center of the circumference of the circular motion (that's why it's called centripetal acceleration).

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2 years ago
Calculate the de Broglie wavelength of an electron accelerated from rest through a potential difference of (a) 100 V, (b) 1.0 kV
forsale [732]

Answer:

(a) \lambda=1.227\ A

(b) \lambda=0.388\ A

(c) \lambda=0.038\ A

Explanation:

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(a) An electron accelerated from rest through a potential difference of 100 V. The De Broglie wavelength in terms of potential difference is given by :

\lambda=\dfrac{h}{\sqrt{2meV} }

Where

m and e are the mass of and charge on an electron

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\lambda=\dfrac{12.27}{\sqrt{V} }\ A

V = 100 V

\lambda=\dfrac{12.27}{\sqrt{100} }\ A

\lambda=1.227\ A

(b) V = 1 kV = 1000 V

\lambda=\dfrac{12.27}{\sqrt{V} }\ A

\lambda=\dfrac{12.27}{\sqrt{1000} }\ A

\lambda=0.388\ A

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