2 years ago

If a force of 40N is applied for 0.2 sec to change the momentum of a volleyball, what is the impulse?

Physics

1 answer:

Montano1993 [528]2 years ago

7 0

Answer:

8ns

Explanation:

40(0.2)

=8ns

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Answer:

The frequency of radiation is $2.61 \times 10^{17} s^{-1}$

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Given:

Wavelength $\lambda = 11.5 \times 10^{-10}$ m

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For finding the frequency of radiation,

$c = f \lambda$

$f = \frac{c}{\lambda}$

$f = \frac{3 \times 10^{8} }{11.5 \times 10^{-10} }$

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