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3 years ago

HELP with this please.

Physics

2 answers:

jasenka [17]3 years ago

6 0

17. C. Isotopes
18. Tracers
19. Alpha Decay

jolli1 [7]3 years ago

4 0

17. C.
18. Tracers
19. Alpha Decay

Hope this helps and please mark as brainliest!

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What is the change in velocity of a 22-kg object that experiences a force of 15 N for

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Answer:

Force = mass × acceleration

Acceleration:

${ \tt{15 = (22 \times a)}} \\ { \tt{a = \frac{15}{22} \: {ms}^{ - 2} }}$

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${ \bf{v = u + at}}$

Change = v - u:

${ \tt{v - u = (a \times t)}} \\ { \tt{v - u = ( \frac{15}{22} \times 1.2) }} \\ { \tt{v - u = 0.82 \: {ms}^{ - 2} }}$

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2 years ago

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3 years ago

Find the fundamental frequency and the next three frequencies that could cause standing-wave patterns on a string that is 30.0 m

maksim [4K]

Answer:

0.786 Hz, 1.572 Hz, 2.358 Hz, 3.144 Hz

Explanation:

The fundamental frequency of a standing wave on a string is given by

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where

L is the length of the string

T is the tension in the string

$\mu$ is the mass per unit length

For the string in the problem,

L = 30.0 m

$\mu=9.00\cdot 10^{-3} kg/m$

T = 20.0 N

Substituting into the equation, we find the fundamental frequency:

$f=\frac{1}{2(30.0)}\sqrt{\frac{20.0}{(9.00\cdot 10^{-3}}}=0.786 Hz$

The next frequencies (harmonics) are given by

$f_n = nf$

with n being an integer number and f being the fundamental frequency.

So we get:

$f_2 = 2 (0.786 Hz)=1.572 Hz$

$f_3 = 3 (0.786 Hz)=2.358 Hz$

$f_4 = 4 (0.786 Hz)=3.144 Hz$

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2 years ago

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