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Mamont248 [21]
2 years ago
15

3. Si usted duplicara la amplitud de un M.A.S. ¿cómo cambiaría la frecuencia, velocidad máxima, la aceleración máxima y la energ

ía total mecánica?
Physics
1 answer:
WITCHER [35]2 years ago
8 0

Answer:

Explanation:

la frecuencia = ω/2π, nada cambio

v(max) = ωA → ω2Α = 2ωA  duplicara velocidad máxima

a(max) = ω²Α → ω²2Α = 2ω²Α duplicara la aceleración máxima

la energía total ½kA² → ½k(2Α)² = 4(½kA²) cuatro veces la energía

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Answer: D and friction

Explanation:

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3 years ago
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10. A hockey puck with mass 0.3 kg is sliding along ice that can be considered frictionless. The puck’s velocity is 20 m/s when
SVEN [57.7K]

Answer:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

Explanation:

For this case  we can use the second law of Newton given by:

\sum F = ma

The friction force on this case is defined as :

F_f = \mu_k N = \mu_k mg

Where N represent the normal force, \mu_k the kinetic friction coeffient and a the acceleration.

For this case we can assume that the only force is the friction force and we have:

F_f = ma

Replacing the friction force we got:

\mu_k mg = ma

We can cancel the mass and we have:

a = \mu_k g = 0.35 *9.8 \frac{m}{s^2}= 3.43 \frac{m}{s^2}

And now we can use the following kinematic formula in order to find the distance travelled:

v^2_f = v^2_i - 2ad

Assuming the final velocity is 0 we can find the distance like this:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

5 0
3 years ago
How many carbon-14 atoms are left after 1 half-life
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Answer:

The time it takes for 14C to radioactively decay is described by its half-life. C has a half-life of 5,730 years. In other words, after 5,730 years, only half of the original amount of 14C remains in a sample of organic material. After an additional 5,730 years–or 11,460 years total–only a quarter of the 14C remains.

Explanation:

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2 years ago
If you poured yourself a drink and then you started to see layers what phenomenon is happening?
arsen [322]

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Spiders may "tune" strands of their webs to give an enhanced response at frequencies corresponding to the frequencies at which d
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Answer:

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f = Frequency = 200 Hz

Mass of the string is

m=\rho V\\\Rightarrow m=\rho Al\\\Rightarrow m=1300\times \pi (0.001\times 10^{-3})^2\times 0.16\\\Rightarrow m=6.5345127195\times 10^{-10}\ kg

Frequency is given by

f=\dfrac{1}{2l}\sqrt{\dfrac{T}{m/l}}\\\Rightarrow T=4lf^2m\\\Rightarrow T=4\times 0.16\times 200^2\times 6.5345127195\times 10^{-10}\\\Rightarrow T=0.0000167283525619\ N

The tension on the string is 0.0000167283525619 N

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3 years ago
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