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snow_tiger [21]
3 years ago
15

The jumping gait of the kangaroo is efficient because energy is stored in the stretch of stout tendons in the legs; the kangaroo

literally bounces with each stride. We can model the bouncing of a kangaroo as the bouncing of a mass on a spring. A 70 kg kangaroo hits the ground, the tendons stretch to a maximum length, and the rebound causes the kangaroo to leave the ground approximately 0.10 s after its feet first touch.
a. Modeling this as the motion of a mass on a spring, what is the period of the motion?
b. Given the kangaroo mass and the period you’ve calculated, what is the spring constant?
c. If the kangaroo speeds up, it must bounce higher and farther with each stride, and so must store more energy in each bounce. How does this affect the time and the amplitude of each bounce?
Physics
1 answer:
Anna [14]3 years ago
6 0

Answer:

the period T of whole motion should be twice the value for half at he bottom so T is 0.2sec.

w is angular frequency

formula:2π/T

now k is spring constant

F/R-->mw²

putting values:70*(2π/0.2)²

=4.9x10⁶

so we can say that SHM is not affected by the amplitude of the bounce.

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A

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A student walks (2.9±0.1)m, stops and then walks another (3.9 ±0.2)m in the same directionWith the given uncertainties, what is
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Answer:

The smallest distance the student that the student could be possibly be from the starting point is 6.5 meters.

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Sum=(A+B)\pm (\Delta A+\Delta B)

For the the values given to us the sum is calculated as

Sum=(2.9+3.9)\pm (0.1+0.2)

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The closest possible distance at which the student can be is obtained by taking the negative sign in the uncertainity

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3 0
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Which of the following metals require ultraviolet light to exhibit the photoelectric effect?The options available: a. Cs, work f
Eddi Din [679]

Answer:

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This energy converted into electronvolts is

E=\frac{5.23\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=3.27 eV

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7 0
3 years ago
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