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Ray Of Light [21]
3 years ago
7

A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?

Physics
1 answer:
dexar [7]3 years ago
3 0

Answer:

1.08 s

Explanation:

From the question given above, the following data were obtained:

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

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An object is moving along a straight line, and the uncertainty in its position is 1.90 m.
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Answer:

2.78\times 10^{-35}\ \text{kg m/s}

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\Delta x = Uncertainty in position = 1.9 m

\Delta p = Uncertainty in momentum

h = Planck's constant = 6.626\times 10^{-34}\ \text{Js}

m = Mass of object

From Heisenberg's uncertainty principle we know

\Delta x\Delta p\geq \dfrac{h}{4\pi}\\\Rightarrow \Delta p\geq \dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p\geq \dfrac{6.626\times 10^{-34}}{4\pi\times 1.9}\\\Rightarrow \Delta p\geq 2.78\times 10^{-35}\ \text{kg m/s}

The minimum uncertainty in the momentum of the object is 2.78\times 10^{-35}\ \text{kg m/s}

Golf ball minimum uncertainty in the momentum of the object

m=0.045\ \text{kg}

Uncertainty in velocity is given by

\Delta p\geq m\Delta v\geq 2.78\times 10^{-35}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{0.045}\\\Rightarrow \Delta v\geq 6.178\times 10^{-34}\ \text{m/s}

The minimum uncertainty in the object's velocity is 6.178\times 10^{-34}\ \text{m/s}

Electron

m=9.11\times 10^{-31}\ \text{kg}

\Delta v\geq \dfrac{\Delta p}{m}\\\Rightarrow \Delta v\geq \dfrac{2.78\times 10^{-35}}{9.11\times 10^{-31}}\\\Rightarrow \Delta v\geq 0.31\times 10^{-4}\ \text{m/s}

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Read 2 more answers
Please help, thank you,
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3. Suppose that an airplane flying 60 m/s, at a height of 300m, dropped a sack of flour (pere the effect
Arlecino [84]

469.24m. An airplane flying 60m/s at a height of 300m dropped a sack of flour that stack the ground 469.24m from the point of release.

This is a example of horizontal parabolic projectile motion,and we represents this motion in the coordinate axis, which means that the velocity has components in x axis and y axis.

The equation of components on the x axis.

v_{0}x=\frac{x}{t}, where x is the distance and Vox the initial velocity before the drop

The equation of components on the y axis.

y = v_{0}yt+\frac{gt^{2} }{2}, where y is the height, and the velocity in y component before the drop is 0, reducing the equation to y = \frac{gt^{2} }{2}

Clear t from both the equation of components on the x axis and the y axis:

t=\frac{x}{v_{0} x} and t=\sqrt{\frac{2h}{g}}

Equating both equations and clearing the distance x:

\frac{x}{v_{0} x}=\sqrt{\frac{2h}{g}}\\x={v_{0} x}\sqrt{\frac{2h}{g}}

Substituting the values:

x=60\frac{m}{s} \sqrt{\frac{2(300m)}{9.81\frac{m}{s^{2} } }}=469.24m

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