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stira [4]
3 years ago
8

The average height of an apple tree is 4.00 meters. How long would it take an apple falling from that height to reach the ground

? Given: g = -9.8 meters/second2
Physics
2 answers:
DENIUS [597]3 years ago
5 0

Answer:

The time is 0.90 sec.

Explanation:

Given that,

Height = 4.00 m

We need to calculate the time

Using equation of motion

s = ut+\dfrac{1}{2}gt^2

Where, s = height

g = acceleration due to gravity

t = time

Put the value into the formula

4.00=0+\dfrac{1}{2}\times9.8\times t^2

t^2=\dfrac{4.00\times2}{9.8}

t=\sqrt{\dfrac{4.00\times2}{9.8}}

t=0.90\ sec

Hence, The time is 0.90 sec.

denpristay [2]3 years ago
3 0
(9.8m/s) divided by 4m. the m's cancel out leaving you with s so 2.45s
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At time t=0, a particle is located at the point (3,6,9). It travels in a straight line to the point (5,2,7), has speed 8 at (3,6
Elis [28]

The particle has constant acceleration according to

\vec a(t)=2\,\vec\imath-4\,\vec\jmath-2\,\vec k

Its velocity at time t is

\displaystyle\vec v(t)=\vec v(0)+\int_0^t\vec a(u)\,\mathrm du

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\displaystyle\vec r(t)=\vec r(0)+\int_0^t\vec v(u)\,\mathrm du

\vec r(t)=(3+v_{0x}t+t^2)\,\vec\imath+(6+v_{0y}t-2t^2)\,\vec\jmath+(9+v_{0z}t-t^2)\,\vec k

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\|\vec v(0)\|=8\iff{v_{0x}}^2+{v_{0y}}^2+{v_{0z}}^2=64

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\begin{cases}3+v_{0x}T+T^2=5\\6+v_{0y}T-2T^2=2\\9+v_{0z}T-T^2=7\end{cases}\implies\begin{cases}v_{0x}=\dfrac{2-T^2}T\\\\v_{0y}=\dfrac{2T^2-4}T\\\\v_{0z}=\dfrac{T^2-2}T\end{cases}

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v_{0y}=-2v_{0x}

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{v_{0x}}^2+(-2v_{0x})^2+(-v_{0x})^2=6{v_{0x}}^2=64\implies v_{0x}=\pm\dfrac{4\sqrt6}3

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\implies v_{0z}=\mp\dfrac{4\sqrt6}3

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\vec r(t)=\left(3+\dfrac{4\sqrt6}3t+t^2\right)\,\vec\imath+\left(6-\dfrac{8\sqrt6}3t-2t^2\right)\,\vec\jmath+\left(9-\dfrac{4\sqrt6}3t-t^2\right)\,\vec k

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Answer:

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Explanation:

Beat: This is the phenomenon obtained when two notes of nearly equal frequency are sounded together. The S.I unit of beat is Hertz (Hz).

From the question,

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Note: The frequency of the other instrument is either f₁ or f₂.

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