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Maurinko [17]
2 years ago
15

One species of eucalyptus tree, Eucalyptus regnans, grow to heights similar to those attained by California redwoods. Suppose a

bird sitting on top of one specimen of eucalyptus tree drops a nut that is 1.7 ounces. If the speed of the falling nut at the moment it is 50.3 m above the ground is 42.7 m/s, how tall is the tree
Physics
1 answer:
mote1985 [20]2 years ago
4 0

Answer:

The tree is 143.325 meters tall

Explanation:

The given parameters of the eucalyptus tree are;

The mass of the eucalyptus tree nut = 1.7 ounces

The speed of the nut at 50.3 m above the ground, v = 42.7 m/s

The equation for free fall is given as follows;

v² = 2·g·h

Where;

v = The velocity after falling through a height, h

g = The acceleration due to gravity = 9.8 m/s²

h = The height through which the seed has already fallen

Therefore, we have;

h = v²/(2·g) = (42.7 m/s)²/(2 × 9.8 m/s²) = 93.025 m

The height through which the seed has already fallen, h = 93.025 m

The height of the tree = h + The height of the seed above ground at the moment it was falling at 42.7 m/s

The height of the tree = 93.025 m + 50.3 m = 143.325 m

The height of the tree = 143.325 m.

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Given:
u = 0, initial velocity
s 0.9 m, distance traveled.
t = 3 s, the time taken.

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s = ut + (1/2)*a*t²
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0.9 = 4.5a
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2 years ago
The capacitors in each circuit are fully charged before the switch is closed. Rank, from longest to shortest, the length of time
IgorLugansk [536]

Answer:

C has the greatest capacitance and the lightest load. It will provide current to the load for the longest.

A & E are the same

B is next, and finally

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Curious that the pictures are out of alphabetic order A B C E D.

Explanation:

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Find an expression for the torsional constant k in terms of the moment of inertia I of the disk and the angular frequency ω of s
sasho [114]

Answer:

Explanation:

The general equation for the disk with moment of inertia I when given small angular displacement  \theta is given by

I\frac{\mathrm{d^2} \theta }{\mathrm{d} t^2}=-k\theta

\frac{\mathrm{d^2} \theta }{\mathrm{d} t^2}+\frac{k\theta }{I}=0

Replacing

\frac{k\theta }{I}=\omega ^2

where \omega is the angular frequency of oscillation

General solution for this Equation is given by

\theta =\theta _{max}\sin \left ( \omega t+\phi \right )

where \theta _{max}=maximum\ angular\ displacement

\phi =Phase\ difference

Thus K can be written as

k=I\omega ^2

5 0
2 years ago
If a current of 2.4 a is flowing in a cylindrical wire of diameter 2.0 mm, what is the average current density in this wire?
Gnom [1K]

The average current density in the wire is given by:

J=\frac{I}{A}

where I is the current intensity and A is the cross-sectional area of the wire.


The cross-sectional area of the wire is given by:

A=\pi r^2

where r is the radius of the wire. In this problem, r=\frac{d}{2}=\frac{2.0 mm}{2}=1.0 mm=0.001 m, so the cross-sectional area is

A=\pi (0.001 m)^2=3.14 \cdot 10^{-6} m^2


and the average current density is

J=\frac{I}{A}=\frac{2.4 A}{3.14 \cdot 10^{-6} m^2}=7.64 \cdot 10^5 A/m^2

8 0
2 years ago
Read 2 more answers
You and your family take a trip to see your aunt who lives 100 miles away along a straight highway. The first 60 miles of the tr
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Answer:

51.2 mi/h

Explanation:

Total distance, d = 100 miles

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Next 40 miles with speed 75 mi/h

Time taken for first 60 miles, t1 = 60 / 55 = 1.09 h

Time taken for 40 miles, t2 = 40 / 75 = 0.533 h

Time spent to get stuck, t3 = 20 min = 0.33 h

Total time, t = t1 + t2 + t3 = 1.09 + 0.533 + 0.33 = 1.953 h

The average speed is defined as the ratio of total distance traveled to the total time taken.

Average speed = =\frac{100}{1.953}=51.2 mi/h

Thus, the average speed of the journey is 51.2 mi/h.

4 0
2 years ago
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