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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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Answer:

Q = ba⁴ * ε₀

Explanation:

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Since A = a² for the cube face, we have

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5 0
2 years ago
A projectile follows a straight-line path instead of a parabolic trajectory. Which could be the launch angle? would it be 90 or
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7 0
2 years ago
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A system delivers 1275 j of heat while the surroundings perform 855 j of work on it. calculate ∆esys in j.
kakasveta [241]
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We must be careful with the signs here. The sign convention generally used is:
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while the work is positive because it is performed by the surrounding on the system:
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6 0
2 years ago
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mart [117]

We use the equation of motion,

S= ut+\frac{1}{2}at^{2}

Here, S is the height, u is initial velocity and a is acceleration.

Given, S = 20 \ ft S = 20 \ ft = 20 \times\frac{1 \ m}{3.2808399 ft}  = 6.096 \ m

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Thus, the time taken by the acorn to fall 20  feet ( 6.096 m ) is 1.12 s.

5 0
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