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Juliette [100K]

3 years ago

12

When red-headed woodpeckers (melanerpes erythrocephalus) strike the trunk of a tree, they can experience an acceleration ten tim

es greater than the acceleration of gravity, or about 98.1 m/s2. what is this acceleration in in/s2?\?

1 answer:

kifflom [539]3 years ago

8 0

98.1 m/s² × (1 in)/(0.0254 m) ≈ 3862 in/s²

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Which term refers to the upper part of a glacier that receives the most snowfall?

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2 years ago

An emf is induced in a conducting loop of wire 1.22 m long as its shape is changed from square to circular. Find the average mag

Afina-wow [57]

Answer:

The induced emf in the loop is $7.35\times 10^{-4}\ V$

Explanation:

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a = 0.305 m

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$A'=\pi r^2\\A'=\pi (0.194)^2\\\\A'=0.118\ m^2$

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3 years ago

Consider the f(x) = cos(x-C) function shown in the figure in blue color, where 0 ≤ C ≤ 2π. What is the value of parameter C for

klemol [59]

The value of parameter C for the function in the figure is 2.

<h3>What is amplitude of a wave?</h3>

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From the blue colored graph; at y = 1, x = -2 cm

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Learn more about phase angle here: brainly.com/question/16222725

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1 year ago

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

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Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

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Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

8 0

1 year ago

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