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4 years ago

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A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.3 m/s rel

ative to the ground.

1 answer:

Sonja [21]4 years ago

4 0

consider east-west direction along X-axis and north-south direction along Y-axis

$V_{ra}$ = velocity of migrating robin relative to air = 12 j m/s

(where "j" is unit vector in Y-direction)

$V_{ag}$ = velocity of air relative to ground = 6.3 i m/s

(where "i" is unit vector in X-direction)

$V_{rg}$ = velocity of migrating robin relative to ground = ?

using the equation

$V_{rg}$ = $V_{ra}$ + $V_{ag}$

$V_{rg}$ = 12 j + 6.3 i

$V_{rg}$ = 6.3 i + 12 j

magnitude : sqrt((6.3)² + (12)²) = 13.6 m/s

direction : tan⁻¹(12/6.3) = 62.3 deg north of east

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The velocity of the transverse waves produced by an earthquake is 5.05 km/s, while that of the longitudinal waves is 8.585 km/s.

sattari [20]

Answer:

$d=691.71km$

Explanation:

The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:

$t=\frac{d}{v_t}-\frac{d}{v_l}$

Here d is the distance at which the earthquake take place and $v_t, v_l$ is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:

$t=d(\frac{1}{v_t}-\frac{1}{v_l})\\d=\frac{t}{\frac{1}{v_t}-\frac{1}{v_l}}\\d=\frac{56.4s}{\frac{1}{5.05\frac{km}{s}}-\frac{1}{8.585\frac{km}{s}}}\\d=691.71km$

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3 years ago

The maximum displacement in an oscillatory motion is A = 0.49 m. Determine the position x at which the kinetic energy of the par

amid [387]

Answer:

x = 0.40 m

Explanation:

When the displacement is maximum, the particle is momentarily at rest, which means that at this point (assuming no friction present) all the mechanical energy is elastic potential, which can be written as follows:

$E_{tot} = U_{o} = \frac{1}{2} *k*A^{2} (1)$

Since in absence of friction, total mechanical energy must keep constant, this means that at any time, the sum of the kinetic and potential energy, must be equal to (1), as follows:

$E_{tot} = U_{o} = \frac{1}{2} *k*A^{2} = (KE)_{f} + U_{f} (2)$

If KEf = U/2f, replacing in (2), we get:

$E_{tot} = U_{o} = \frac{1}{2} *k*A^{2} = (U/2)_{f} + U_{f} = \frac{3}{2} *U_{f} (3)$

At any point, the elastic potential energy is given by the following expression:

$U_{f} = \frac{1}{2} *k*x^{2} (4)$

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Replacing (4) in (3), simplifying and rearranging, we get:

$E_{tot} = U_{o} = \frac{1}{2} *A^{2} = \frac{3}{4} *x^{2} (5)$

Finally, solving for x, we get:

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3 years ago

What is the square root of 25x10^12

kirill115 [55]

$\sqrt{25 \times 10^{12} }$

In order to find it's square root, we could make it into two square roots.

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Each pair of a number inside square root gives a number out .

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$\sqrt{10*10*10*10*10*10*10*10*10*10*10*10} = 10*10*10*10*10*10$

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Therefore,

$\sqrt{25 \times 10^{12} }=\sqrt{25} \times \sqrt{10^{12}}={5 \times 10^6}$

$\sqrt{25 \times 10^{12} } = {5 \times 10^6}$

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3 years ago

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yaroslaw [1]

Answer:

I think it is acute angle.

Explanation:

Because it is an angle between 0° and 90°. Hope this answer wil help you.

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3 years ago