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2 years ago

What is the relationship between wave amplitude and wave energy?

Physics

1 answer:

Nataliya [291]2 years ago

3 0

Answer:

waves

Explanation:

they are both waves

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Which of the following are examples of acceleration?

attashe74 [19]

Answer:

cough

Explanation:

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3 years ago

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What is the average speed of a train that travels at 100km/hr for 6 hours and then 120km/hr for 5 hours

Sindrei [870]

Answer:

Explanation:

Average speed = Total distance / Total time.

100 km/hr

r = 100 km / hr

t = 6 hours

d = 6 * 100 = 600 km

120 km / hr

r = 120 km / hr

t = 5 hour

d = 120 * 5

d = 600 km

Total distance = 600 + 600 = 1200 km

Total time = 5 hour + 6 hours = 11 hours.

Average speed = 1200 km / 11 hours = 109.1

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3 years ago

Which information about the earth would not be represented when a globe is used as a model?

egoroff_w [7]

A globe sitting on the desk can't demonstrate the speed of axial rotation
or the speed of orbital revolution.

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3 years ago

Light is incident on the left face of an isosceles prism; with an apex angle of 49o, such that the light exiting the right face

Sunny_sXe [5.5K]

Answer:

$\mu = 1.645$

Explanation:

By Snell's law we know at the left surface

$\theta_i = 19^o$

$\theta_r = ?$

$\mu_1 = 1$

$\mu_2 = \mu$

now we have

$1 sin19 = \mu sin\theta_r$

$0.33 = \mu sin\theta_r$

now on the other surface we know that

angle of incidence = $\theta_r'$

$\theta_e = 90$

so again we have

$\mu sin\theta_r' = 1 sin90$

so we have

$\theta_r = sin^{-1}\frac{0.33}{\mu}$

$\theta_r' = sin^{-1}\frac{1}{\mu}$

also we know that

$\theta_r + \theta_r' = 49$

$sin^{-1}\frac{0.33}{\mu} + sin^{-1}\frac{1}{\mu} = 49$

By solving above equation we have

$\mu = 1.645$

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3 years ago

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Cho lực F ⃗=6x^3 i ⃗-4yj ⃗ tác dụng lên vật làm vật chuyển động từ A(-2,5) đến B(4,7). Vậy công của lực là:

Natasha2012 [34]

The work done by $\vec F$ along the given path C from A to B is given by the line integral,

$\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r$

I assume the path itself is a line segment, which can be parameterized by

$\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath$

with 0 ≤ t ≤ 1. Then the work performed by F along C is

$\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}$

7 0

2 years ago