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2 years ago

Determine the magnitude of the effective value of g⃗ at a latitude of 60 ∘ on the earth. assume the earth is a rotating sphere.

1 answer:

5 0

In addition to acceleration of gravity we experience centrifugal acceleration away from the axis of rotation of the earth. this additional acceleration has value ac = r w^2 where w = angular velocity and r is distance from your spot on earth to the earth's axis of rotation so r = R cos(l) where l = 60 deg is the lattitude and R the earth's radius and w = 1 / (24hr x 3600sec/hr)
<span>now you look up R and calculate ac then you combine the centrifugal acc. vector ac with the gravitational acceleration vector ag = G Me/R^2 to get effective ag' = ag -</span>

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Two microwave signals of nearly equal wavelengths can generate a beat frequency if both are directed onto the same microwave det

alekssr [168]

Answer:

1.5106 cm

Explanation:

The beat frequency is equal to the absolute value of the difference between the frequencies of the two signals:

$f_B = |f_1 - f_2|$

using the wave equation, we can re-write each frequency as

$f=\frac{c}{\lambda}$

where c is the speed of light and $\lambda$ is the wavelength. Therefore,

$f_B = |\frac{c}{\lambda_1}-\frac{c}{\lambda_2}|$

where:

$f_B = 140 MHz = 140\cdot 10^6 Hz$ is the beat frequency

$\lambda_1 = 1.50 cm = 0.015 m$ is the wavelength of the first generator

$\lambda_2$ is the wavelength of the second generator

We also know that the second generator emits the longer wavelength, so we already know that the term inside the module is positive. Therefore, we can now solve for $\lambda_2$ :

$f_B = c(\frac{1}{\lambda_1}-\frac{1}{\lambda_2})\\\lambda_2=(\frac{1}{\lambda_1}-\frac{f_B}{c})^{-1}=(\frac{1}{0.015}-\frac{140\cdot 10^6}{3\cdot 10^8})^{-1}=0.015106 m = 1.5106 cm$

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2 years ago

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Verizon [17]

Answer: Protons are positively charged particles

Explanation:

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3 years ago

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