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Kipish [7]
4 years ago
14

A skier of mass 70 kg is pulled up a slope by a motor-driven cable. The acceleration of gravity is 9.8 m/s 2 . How much work is

required to pull him 62.5 m up a 38.4 ◦ slope (assumed to be frictionless) at a constant speed of 3.77 m/s.
Physics
1 answer:
mote1985 [20]4 years ago
4 0

Answer:26.65 kJ

Explanation:

Given

mass of skier=70 kg

length of ramp=62.5 m

slope=38.4 ^{\circ}

speed=3.77 m/s

so net force on skier must be zero

Net work done will be change in Potential energy of skier

Potential Energy=mg(\Delta H)

\Delta H=62.5sin38.4=38.821 m

P.E.=70\times 9.81\times 38.82=26,658.88 J

P.E.=26.65 kJ

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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
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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}

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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

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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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