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jeyben [28]
3 years ago
8

3. A 70 kg person climbs a 6 m ladder. How much work is required by the person?

Physics
1 answer:
Lina20 [59]3 years ago
6 0

Answer:

4116J

Explanation:

Given parameters:

Mass of the person  = 70kg

Height of the ladder  = 6m

Unknown:

Work done  = ?

Solution:

The work done by the person climbing is the same as the potential energy.

Work done is the force applied to move a body through a distance;

So;

   Potential energy  = mass x acceleration due to gravity x height

Therefore;

    Potential energy  = 70 x 9.8 x 6   = 4116J

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Write an expression for a harmonic wave with an amplitude of 0.19 m, a wavelength of 2.6 m, and a period of 1.2 s. The wave is t
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Answer:

y = 0.19 sin(5.23 t - 2.42x + \frac{\pi}{2})

Explanation:

As we know that the wave equation is given as

y = A sin(\omega t - k x + \phi_0)

now we have

A = 0.19 m

\lambda = 2.6 m

so we have

k = \frac{2\pi}{\lambda}

k = \frac{2\pi}{2.6}

k = 2.42  per m

also we have

T = 1.2 s

so we have

\omega = \frac{2\pi}{T}

\omega = \frac{2\pi}{1.2}

\omega = 5.23 rad/s

now we know that at t = 0 and x = 0 wave is at y = 0.19 m

so we have

\phi_0 = \frac{\pi}{2}

so we have

y = 0.19 sin(5.23 t - 2.42x + \frac{\pi}{2})

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