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IgorC [24]
4 years ago
5

To hoist himself into a tree, a 72.0-kg man ties one end of a nylon rope around his waist and throws the other end over a branch

of the tree. he then pulls downward on the free end of the rope with a force of 371 n. neglect any friction between the rope and the branch, and determine the manâs upward acceleration. use g =9.80 m/sec2.
Physics
1 answer:
taurus [48]4 years ago
4 0
<span>0.506 m/s^2 Given the arrangement of rope and the man, he's effectively suspended by a rope from a frictionless pulley. So he's pulling the rope with a force of 371 n causing him to be pulled upwards by his arms. Additionally, the rope goes up to the pulley and back down to the man causing an additional 371 n of force pulling him upwards. So the total force being used to lift the man is 742 n. Additionally, the man is being pulled downwards by gravity at 9.80 m/s^2. And since he masses 72.0 kg, the downward force in newtons is 72.0 kg * 9.80 m/s^2 = 705.6 n downward. So the total force being exerted on the man is 742 n - 705.6 n = 36.4 n To calculate his acceleration, simply divide the number of newtons applied by his mass. 36.4 kg m/s^2 / 72.0 kg = 0.505556 m/s^2 And round to 3 significant figures. 0.505556 m/s^2 = 0.506 m/s^2</span>
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Equate (1) and (2).
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From (1), obtain the required expression for y.

Answer:
y= \frac{d \, tan(\phi) \, tan(\theta)}{tan(\phi)-tan(\theta)}

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1 year ago
A motorcycle is travelling at a constant velocity of 30ms. The motor is in high gear and emits a hum of 700Hz. The speed of soun
timurjin [86]

Answer:

a) T=1.43\times 10^{-3}\ s

b) d=0.0429\ m

c) \lambda=0.4857\ m

d) f_o=767.7\ Hz

Explanation:

Given:

  • velocity of the sound from the source, S=340\ m.s^{-1}
  • original frequency of sound from the source, f_s=700\ Hz
  • speed of the source, v_s=30\ m.s^{-1}

(a)

We know time period is inverse of frequency:

Mathematically:

T=\frac{1}{f}

T=\frac{1}{700}

T=1.43\times 10^{-3}\ s

(b)

Distance travelled by the motorcycle during one period of sound oscillation:

d=v_s.T

d=30\times 1.43\times 10^{-3}

d=0.0429\ m

(c)

The distance travelled by the sound during the period of one oscillation is its wavelength.

\lambda=\frac{S}{f}

\lambda=\frac{340}{700}

\lambda=0.4857\ m

(d)

observer frequency with respect to a stationary observer:

<u>According to the Doppler's effect:</u>

\frac{f_o}{f_s}= \frac{S+v_o}{S-v_s} ...........................(1)

where:

f_o\ \&\ v_o are the observed frequency and the velocity of observer respectively.

Here, observer is stationary.

\therefore v_o=0\ m.s^{-1}

Now, putting values in eq. (1)

\frac{f_o}{700}= \frac{340+0}{340-30}

f_o=767.7\ Hz

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