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N76 [4]
3 years ago
11

Spiderman, whose mass is 74.0 kg, is dangling on the free end of a 11.0-m-long rope, the other end of which is fixed to a tree l

imb above. By repeatedly bending at the waist, he is able to get the rope in motion, eventually getting it to swing enough that he can reach a ledge when the rope makes a θ = 60.6° angle with the vertical. How much work was done by the gravitational force on Spiderman in this maneuver?
Physics
1 answer:
Anestetic [448]3 years ago
7 0

Answer:

W = -1844.513 J

Explanation:

GIVEN DATA:

mass of spider man is m  74 kg

vertical displacement if spider is 11 m

final displacement  =  11 cos 60.6 =  - 6.753 m

change in displacement is  = -6.753 - (-11) = 4.25 m

gravity force act on spiderman is f = mg = 74 × 9.8 = 725.2 N

work done by gravity is W = F \delta r cos\theta

W = 725.2 \times 4.25 \times cos 180

where 180 is the angle between spiderman weight and displacement

W = -1844.513 J

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k) A stream of warm water is produced in a steady-flow mixing process by combining 1.0 kg s-1of cool water at 25°C with 0.8 kg s
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Answer:

T_ww = 43,23°C

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To solve this question, we use energy balance and we state that the energy that enters the systems equals the energy that leaves the system plus losses. Mathematically, we will have that:

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m_CW*C_p*T_CW+m_HW*C_p*T_HW=m_WW*C_p*T_WW+E_loss

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1.0 kg/s*4,18 kJ/(kg°C)*25°C+0.8  kg/s*4,18 kJ/(kg°C)*75°C=1.8 kg/s*4,18 kJ/(kg°C)*T_WW+30  kJ/s

Solving for the temperature Tww, we have:

(1.0 kg/s*4,18 kJ/(kg°C)*25°C+0.8 kg/s*4,18 kJ/(kg°C)*75°C-30 kJ/s)/(1.8 kg/s*4,18 kJ/(kg°C))=T_WW

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Have a nice day! :D

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Please help me with this (with explanation)
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Suppose the cyclist travels for a total time of <em>t</em> hours.

For 20 min = 1/3 hr, the cyclist does not move.

Over the remaining (<em>t</em> - 1/3) hr, the cyclist is moving at a constant speed of 22.0 km/hr, so that the cyclist would travel a distance of

<em>x</em> = (22.0 km/hr) • ((<em>t</em> - 1/3) hr) ≈ (22.0 km/hr) <em>t</em> - 7.33 km

If the cyclist's average speed over the total time <em>t</em> was 17.5 km/hr, then by the definition of average speed,

17.5 km/hr = <em>x</em> / <em>t</em>

Replace <em>x</em> with the distance expression from earlier:

17.5 km/hr = ((22.0 km/hr) <em>t</em> - 7.33 km) / <em>t</em>

Solve for <em>t</em> :

17.5 km/hr = 22.0 km/hr - (7.33 km) / <em>t</em>

(7.33 km) / <em>t</em> = 4.5 km/hr

<em>t</em> = (7.33 km) / (4.5 km/hr)

<em>t</em> ≈ 1.62963 hr

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<em>x</em> ≈ (22.0 km/hr) (1.62963 hr) - 7.33 km ≈ 28.5 km

and so the answer is A.

Alternatively, as soon as you arrive at

17.5 km/hr = <em>x</em> / <em>t</em>

you can instead solve for <em>t</em> in terms of <em>x</em>, then plug that into the distance equation.

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<em>x</em> ≈ 1.25714 <em>x</em> - 7.33 km

0.25714<em>x</em> ≈ 7.33 km

<em>x</em> = (7.33 km) / 0.25714 ≈ 28.5 km

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