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Readme [11.4K]
3 years ago
14

To tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.50•10^6 N, one at an angle 19.0° west of north, a

nd the other at an angel 19.0° east of north, as they pull the tanker a distance 0.710 km toward the north

Physics
1 answer:
irina1246 [14]3 years ago
6 0

Answer:

Work done by a tug boat, W = 1.735 x 10⁸ J

Explanation:

Given,

The of each tugboat, F = 1.5 x 10⁶ N

The angle of each tugboat forms with the resultant force, θ = 19°

The displacement of the supertanker, s = 710 m

The individual tugboat will be responsible for the displacement, d = 710/2

                                                                                                               = 355 m

The displacement component in each tugboat direction = 355 · sin θ meter

Therefore, the work done by each tugboat is

                                           W = F x S    joules

Substituting the values in the above equation

                                            W = 1.5 x 10⁶  x  355 · sin θ

                                                = 1.735 x 10⁸ J

Hence, the work done by each tugboat is, W = 1.735 x 10⁸ J                        

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The kinetic energy of a moving object is E=12mv2. A 61 kg runner is moving at 10kmh. However, her speedometer is only accurate t
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Answer:

e=3367.2J

%e=1.43%

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From the exercise we know two information. The real speed and the experimental measured by the speedometer

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Since the speedometer is only accurate to within 0.1km/h the experimental speed is

v_{e}=10km/h-0.1km/h=9.9km/h=2.75m/s

Knowing that we can calculate Kinetic energy for the real and experimental speed

E_{r}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.77m/s)^2=234023J

E_{e}=\frac{1}{2}mv^2=\frac{1}{2}(61000g)(2.75m/s)^2=230656J

Now, the potential error in her calculated kinetic energy is:

e=E_{r}-E_{e}=(234023-230656)J=3367.2J

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Answer : The final temperature is, 25.0^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

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where,

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c_2 = specific heat of water = 4.18J/g^oC

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T_f = final temperature = ?

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Now put all the given values in the above formula, we get:

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