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OlgaM077 [116]
4 years ago
11

A 216-m-wide river flows due east at a uniform speed of 2.2 m/s. A boat with speed of 7.9 m/s relative to the water leaves the s

outh bank of the river pointed in a direction 34 degrees west of north. What is the magnitude of the boat's velocity relative to the ground (in m/s)
Physics
1 answer:
Mademuasel [1]4 years ago
7 0

Answer:

The magnitude of the velocity of boat w.r.t ground is 6.2015 m/s

Solution:

As per the question;

Width of the river, w = 216 m

Uniform speed of boat due east, u_{x} = 2.2\hat{i} m/s

Angle in north west direction, \theta = 34^{\circ}

Velocity of the boat relative to water, v_{bw} = 7.9m/s

Now,

Velocity of the boat relative to the water is given by:

v_{bw} = -7.9cos34^{\circ}\hat{i} + 7.9sin34^{\circ}\hat{j}

v_{bw} = - 6.55\hat{i} + 4.42\hat{j} m/s

Also, velocity of water w.r.t ground is v_{wg} = u_{x} = 2.2\hat{i} m/s

Now,

Magnitude of the velocity of boat w.r.t ground is given by:

v_{bg} = v_{bw} + v_{wg}

v_{bg} = - 6.55\hat{i} + 4.42\hat{j} + 2.2\hat{i}

v_{bg} = - 4.35\hat{i} + 4.42\hat{j} m/s

|v_{bg}| = \sqrt{(- 4.35)^{2} + (4.42)^{2}} = 6.2015\ m/s

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Answer:

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We will use law of conservation of energy to find out the velocity

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                       Potential energy = Kinetic energy

                                  m×g×h      = (1/2) m×v²

m stands  for mass

g stands for gravitational constant

h stands for height

v stands for velocity

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                      0.3×9.8×2.3 = (1/2) .3×v²

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Part B

 The potato is in uniform circular motion so the acceleration is given by

                          a = v² / r

a stands for centripetal acceleration

According to Newton second law the acceleration  is  directly proportional to applied force and inversely proportional to the mass of object

     So                     ∑ F = m × a

Two forces act on potato 1. weight due to gravitation in downward direction 2. tension in upward direction So the above equation become

                     Tension force - Force due to weight = m×a

                     T-m×g     = m×a

           centripetal acceleration formula =v²/r    ,   r = l = length of the string

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  v = 6.714 m/s ,  l = 2.30 m ,   m = .3 kg Now put values in above equation

                         T = .3 × (6.714²/2.30) + .3 × 9.8

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