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kirill115 [55]
3 years ago
6

An 0.80-m aluminum bar is held with its length parallel to the east-west direction and dropped from a bridge. Just before the ba

r hits the river below, its speed is 23 m/s, and the emf induced across its length is 5.5 x 10-4 V. Assuming the horizontal component of the Earth's magnetic field at the location of the bar points directly north, (a) determine the magnitude of the horizontal component of the Earth's magnetic field.(b) state whether the east end or the west end of the bar is positive.
Physics
1 answer:
Dimas [21]3 years ago
4 0

Answer:

a) B=2.9891\times 10^{-5}\,T

b)  west end of the bar is positive.

Explanation:

Given:

  • emf induced in the bar, \epsilon=5.5\times 10^{-4}\,V
  • length of the bar, l=0.8\,m
  • velocity of the bar at the given instant, v=23\,m.s^{-1}

(a)

The magnitude of the horizontal component of the Earth's magnetic field(B):

We know:

\epsilon=B.l.v

5.5\times 10^{-4}=B\times 0.8\times 23

B=2.9891\times 10^{-5}\,T

(b)

Using Fleming's left hand rule we determine that the current is flowing towards east end of the bar i.e. west end of the bar is positive.

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Answer:

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Explanation:

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Find:

a) Find V_o,y, the vertical component of the velocity of the ball when the quarterback releases it.  Express V_o,y in terms of t_c and g.

b) Find V_o,x, the initial horizontal component of velocity of the ball.   Express your answer for V_o,x in terms of D, t_c, and v_r.

c) Find the speed V_o with which the quarterback must throw the ball.  

   Answer in terms of D, t_c, v_r, and g.

d) Assuming that the quarterback throws the ball with speed V_o, find the angle Q above the horizontal at which he should throw it.

Solution:

- The vertical component of velocity V_o,y can be calculated using second kinematics equation of motion:

                               y = y(0) + V_o,y*t_c - 0.5*g*t_c^2

                              0 = 0 + V_o,y*t_c - 0.5*g*t_c^2

                               V_o,y = 0.5*g*t_c

- The horizontal component of velocity V_o,x witch which velocity is thrown can be calculated using second kinematics equation of motion:

- We know that V_i, x = V_o,x + v_r. Hence,

                               x = x(0) + V_i,x*t_c

                               D = 0 + V_i,x*t_c

                               V_o,x + v_r = D/t_c

                                V_o,x = D/t_c - v_r

- The speed with which the ball was thrown can be evaluated by finding the resultant of V_o,x and V_o,y components of velocity as follows:

                           V_o = sqrt ( V_o,x^2 + V_o,y^2)

                          V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

       

- The angle with which it should be thrown can be evaluated by trigonometric relation:

                            tan(Q) = ( V_o,y / V_o,x )

                            tan(Q) = ( (0.5*g*t_c)/ (D/t_c - v_r) )

                                   Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

                           

                               

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