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swat32
3 years ago
8

The idea is to get as much EMF produced from the sprinter running through it. If you were the Olympic coach on a year when there

happens to be a global energy crisis, and medals were assigned based on how much EMF (or current) were produced by the sprinters, what 3 pieces of advice (one is quite obvious; the other two involve the fixed orientation of the baton and the maintained position of the baton within the circular solenoid cross-section) would you give your sprinters in order to win?
Physics
1 answer:
never [62]3 years ago
3 0

Answer:

the greater the speed, the greater the electromotive force

* The metal pole must be parallel to the field

* you must keep the ball of the field

Explanation:

To determine the advice to the runners, let's use the Farad equation to and

           fem = -N \frac{ d \phi}{dt}  = -N \frac{ B A Cos \theta }{dt}

how the runners are moving

                 fi = B l x

            fem = -N B l v

therefore the advice we can give are:

* the greater the speed, the greater the electromotive force

* The metal pole must be parallel to the field

* you must keep the ball of the field

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A motorcyclist drove 7 km at 57km/h and then another 7 km at 81 km/h. What was the average speed? ​
alex41 [277]

<u>Answer:</u>

<em>The average speed of the car is 66.9 km/h</em>

<u>Explanation:</u>

Here distance covered with the speed <em>57 km/h=7 km  </em>

distance covered with the speed of <em>81 km/h=7 km</em>

<em>Average speed is equal to the ratio of total distance to the total time. </em>

<em>total distance= 7 + 7= 14 km  </em>

<em>time= \frac{distance}{speed} </em>

<em>time taken to cover the first 7 km= 7/57 h  </em>

<em>time taken to cover the second part of the journey = 7/81 h </em>

<em>average speed =  14/(7/57+7/81)=(14 \times 57 \times 81)/945=66.9 km/h</em>

<u><em>Shortcut: </em></u>

<em>When equal distances are covered with different speeds average speed=2 ab/(a+b) where a and b are the variable speeds in the phases. </em>

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Calculate the acceleration of a 300,000-kg jumbo jet just before takeoff when the thrust on the aircraft is 120,000 n.
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7 0
3 years ago
A radar station sends out a 250000 Hz sound wave at a speed of 340 m/s. The sound wave bounces off a weather ballon and returns
Eddi Din [679]

Answers:

a)The balloon is 68 m away of the radar station

b) The direction of the balloon is towards the radar station

Explanation:

We can solve this problem with the Doppler shift equation:

f'=\frac{V+V_{o}}{V-V_{s}} f  (1)

Where:

f=250,000 Hz is the actual frequency of the sound wave

f'=240,000 Hz is the "observed" frequency

V=340 m/s is the velocity of sound

V_{o}=0 m/s is the velocity of the observer, which is stationary

V_{s} is the velocity of the source, which is the balloon

Isolating V_{s}:

V_{s}=\frac{V(f'-f)}{f'}  (2)

V_{s}=\frac{340 m/s(240,000 Hz-250,000 Hz)}{240,000 Hz}  (3)

V_{s}=-14.16 m/s (4) This is the velocity of the balloon, note the negative sign indicates the direction of motion of the balloon: It is moving towards the radar station.

Now that we have the velocity of the balloon (hence its speed, the positive value) and the time (t=4.8 s) given as data, we can find the distance:

d=V_{s}t (5)

d=(14.16 m/s)(4.8 s) (6)

Finally:

d=68 m (8) This is the distance of the balloon from the radar station

6 0
3 years ago
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