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ale4655 [162]
2 years ago
11

A diver stands on a diving platform 10.0 m above the surface of a pool and leaps upward with an initial speed of 2.5 m/s. how fa

st is the diver moving while falling past a diving board that is 3.0 m above the surface of the pool?
Physics
1 answer:
Alexus [3.1K]2 years ago
7 0
<span>The diver is heading downwards at 12 m/s Ignoring air resistance, the formula for the distance under constant acceleration is d = VT - 0.5AT^2 where V = initial velocity T = time A = acceleration (9.8 m/s^2 on Earth) In this problem, the initial velocity is 2.5 m/s and the target distance will be -7.0 m (3.0 m - 10.0 m = -7.0 m) So let's substitute the known values and solve for T d = VT - 0.5AT^2 -7 = 2.5T - 0.5*9.8T^2 -7 = 2.5T - 4.9T^2 0 = 2.5T - 4.9T^2 + 7 We now have a quadratic equation with A=-4.9, B=2.5, C=7. Using the quadratic formula, find the roots, which are -0.96705 and 1.477251164. Now the diver's velocity will be the initial velocity minus the acceleration due to gravity over the time. So V = 2.5 m/s - 9.8 m/s^2 * 1.477251164 s V = 2.5 m/s - 14.47706141 m/s V = -11.97706141 m/s So the diver is going down at a velocity of 11.98 m/s Now the negative root of -0.967047083 is how much earlier the diver would have had to jump at the location of the diving board. And for grins, let's compute how fast he would have had to jump to end up at the same point. V = 2.5 m/s - 9.8 m/s^2 * (-0.967047083 s) V = 2.5 m/s - (-9.477061409 m/s) V = 2.5 m/s + 9.477061409 m/s V = 11.97706141 m/s And you get the exact same velocity, except it's the opposite sign. In any case, the result needs to be rounded to 2 significant figures which is -12 m/s</span>
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If a rock is thrown upward on the planet mars with a velocity of 11 m/s, its height (in meters) after t seconds is given by h =
Butoxors [25]
(a) 3.56 m/s 
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(c) t = 5.9 s 
(d) -11 m/s  
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule. 
y = 11t - 1.86t^2 
y' = 11 - 3.72t 
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(b) Velocity after a seconds. 
y' = 11 - 3.72t 
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So the answer is 11 - 3.72a  
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.  
(d) Plug in the value of t calculated for (c) into the velocity function, so: 
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2 years ago
How would you solve for x I cant remember right now 4x+6x=9x-10
-Dominant- [34]
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Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t
yarga [219]

Answer

The rate at which the magnetic field is changing is  [\frac{dB}{dt} ] =  0.000467 T/s

Explanation

From the question we are told that

   The electric field strength is E =  3.5mV/m =  3.5 *10^{-3} \ V/m

    The radius is  r =  1.5 \ m

The rate of change of the  magnetic  field  is mathematically represented as

        \frac{d \phi }{dt}  =  \int\limits^{} {E \cdot dl}

Where dl is change of a unit length

     \frac{d \phi}{dt}  =  A *  \frac{dB}{dt}

Where A is the area which is mathematically represented as

     A = \pi r^2

    So

    E \int\limits^{} {  dl} =  ( \pi r^2) (\frac{dB}{dt} )  

  E L  =  ( \pi r^2) (\frac{dB}{dt} )  

where L is the circumference of the circle which is mathematically represented as

     L = 2 \pi r

So

     E (2 \pi r ) =  (\pi r^2 ) [\frac{dB}{dt} ]

      E  =   \frac{r}{2}  [\frac{dB}{dt} ]

       [\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }

substituting values

      [\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }

      [\frac{dB}{dt} ] =  0.000467 T/s    

8 0
2 years ago
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MrRa [10]

Explanation :

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3 years ago
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Ilia_Sergeevich [38]

Answer:

The lighter frog goes higher than the heavier frog.

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If both frogs have the same kinetic energy when they leave the ground, the following equality applies:

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hmax =\frac{vo^{2} }{2*g}

As the lighter frog, in order to have the same kinetic energy than the heavier one, has a greater initial velocity, it will go higher than the other.

As a consequence of both having the same kinetic energy, the lighter frog will be moving faster than the heavier frog.

5 0
2 years ago
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