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Lina20 [59]
3 years ago
10

Imagine that you are in a bleachers watching a swim meet in which your friend is competing in the freestyle event. At the instan

t the starting gun fires, the lights go out! When the lights come back on, the timer on the scoreboard records 86s. You observe your friend is now about halfway along the length of pool, swimming in a direction opposite to that in which he started. The pool is 50 m long.
what are two possible distances that you might infer your friends swam while the lights were out, given the average velocity is 0.29 meters per second
Physics
1 answer:
kramer3 years ago
7 0

The two possible distances that you might infer your friends swam while the lights were out are 25 m and 75 m.

Answer:

Explanation:

So you have to measure the distance covered by your friend in a time gap of 86 s. And the average velocity is given as 0.29 m/s.

Then as per the mathematical calculation of velocity, distance can be measured as the product of velocity with time interval.

Distance = Velocity × Time Interval

Distance = 0.29 m/s×86 s = 24.94 m.

So based on this calculation, one of the possible distance inferred by you will be 24.94 m.

Another possible distance can be guessed from the statements provided. So if the length of pool is 50 m, then covering halfway in opposite direction to his starting direction means completion of one full length i.e., 50 m and then halfway of that 50 m which is 25m, so totally 50 +25 = 75 m.

So in other way, we can assume that your friend has covered 75 m distance during the light out.

Thus, the two possible distances that you might infer your friends swam while the lights were out are 25 m and 75 m.

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The reaction is at dynamic equilibrium.
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Answer:

Nitrogen and hydrogen combine at the same rate that ammonia breaks down.

Explanation:

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2 years ago
A major-league pitcher can throw a ball in excess of 40.1 m/s. If a ball is thrown horizontally at this speed, how much will it
mote1985 [20]

Answer:

The ball will drop 0.881 m by the time it reaches the catcher.

Explanation:

The position of the ball at time "t" is described by the position vector "r":

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

Where:

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

When the ball reaches the catcher, the position vector will be "r final" (see attached figure).

The x-component of the vector "r final", "rx final", will be 17.0 m. We have to find the y-component.

Using the equation of the x-component of the position vector, we can calculate the time it takes the ball to reach the catcher (notice that the frame of reference is located at the throwing point so that x0 and y0 = 0):

x = x0 + v0x · t

17.0 m = 0 m + 40.1 m/s · t

t = 17.0 m/ 40. 1 m/s = 0.424 s

With this time, we can calculate the y-component of the vector "r final", the drop of the ball:

y = y0 + v0y · t + 1/2 · g · t²

Initially, there is no vertical velocity, then, v0y = 0.

y = 1/2 · g · t²

y = -1/2 · 9.8 m/s² · (0.424 s)²

y = -0.881 m

The ball will drop 0.881 m by the time it reaches the catcher.

8 0
3 years ago
A ball is thrown upwards at an unknown speed. in a time of
alexandr402 [8]

Answer:

a)  Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

c) Maximum height reached = 10.65 m

Explanation:

a)  We have equation of motion s=ut+\frac{1}{2} at^2, where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8m/s^2

 Substituting

    6=u*0.5-\frac{1}{2} *9.8*0.5^2\\ \\ u=14.45m/s

 Initial speed of the ball = 14.45 m/s

 b) We have equation of motion v^2=u^2+2as, where v is the final velocity

   s = 6 m, u = 14.45 m/s, a = -9.8m/s^2

    Substituting

        v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s

  So at height 6 m speed of ball = 9.55 m/s

c) We have equation of motion v^2=u^2+2as, where v is the final velocity

   u = 14.45 m/s, v =0 , a = -9.8m/s^2

   Substituting

     0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m

  Maximum height reached = 10.65 m

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