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Genrish500 [490]
3 years ago
6

World class swimmers can swim the 100 meter in about 55 seconds what is their approximate average speed

Physics
2 answers:
Mice21 [21]3 years ago
3 0

About 1.8181 meters per second

goldfiish [28.3K]3 years ago
3 0

Answer:

v = 1.82 m/s

Explanation:

As we know that average speed is defined as the total distance covered in total interval of time.

So here given that

total distance = 100 m

total time = 55 s

so here from above formula

v_{avg} = \frac{d}{t}

v_{avg} = \frac{100}{55}

v_{avg} = 1.82 m/s

so the average speed of the swimmer will be 1.82 m/s

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Answer:

Explanation:

Energy of signal being radiated per second on all sides = 71 x 10³ J .

At a distance of 220 m it is spread over an area of 4 π x (220)² because it is spreading uniformly on all sides.

So energy crossing per unit area

= \frac{71\times10^3}{4 \times \pi\times(220)^2}

= 11.67 x 10⁻² Wm⁻²s⁻¹.

This is the intensity of the signal.

At 2200 m this intensity will further reduce by 100 times

So there it becomes equal to

11.67 x 10⁻⁴ Wm⁻² s⁻¹.

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The relative uncertainty gives the uncertainty as a percentage of the original value. Work this out with: Relative uncertainty = (absolute uncertainty ÷ best estimate) × 100%. So in the example above: Relative uncertainty = (0.2 cm ÷ 3.4 cm) × 100% = 5.9%. The value can therefore be quoted as 3.4 cm ± 5.9%.

Explanation:

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An object is launched from the top of a building which is 100 m tall (relative to the ground) at a speed of 22 m/s at an angle o
klio [65]

Answer:

3.58\:\mathrm{s}

Explanation:

We can use the kinematics equation \Delta y=v_it+\frac{1}{2}at^2 to solve this problem. To find the initial vertical velocity, find the vertical component of the object's initial velocity using basic trigonometry for right triangles:

\sin28^{\circ}=\frac{y}{22},\\y=22\sin28^{\circ}=10.3283743813\:\mathrm{m/s}

Now we can substitute values in our kinematics equation:

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-100=-10.3283743813t+\frac{1}{2}\cdot -9.8\cdot t^2,\\\\-4.9t^2-10.3283743813t+100=0,\\\\\boxed{t=3.5849312673637455}, t=-5.692762773751501\:\text{(Extraneous)}

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