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3 years ago

13

Olympic gold medalist Michael Johnson runs one time around the track 400 meters in 38 seconds what is his displacement what is h

is average velocity

1 answer:

Sveta_85 [38]3 years ago

8 0

Displacement = 0, assuming that he runs back to original position
Average velocity is displacement/ time, since displacement =0, velocity is also 0

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A bicycle takes 8.0 seconds to accelerate at a constant rate from rest to a speed of 4.0 m/s. If the mass of the bicycle and rid

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11.0 second hope i helped

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3 years ago

Water (density = 1x10^3 kg/m^3) flows at 15.5 m/s through a pipe with radius 0.040 m. The pipe goes up to the second floor of th

RUDIKE [14]

Answer:

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

Explanation:

By assuming that fluid is incompressible and there are no heat and work interaction through the line of current corresponding to the pipe, we can calculate the speed of the water floor in the pipe on the second floor by Bernoulli's Principle, whose model is:

$P_{1} + \frac{\rho\cdot v_{1}^{2}}{2}+\rho\cdot g\cdot z_{1} = P_{2} + \frac{\rho\cdot v_{2}^{2}}{2}+\rho\cdot g\cdot z_{2}$ (1)

Where:

$P_{1}$ , $P_{2}$ - Pressures of the water on the first and second floors, measured in pascals.

$\rho$ - Density of water, measured in kilograms per cubic meter.

$v_{1}$ , $v_{2}$ - Speed of the water on the first and second floors, measured in meters per second.

$z_{1}$ , $z_{2}$ - Heights of the water on the first and second floors, measured in meters.

Now we clear the final speed of the water flow:

$\frac{\rho\cdot v_{2}^{2}}{2} = P_{1}-P_{2}+\rho \cdot \left[\frac{v_{1}^{2}}{2}+g\cdot (z_{1}-z_{2}) \right]$

$\rho\cdot v_{2}^{2} = 2\cdot (P_{1}-P_{2})+\rho\cdot [v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})]$

$v_{2}^{2}= \frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})$

$v_{2} = \sqrt{\frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2}) }$ (2)

If we know that $P_{1}-P_{2} = 0\,Pa$ , $\rho=1000\,\frac{kg}{m^{3}}$ , $v_{1} = 15.5\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $z_{1}-z_{2} = -3.5\,m$ , then the speed of the water flow in the pipe on the second floor is:

$v_{2}=\sqrt{\left(15.5\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (-3.5\,m)}$

$v_{2} \approx 13.100\,\frac{m}{s}$

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

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3 years ago

What is the farthest distance at which a typical "nearsighted" frog can see clearly in air?

Umnica [9.8K]

Answer: the correct option is D (17m).

Explanation: The farthest distance at which a typical "nearsighted" frog can see clearly in air is 17m.

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3 years ago

A bicyclist passing through a city accelerates after he passes the sign post marking the city limits at x=0. His acceleration is

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Answer:

A. 24 m, 14 m/s

B. 8.0 m

Explanation:

Given:

x₀ = 6.0 m

v₀ = 4.0 m/s

a = 5.0 m/s²

t = 2.0 s

A. Find: x and v

x = x₀ + v₀ t + ½ at²

x = (6.0 m) + (4.0 m/s) (2.0 s) + ½ (5.0 m/s²) (2.0 m/s)²

x = 24 m

v = at + v₀

v = (5.0 m/s²) (2.0 s) + (4.0 m/s)

v = 14 m/s

B. Find x when v = 6.0 m/s.

v² = v₀² + 2a (x − x₀)

(6.0 m/s)² = (4.0 m/s)² + 2 (5.0 m/s²) (x − 6.0 m)

x = 8.0 m

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3 years ago

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