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valkas [14]
3 years ago
12

A man stands still on a moving walkway that is going at a speed of 0.3 m/s to the south. What is the velocity of the man accordi

ng to a stationary observer?
Physics
2 answers:
allochka39001 [22]3 years ago
6 0

Answer:

With respect to stationary observer the velocity of man is 0.3 m/s towards South

Explanation:

As we know that here man is standing still on the moving walkway

so here we can say that velocity of man with respect to walkway is zero

\vec v_{mw} = 0

now as per relative velocity concept we know

\vec v_{mw} = \vec v_m - \vec v_w

here we know

\vec v_m = velocity of man with respect to stationary observer

\vec v_w = velocity of walkway

now from above equation we have

\vec v_m = \vec v_{mw} + \vec v_w

now plug in all values in it

\vec v_m = 0 + 0.3 m/s = 0.3 m/s towards South

RSB [31]3 years ago
3 0
The velocity is 0.3 m/s South.
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A student runs up a flight of stairs which info is not needed to calculate the rate of the student is doing work against gravity
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Answer:

B.

Explanation:

Given that a student runs up a flight of stairs which info is not needed to calculate the rate of the student is doing work against gravity A the height of the flight of stairs B the length of flight of stairs C the time taken to run up the stairs D the weight of the student

The rate of doing work is known as power.

Power = work done/time

Work done = energy = mgh

Since energy = mgh

Where mg = weight

And h = height

Time = time taken to run up the stairs.

Time, weight mg and height h will be needed while the Length of the flight is therefore not needed.

The correct answer is B.

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3 years ago
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6 0
3 years ago
In an incompressible three-dimensional flow field, the velocity components are given by u = ax + byz; υ = cy + dxz. Determine th
Ksivusya [100]

An incompressible flow field F in a 3D cartesian grid with components u,v,w:

F = u + v + w

where u,v,w are functions of x,y,z

Must satisfy:

∇·F = du/dx + dv/dy + dw/dz = 0

We have a field F defined:

F = u+v+w, u = ax+byz, v = cy+dxz

du/dx = a, dv/dy = c

Recall ∇·F = 0:

∇·F = du/dx + dv/dy + dw/dz = 0

a + c + dw/dz = 0

dw/dz = -a-c

Solve for w by separation of variables:

w = ∫(-a-c)dz

w = -az - cz + f(x,y)

f(x,y) is some undetermined function of x and y

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w = -az - cz

8 0
3 years ago
A 5 m3 tank containing 5kg of an unknown ideal gas at 500 kPa is connected through a valve to another tank containing 10 kg of t
Ivan

Answer:

a) V_{T} = 9\,m^{2}, b) m_{T} = 15\,kg, c) P_{T} = 416.667\,kPa

Explanation:

a) The equation of state for ideal gas is:

P \cdot V = \frac{m}{M}\cdot R_{u}\cdot T

Given the existence of an isothermal process, the following relation is derived:

\frac{P_{1}\cdot V_{1}}{m_{1}} = \frac{P_{2}\cdot V_{2}}{m_{2}}

The volume of the other tank is:

V_{2} = \left(\frac{m_{2}}{m_{1}} \right)\cdot \left(\frac{P_{1}}{P_{2}}\right)\cdot V_{1}

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V_{2} = 4\,m^{3}

The total volume is:

V_{T} = V_{1} + V_{2}

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c) The pressure of the gas in the two tanks is:

P_{2} = \left(\frac{m_{2}}{m_{1}} \right)\cdot \left(\frac{V_{1}}{V_{2}}\right)\cdot P_{1}

P_{T} = \left(\frac{15\,kg}{5\,kg}\right)\cdot \left(\frac{5\,m^{2}}{9\,m^{2}} \right)\cdot (500\,kPa)

P_{T} = 416.667\,kPa

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Answer:

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Explanation:

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