What is the hardest rock to break with a hammer

Water flows past a flat plate that is oriented parallel to the flow with an upstream velocity of 0.4 m/s. (a) Determine the appr

Trava [24]

Answer:

1.12 m

0.08291 m

Explanation:

u = Upstream velocity = 0.4 m/s

Re = Reynold's number = $5\times 10^5$ (turbulent)

$\nu$ = Viscosity of water = $1.12\times 10^{-6}\ Pas$

Here the flow is turbulent so we have the relation

$Re_{xcr}=\frac{ux_{cr}}{\nu}\\\Rightarrow x_{cr}=\frac{Re_{xcr}\nu}{u}\\\Rightarrow x_{cr}=\frac{5\times 10^5\times 1.12\times 10^{-6}}{0.4}\\\Rightarrow x_{cr}=1.4\ m$

The approximate location downstream from the leading edge where the boundary layer becomes turbulent is 1.4 m

Boundary layer thickness relation is given by

$\delta={\frac{\nu x}{u}}^{\frac{1}{5}}\\\Rightarrow \delta={\frac{1.12\times 10^{-6}\times 1.4}{0.4}}^{\frac{1}{5}}\\\Rightarrow \delta=0.08291\ m$

The boundary layer thickness is 0.08291 m

4 0

4 years ago

The term used to describe the quantity of matter that a body possesses is

Free_Kalibri [48]

Mass - the answer is mass. Since I need to write a bit more, the answers are not density, weight, gravity, size or anything like that!

4 0

4 years ago

Read 2 more answers

A 450.0 N, uniform, 1.50 m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tensi

bogdanovich [222]

Answer:

1) $W_{object} = 400 N$

2) x = 0.28 m from cable A.

Explanation:

1 ) Let's use the first Newton to find the , because bar is in equilibrium.

$\sum F_{Tot} = 0$

In this case we just have y-direction forces.

$\sum F_{Tot} = T_{A}+T_{B}-W_{bar}-W_{object} = 0$

Now, let's solve the equation for W(object).

$W_{object} = T_{A}+T_{B}-W_{bar} = 550 +300 - 450 = 400 N$

2 ) To find the position of the heaviest weight we need to use the torque definition.

$\sum \tau = 0$

The total torque is evaluated in the axes of the object.

Let's put the heaviest weight in a x distance from the cable A. We will call this point P for instance.

First let's find the positions from each force to the P point.

$L = 1.50 m$ ; total length of the bar.

$D_{AP} = x$ ; distance between Tension A and P point.

$D_{BP} = L-x$ ; distance between Tension B and P point.

$D_{W_{bar}P} = \frac{L}{2}-x$ ; distance between weight of the bar (middle of the bar) and P point.

Now, let's find the total torque in P point, assuming counterclockwise rotation as positive.

$\sum \tau = T_{B}(L-x)-T_{A}(x)-W_{bar}(\frac{L}{2}-x) = 0$

Finally we just need to solve it for x.

$x = \frac{T_{B}L-W_{bar}(L/2)}{T_{B}+W_{bar}+T_{A}}$

$x = 0.28 m$

So the distance is x = 0.28 m from cable A.

Hope it helps!

Have a nice day! :)

8 0

4 years ago

What was the name of the first man made satellite launched by the Soviet Union in 1957.

OverLord2011 [107]

Answer:

In 1957 the Soviet union launched Sputnik , the first satellite in space...

6 0

4 years ago

Give an example of something moving that is not accelerating

Anastaziya [24]

Anything that doesnt have a change in magnitude or direction is something thats not accelerating. so, for example, a car moving at a constant speed of 5mph on a flat road is an example of an object moving but not accelerating

6 0

3 years ago