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Tasya [4]
3 years ago
15

Water flows steadily through a pipe of length L and radius R=75mm. The velocity distribution across the outlet is given by u=uma

x 1− r2 R2 and umax =3m=s. Evaluate the ratio of the x-direction momentum flux at the pipe outlet to that at the inlet.

Physics
1 answer:
ruslelena [56]3 years ago
8 0

Answer:

4/3

Explanation:

Shown in the picture attached

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a block of wood has a length of 4 cm a width of 5 cm and a height of 10 cm what is the volume of the wood
geniusboy [140]

Volume= Length X width X height.

Plug in the values for each and solve for the volume.

V= (L)(W)(H)

V=(4cm)(5cm)(10cm).


8 0
3 years ago
Can anyone help me on the summer and winter Gizmo questions. Activities A and B?
Anuta_ua [19.1K]

Answer:

sure, do you have more detailed of the assignments...?

Explanation:

6 0
2 years ago
A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless, horizontal axles. A light cord wrapped around
Alex_Xolod [135]

Answer:

\alpha =\frac{m*g*R}{I-m*R^2}

a = \frac{m*g*R^2}{I-m*R^2}

T=\frac{I*m*g}{I-m*R^2}

Explanation:

By analyzing the torque on the wheel we get:

T*R=I*\alpha    Solving for T:   T=I/R*\alpha

On the object:

T-m*g = -m*a    Replacing our previous value for T:

I/R*\alpha-m*g = -m*a

The relation between angular and linear acceleration is:

a=\alpha*R

So,

I/R*\alpha-m*g = -m*\alpha*R

Solving for α:

\alpha =\frac{R*m*g}{I+m*R^2}

The linear acceleration will be:

a =\frac{R^2*m*g}{I+m*R^2}

And finally, the tension will be:

T =\frac{I*m*g}{I+m*R^2}

These are the values of all the variables: α, a, T

8 0
3 years ago
calculate the mass of potassium chlorate (kcio3) required to obtain 10g of oxygen in the following reaction:kclO3-kcl+O2​
igor_vitrenko [27]

First, balance the reaction:

_ KClO₃   ==>   _ KCl + _ O₂

As is, there are 3 O's on the left and 2 O's on the right, so there needs to be a 2:3 ratio of KClO₃ to O₂. Then there are 2 K's and 2 Cl's among the reactants, so we have a 1:1 ratio of KClO₃ to KCl :

2 KClO₃   ==>   2 KCl + 3 O₂

Since we start with a known quantity of O₂, let's divide each coefficient by 3.

2/3 KClO₃   ==>   2/3 KCl + O₂

Next, look up the molar masses of each element involved:

• K: 39.0983 g/mol

• Cl: 35.453 g/mol

• O: 15.999 g/mol

Convert 10 g of O₂ to moles:

(10 g) / (31.998 g/mol) ≈ 0.31252 mol

The balanced reaction shows that we need 2/3 mol KClO₃ for every mole of O₂. So to produce 10 g of O₂, we need

(2/3 (mol KClO₃)/(mol O₂)) × (0.31252 mol O₂) ≈ 0.20835 mol KClO₃

KClO₃ has a total molar mass of about 122.549 g/mol. Then the reaction requires a mass of

(0.20835 mol) × (122.549 g/mol) ≈ 25.532 g

of KClO₃.

7 0
3 years ago
In the movie "gravity" - Why did the escape pod get stuck and pulled back?
victus00 [196]

Answer:

because nono gravity

Explanation:

no gravity

3 0
2 years ago
Read 2 more answers
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