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A 7.5-cmcm-diameter horizontal pipe gradually narrows to 4.5 cmcm . When water flows through this pipe at a certain rate, the ga

tino4ka555 [31]

Answer

given,

diameter,d₁ = 7.5 cm

d₂ = 4.5 cm

P₁ = 32 kPa

P₂ = 25 kPa

Assuming, we have calculation of flow in the pipe

using continuity equation

A₁ v₁ = A₂ v₂

π r₁² v₁ = π r₂² v₂

$v_1= \dfrac{r_2^2}{r_1^2} v_2$

$v_1= \dfrac{2.25^2}{3.75^2} v_2$

$v_1= 0.36 v_2$

Applying Bernoulli's equation

$\Delta P = \dfrac{1}{2}\rho (v_2^2-v_1^2)$

$P_1-P_2 = \dfrac{1}{2}\rho (v_2^2-(0.36 v_2)^2)$

$32-25 = \dfrac{1}{2}1000\times v_2^2 (1 - 0.1269)$

$v_2=\sqrt{\dfrac{2\times 7\times 10^3}{1000\times (0.8704)}}$

$v_2=\sqrt{16.084}$

v₂ = 4.01 m/s

fluid flow rate

Q = A₂ V₂

Q = π (0.0225)² x 4.01

Q = 6.38 x 10⁻³ m³/s

flow in the pipe is equal to 6.38 x 10⁻³ m³/s

4 0

2 years ago

A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24

alex41 [277]

Answer:

h=12.41m

Explanation:

N=392

r=0.6m

w=24 rad/s

$I=0.8*m*r^{2}$

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion

$N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}$

$m=40kg$

moment of inertia

$M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}$

Kinetic energy of the rotation motion

$K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J$

Kinetic energy translational

$K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J$

Total kinetic energy

$K=3317.79J+4147.2J\\K=7464.99J$

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

$K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m$

6 0

3 years ago

Imagine that the apparent weight of the crown in water is Wapparent=4.50N, and the actual weight is Wactual=5.00N. Is the crown

9966 [12]

Answer:

Explanation:

Actual weight, Wo = 5 N

Apparent weight, W = 4.5 N

density of water = 1 g/cm^3 = 1000 kg/m^3

density of gold, = 19.32 g/cm^3 = 19.32 x 1000 kg/m^3

Buoyant force = Actual weight - Apparent weight

Volume x density of water x g = 5 - 4.5

V x 1000 x 9.8 = 0.5

V = 5.1 x 10^-6 m^3

Weight of gold = Volume of gold x density of gold x gravity

W' = 5.1 x 10^-6 x 19.32 x 1000 x 9.8 = 0.966 N

As W' is less than W so, it is not pure gold.

4 0

3 years ago

Arcsin 0.9331 in degrees

lisabon 2012 [21]

Answer:

68.9233231661

Explanation:

Just put it into your calculator, shift sin should do it but it will come up like this: $sin^{-1}$ which is the same as arcsin

6 0

2 years ago

Why is the recruitment and selection of staff the concern of all managers in an organization?

leva [86]

Answer:

Explanation:

It is a concern to managers in an organization because it is important for the organization to get the number and quality of staffs needed to attain the goals and objectives of the organization.

Selection and recruitment is also important in order to ensure the continuity of the company , it will also fulfills the organisations job requirements and also this provide a pool of employees in which the management will have to choose the right or best candidates for the job position.

7 0

2 years ago

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