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3 years ago

• What are some methods currently used to unclog arteries and keep them clear?

1 answer:

5 0

Good fats are also called unsaturated fats. They're found in foods like olives, nuts, avocado, and fish. Cut sources of saturated fat, such as fatty meat and dairy. Choose lean cuts of meat, and try eating more plant-based meals.

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The water of a 14’ × 48’ metal frame pool can drain from the pool through an opening at the side of the pool. The opening is abo

Tresset [83]

Answer:

Explanation:

Height h = 1.03m

Volume v = 3780 gallons = 3780 * 0.0037851m^3 = 14.3073m^3

Time t = 13.5 mins = 13.5 * 60 = 810 seconds

Length of pool L = 14 inch = 14 * 2.54 = 35.56cm

width of pool b = 48 inch = 48 * 2.54 = 121.92 cm

a.) Consider the bernoulli's equation is given as:

$P_1+\rho gh_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gh_2 + \frac{1}{2}\rho v_2^2 ...(1)$

consider the equation of bernoulli at the top of the pool

$P_0+\rho gh_1 + \frac{1}{2}\rho v_1^2 =constant ...(2)$

where $P_1=P_0$ atm pressure

At the top of the pool $v_1=0m/s$ , substitute in V_1 in equation (2)

$P_0+\rho gh_1 =constant ...(3)$

Hence equation (3) serves as the bernoullis equation at the top.

b.) Consider the equation of bernoulli's at the opening of the pool

$P_2+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(4)\\P_0+\rho gh_2 + \frac{1}{2}\rho v_2^2 =constant ...(5)$

where $P_2=P_0$ atm pressure and $h_2=0m$

$P_0+\rho v_1^2 =constant ...(6)$

Hence equation (6) serves as the bernoullis equation of water at the opening of the pool.

c.) Consider the equation (3) and (4)

$P_0+\rho gh_1 =P_0+\rho v_1^2\\\\\frac{1}{2}\rho v_2^2=\rho gh_1\\v_2^2=2gh_1\\v_2=(\sqrt{2gh_1})m/s...(7)$

Hence velocity is $v_2=(\sqrt{2gh_1})m/s$

d.) consider (7)

$v_2=(\sqrt{2(9.81)(1.03)})=4.4954m/s(approx)$

This is the norminal value of velocity

e.) consider the equation of flow rate interval of v and t

flow(t)= $\frac{dv}{dt}(m^3/s)$ hence this is the flow rate

f.) Consider the equation cross sectional area in terms of V,v2 and t

$AV_2=\frac{v}{t}\\\\A=\frac{v}{v_2t}(m^2)...8$

hence this serves as the cross sectional area.

g.) Consider the equation of area from equation (8)

$A=\frac{v}{v_2t}\\=\frac{14.3073}{4.4954\times 810}=0.003929=0.00393m^2=39.3cm^2$

6 0

3 years ago

A socket can be driven using any of the following except for (A) a socket ratchet

AfilCa [17]

Answer:

a socket can be driven by a flexible head socket wrench, I believe

4 0

3 years ago

Calculate the load, PP, that would cause AA to be displaced 0.01 inches to the right. The wires ABAB and ACAC are A36 steel and

Nataly [62]

Answer:

P = 4.745 kips

Explanation:

Given

ΔL = 0.01 in

E = 29000 KSI

D = 1/2 in

LAB = LAC = L = 12 in

We get the area as follows

A = π*D²/4 = π*(1/2 in)²/4 = (π/16) in²

Then we use the formula

ΔL = P*L/(A*E)

For AB:

ΔL(AB) = PAB*L/(A*E) = PAB*12 in/((π/16) in²*29*10⁶ PSI)

⇒ ΔL(AB) = (2.107*10⁻⁶ in/lbf)*PAB

For AC:

ΔL(AC) = PAC*L/(A*E) = PAC*12 in/((π/16) in²*29*10⁶ PSI)

⇒ ΔL(AC) = (2.107*10⁻⁶ in/lbf)*PAC

Now, we use the condition

ΔL = ΔL(AB)ₓ + ΔL(AC)ₓ = ΔL(AB)*Cos 30° + ΔL(AC)*Cos 30° = 0.01 in

⇒ ΔL = (2.107*10⁻⁶ in/lbf)*PAB*Cos 30°+(2.107*10⁻⁶ in/lbf)*PAC*Cos 30°= 0.01 in

Knowing that PAB*Cos 30°+PAC*Cos 30° = P

we have

(2.107*10⁻⁶ in/lbf)*P = 0.01 in

⇒ P = 4745.11 lb = 4.745 kips

The pic shown can help to understand the question.

5 0

3 years ago

Working with which of these systems requires a technician that has been certified in an EPA-approved course?

makvit [3.9K]

EPA Regulations provides a certified course for the technicians involved in the Air-conditioning system.

Answer: Option (b)

<u>Explanation:</u>

The EPA regulation has implemented an act called the "Clean Air Act" under the "section of 609".

This act provides some basic requirements for EPA Regulation such as follows;

Refrigerant: This unit must be approved by EPA Regulations before being implemented into the atmosphere.
Servicing: This system provides a certified course for technicians in service and also approve them with proper refrigerant equipment.
Reuse Refrigerants: The use of recycled refrigerants must be properly monitored before it comes in to serve.

6 0

3 years ago

At the instant under consideration, the hydraulic cylinder AB has a length L = 0.75 m, and this length is momentarily increasing

Inessa [10]

Answer:

vB = - 0.176 m/s (↓-)

Explanation:

Given

(AB) = 0.75 m

(AB)' = 0.2 m/s

vA = 0.6 m/s

θ = 35°

vB = ?

We use the formulas

Sin θ = Sin 35° = (OA)/(AB) ⇒ (OA) = Sin 35°*(AB)

⇒ (OA) = Sin 35°*(0.75 m) = 0.43 m

Cos θ = Cos 35° = (OB)/(AB) ⇒ (OB) = Cos 35°*(AB)

⇒ (OB) = Cos 35°*(0.75 m) = 0.614 m

We apply Pythagoras' theorem as follows

(AB)² = (OA)² + (OB)²

We derive the equation

2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB

⇒ (AB)*(AB)' = (OA)*vA + (OB)*vB

⇒ vB = ((AB)*(AB)' - (OA)*vA) / (OB)

then we have

⇒ vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)

⇒ vB = - 0.176 m/s (↓-)

The pic can show the question.

7 0

3 years ago

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