Can someone pls give me the answer to this?

trasher [3.6K]

It is auxillary sorry i couldn’t help it happens to the best of us

5 0

3 years ago

how is friction losses in pipes reduced? a. decrease the pipe diameter b. increase the length of the pipes. c. decrease the leng

Citrus2011 [14]

Friction losses in pipes can be reduced by decreasing the length of the pipes, reducing the surface roughness of the pipes, and increasing the pipe diameter. Thus, options (c),(e), and (f) hold correct answers.

Friction loss is a measure of the amount of energy a piping system loses because flowing fluids meet resistance. As fluids flow through the pipes, they carry energy with them. Unfortunately, whenever there is resistance to the flow rate, it diverts fluids, and energy escapes. These opposing forces result in friction loss in pipes.

Friction loss in pipes can decrease the efficiency of the functions of pipes. These are a few ways by which friction loss in pipes can be reduced and the efficiency of the piping system can be boosted:

Decrease the length of the pipes: By decreasing pipe lengths and avoiding the use of sharp turns, fittings, and tees, whenever possible result in a more natural path for fluids to flow.
Reduce the surface roughness of the pipes: By reducing the interior surface roughness of pipes, a smooth and clearer path is provided for liquids to flow.
Increase the pipe diameter: By widening the diameters of pipes, it is ensured that fluids squeeze through pipes easily.

You can learn more about friction losses at

brainly.com/question/13348561

#SPJ4

3 0

1 year ago

What’s cat plus cat ?

iris [78.8K]

Uhmmmmm, a kitten...? Lol

5 0

3 years ago

Read 2 more answers

. Two rods, with masses MA and MB having a coefficient of restitution, e, move

GarryVolchara [31]

Answer:

a) $V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}$

$V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}$

b) $U_A = 3.66 m/s$

$V_B = 4.32 m/s$

c) Impulse = 0 kg m/s²

d) percent decrease in kinetic energy = 47.85%

Explanation:

Let $U_A$ be the initial velocity of rod A

Let $U_B$ be the initial velocity of rod B

Let $V_A$ be the final velocity of rod A

Let $V_B$ be the final velocity of rod B

Using the principle of conservation of momentum:

$M_AU_A + M_BU_B = M_AV_A + M_BV_B$ ............(1)

Coefficient of restitution, $e = \frac{V_B - V_A}{U_A - U_B}$

$V_A = V_B - e(U_A - U_B)$ ........................(2)

Substitute equation (2) into equation (1)

$M_AU_A + M_BU_B = M_A(V_B - e(U_A - U_B)) + M_BV_B$ ..............(3)

Solving for $V_B$ in equation (3) above:

$V_B = \frac{M_AU_A(1+e) + (M_B - eM_A)U_B}{M_A + M_B}$ ....................(4)

From equation (2):

$V_B = V_A + e(U_A -U_B)$ ......(5)

Substitute equation (5) into (1)

$M_AU_A + M_BU_B = M_AV_A + M_B(V_A + e(U_A -U_B))$ ..........(6)

Solving for $V_A$ in equation (6) above:

$V_A = \frac{(M_A - eM_B)U_A + M_BU_B(1+e)}{M_A + M_B}$ .........(7)

b)

$M_A = 2 kg\\M_B = 1 kg\\U_B = -3 m/s( negative x-axis)\\e = 0.65\\U_A = ?$

Rod A is said to be at rest after the impact, $V_A = 0 m/s$

Substitute these parameters into equation (7)

$0 = \frac{(2 - 0.65*1)U_A - (1*3)(1+0.65)}{2+1}\\U_A = 3.66 m/s$

To calculate the final velocity, $V_B$ , substitute the given parameters into (4):

$V_B = \frac{(2*3.66)(1+0.65) - (1 - (0.65*2))*3}{2+1}\\V_B = 4.32 m/s$

c) Impulse, $I = M_AV_A + M_BV_B - (M_AU_A + M_BU_B)$

$I = (2*0) + (1*4.32) - ((2*3.66) + (1*-3))$

I = 0 $kg m/s^2$

d) % $\triangle KE = \frac{(0.5 M_A V_A^2 + 0.5 M_B V_B^2) - ( 0.5 M_A U_A^2 + 0.5 M_B U_B^2)}{0.5 M_A U_A^2 + 0.5 M_B U_B^2} * 100\%$

% $\triangle KE = \frac{((0.5*2*0) + (0.5 *1*4.32^2)) - ( (0.5 *2*3.66^2) + 0.5*1*(-3)^2))}{ (0.5 *2*3.66^2) + 0.5*1*(-3)^2)} * 100\%$

% $\triangle KE = -47.85 \%$

7 0

3 years ago

Mrs. Pham has 8 apples. She wants to give 3/4 of the

Sergio039 [100]

The answer would be 32/3 or if you want the simplified version it’s 10 2/3.

6 0

2 years ago

Read 2 more answers