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ollegr [7]
4 years ago
8

The same pipe is used to carry both air and water. For the same fluid velocity and friction factor for the air and water flows:

the pressure drop for the air flow is greater than that for the water flow. the pressure drop for the air flow equals the pressure drop for the water flow. the pressure drop for the water flow is greater than that for the air flow.

Physics
1 answer:
VladimirAG [237]4 years ago
4 0

Answer:

the pressure drop for the water flow is greater than that for the air flow.

Explanation:

Detailed analysis of the problem is show below.

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Answer:

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3 years ago
1.A test rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upwar
Charra [1.4K]

Answer:

1. t = 3.27 seconds

2. y = 147.3 m

Explanation:

Newton's Laws of Motions.

y = v₁t + 1/2 at²

a = (v₂-v₁)/t

where

y = the vertical distance travelled

v₁ = the initial velocity

v₂ = the final velocity

t = the time

a = the acceleration

final velocity is equal to 0.

So, v₂ = 0.

a = (v₂-v₁)/t

a = (0-30)/t

a = -30/t

plugin values into the first equation:

y = v₁t + 1/2 at²

49 = 30t + 1/2 (-30/t)t²

49 = 30t -15t

49 = 15 t

t = 49/15

t = 3.27 seconds

2.

y = v₁t + 1/2 at²

a = -30/3.27

a = 9.2

y = 30(3.27) + 1/2(9.2) 3.27²

y = 147.3 m

6 0
4 years ago
What is the wavelength of a 2.2 mhz ultrasound wave traveling through aluminum? assume that the speed of sound in aluminum is 51
Eva8 [605]
The frequency of a wave is equal to the linear speed divided the wavelength. so in equation form.
f = v / l
so the wavlength
l = v / f
where f is the frequency
v iss the linear speed
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7 0
3 years ago
Pleaseeeee help me with b, c, and d. There are no angles.
taurus [48]

Answer:

a. 150 J

b. 150 J

c. 0 J

d. 0 J

Explanation:

The given parameters are;

The horizontal force with which the man pulls the canister, F = 50 N

The distance he moves the vacuum cleaner, d = 3.0 m

a. Work done, W = Force applied, F × Distance moved by the force, d

Therefore, for the work done by the 50 N force on the canister, we have;

W = 50 N × 3.0 m = 150 N·m = 150 J

b. Given that he pulls the canister at a constant speed, we have;

The acceleration of the canister, a = 0 m/s²

Therefore, the net force on the canister, F_{NET} = F - F_{Friction}  = m × a

Where;

m = The mass of the canister

a = The acceleration of the canister

F = The applied force = 50 N

F_{Friction} = The force of friction

∴ F_{NET} = m × a = m × 0 m/s² = 0 N

Therefore;

F_{NET} =  F - F_{Friction} = 0 N

From which we have;

F = F_{Friction} = 50 N (The applied force, F is equal to the force of friction,

The work done by friction = The force of friction × The distance in which the force of friction acts

∴ The work done by friction = F_{Friction} × d - 50 N × 3.0 m = 150 J

The work done by friction = 150 J

c. The normal force, N acts perpendicular to the force of friction

The distance the canister moves in the perpendicular direction, d_p = 0 m

∴ The work done by the normal direction = N × d_p = N × 0 m = 0 J

The work done by the normal direction = 0 J

d. The vacuum weight, W, acts on the same line as the normal force but in the opposite direction to the normal force, N

Therefore, the weight, W, acts perpendicular to the line of motion of canister

The distance the canister moves in the direction of the weight, d_{wieght} = 0 m

Therefore, the work done by the weight = W × d_{wieght} = W × 0 m = 0 J

The work done by the weight = 0 J

7 0
3 years ago
Answer and I will give you brainiliest <br><br><br><br><br>Please I need a surely answer ​
hichkok12 [17]

Answer:

<h2><u>Constant</u></h2>

Explanation:

Please don't comment in this question's comment box

<h2>Thanks</h2>
8 0
3 years ago
Read 2 more answers
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