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jekas [21]
3 years ago
11

12) Water flows through a horizontal pipe of cross-sectional area 10.0 cm2 at a pressure of 0.250 atm with a flow rate is 1.00 L

/s. At a valve, the effective cross-sectional area of the pipe is reduced to 5.00 cm2. What is the pressure at the valve? The density of water is 1000 kg/m3, and treat it as an ideal incompressible fluid

Physics
1 answer:
masha68 [24]3 years ago
5 0

Answer:

The pressure after passing the valve is 23,8 [Kpa] ( 0,234 atm) and the pressure drop is about 1,53 [Kpa]

Explanation:

We need to use the formula of bernoulli, in the attached image we can see the fluid throw the pipe, we also can calculate the velocity inside the pipe using the flow rate and the cross sectional area.

For this case, we don't use the elevation difference and therefore those terms can be cancelled.

When the area has reduced the velocity of the fluid is increased but there is a drop pressure through the valve.

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Likurg_2 [28]

Answer:

The ruling separation must be between 2.2 × 10⁻⁶ m and 3.3 × 10⁻⁶ m

Explanation:

Fringe spacing w = λ·D/s

Diffraction grating, d·sin(θ) = n·λ

n = Number of lines

When θ = 90°, we have, sin(θ) = sin(90°) = 1

Therefore;

d·sin(θ) = d × 1 = d = n·λ

When n = 6 lines and d = 550-nm, we have;

d = 6 × 550-nm = 3,300 nm = 3.3 microns = 3.3 × 10⁻⁶ m

Therefore, the ruling separation must be between 2.2 × 10⁻⁶ m and 3.3 × 10⁻⁶ m.

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3 years ago
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Which property increases as an electromagnetic wave's energy decreases?
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A football kicked in front of a goal post at an angle of 45 degree to the ground just clear the top by of the post 3m high. Calc
Firlakuza [10]

Answer:

A. 10.84 m/s

B. 1.56 s

Explanation:

From the question given above, the following data were obtained:

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

Time (T) taken to hit the ground again =?

A. Determination of the velocity of projection.

Angle of projection (θ) = 45°

Maximum height (H) = 3 m

Acceleration due to gravity (g) = 9.8 m/s²

Velocity of projection (u) =?

H = u²Sine²θ / 2g

3 = u²(Sine 45)² / 2 × 9.8

3 = u²(0.7071)² / 19.6

Cross multiply

3 × 19.6 = u²(0.7071)²

58.8 = u²(0.7071)²

Divide both side by (0.7071)²

u² = 58.8 / (0.7071)²

u² = 117.60

Take the square root of both side

u = √117.60

u = 10.84 m/s

Therefore, the velocity of projection is 10.84 m/s.

B. Determination of the time taken to hit the ground again.

Angle of projection (θ) = 45°

Velocity of projection (u) = 10.84 m/s

Time (T) taken to hit the ground again =?

T = 2uSine θ /g

T = 2 × 10.84 × Sine 45 / 9.8

T = 21.68 × 0.7071 / 9.8

T = 1.56 s

Therefore, the time taken to hit the ground again is 1.56 s.

7 0
3 years ago
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