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katen-ka-za [31]
3 years ago
7

A 1.0-cm-diameter pipe widens to 2.0, then narrows to 0.5. Liquid flows through the first segment at a speed of 4.0. What is the

volume flow rate through the pipe?
Physics
1 answer:
uysha [10]3 years ago
3 0

Answer:

3.14 ×  10⁻⁴  m³  /s

Explanation:

The flow rate (Q) of a fluid is passing through different cross-sections remains of pipe always remains the same.

Q = Area x velocity

Given:

Diameters of 3 sections of the pipe are given as  

d1  =  1.0  cm,  d2  =  2.0  cm  and  d3  =  0.5  cm.

Speed in the first segment of the pipe is  

v1  =  4  m/s.

From the equation of continuity the flow rate through different cross-sections remains the same.

Flow  rate  =  Q  =  A1  v1  =  A2  v2  =  A3  v3.

Q = A1v1

   =π/4  d²1  v1  =  π/4  * 0.01² ×4.0 m³/s  =  3.14 ×  10⁻⁴  m³  /s

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a hotgas is injected into at engine at 573 k and exhausts at 343 k. what is the highest efficiency of thisengine?
Dovator [93]

Answer:

230k

Explanation:

energy capacity maximum is 573 total available. 343k of that energy is present at release and subsequent loss leaving max used energy of 230k.

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A tornado warning means that the current weather conditions could produce a tornado. True or False
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4 0
3 years ago
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A series RLC circuit with R = 15.0 ohms, C = 4.70 uF, and L = 25.0mH is connected to an AC voltage source with V(t) = 75.0 Vrms
Brilliant_brown [7]

(a) The rms current in the circuit is 2.58 A.

(b) The rms voltage of Vab is 38.7 V, Vbc is 158.83 V, Vcd is 222.93 V, Vbd is 64.11 V, and Vad is 75 V.

(c) The average rate at which energy is dissipated in each of the 3 circuit elements is 193.23 W.

<h3>Capacitive reactance of the circuit</h3>

The capacitive reactance of the circuit is calculated as follows;

X_c = \frac{1}{\omega C} = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 550 \times 4.7 \times 10^{-6}} \\\\X_c = 61.56 \ ohms

<h3>Inductive reactance of the circuit</h3>

Xl = ωL

Xl = 2πfL

Xl = 2π x 550 x 25 x 10⁻³

Xl = 86.41 ohms

<h3>Impedance of the circuit</h3>

Z = \sqrt{R^2 + (X_L - X_C)^2} \\\\Z = \sqrt{15^2 + (86.41 -61.56)^2 } \\\\Z = 29.03 \ ohms

<h3>rms current in the circuits</h3>

I_{rms} = \frac{V_{rms}}{Z} \\\\I_{rms} = \frac{75}{29.03} \\\\I_{rms} = 2.58 \ A

<h3>rms voltage in resistor (Vab)</h3>

V_{ab} = I_{rms} R\\\\V_{ab} = 2.58 \times 15\\\\V_{ab} = 38.7 \ V

<h3>rms voltage in capacitor (Vbc)</h3>

V_{bc} = I_{rms} X_c\\\\V_{bc} = 2.58 \times 61.56\\\\V_{bc} = 158.83 \ V

<h3>rms voltage in inductor (Vcd)</h3>

V_{cd} = I_{rms} X_l\\\\V_{cd} = 2.58 \times 86.41\\\\V_{cd} = 222.93\ V

<h3>rms voltage in capacitor and inductor (Vbd)</h3>

V_{bd} = I_{rms} \times (X_l - X_c)\\\\V_{bd} = 2.58 \times (86.41 - 61.56)\\\\V_{bd} = 64.11 \ V

<h3>rms voltage in resistor, capacitor and inductor (Vad)</h3>

V_{ad} = I_{rms} \times \sqrt{R^2 + (X_l- X_c)^2} \\\\V_{ad} = 2.58 \times Z \\\\V_{ad} = 2.58 \times 29.03\\\\V_{ad} = 75 \ V

<h3>Average rate of energy dissipation in the 3 circuit element</h3>

P = I_{rms}^2 Z\\\\P = (2.58)^2 \times 29.03\\\\P = 193.23 \ W

Learn more about RLC series circuit here: brainly.com/question/15595203

6 0
2 years ago
A cat chases a mouse across a 0.63 m high
Ber [7]

Answer:

6.7 m/s

Explanation:

In the vertical direction:

y₀ = 0.63 m

y = 0 m

v₀ᵧ = 0 m/s

aᵧ = -9.81 m/s²

In the horizontal direction:

x₀ = 0 m

x = 2.4 m

aₓ = 0 m/s²

Find: v₀ₓ

First, find the time:

y = y₀ + v₀ᵧ t + ½ aᵧt²

0 = 0.63 + (0) t + ½ (-9.81) t²

t = 0.358

Now, find the velocity:

x = x₀ + v₀ₓ t + ½ aₓt²

2.4 = 0 + v₀ₓ (0.358) + ½ (0) (0.358)²

v₀ₓ = 6.70

Rounded to two significant figures, the cat's velocity when it slides off the table is 6.7 m/s.

3 0
3 years ago
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