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artcher [175]
3 years ago
11

Which fallacy is committed in the following argument?

Physics
1 answer:
Sveta_85 [38]3 years ago
4 0
I think the answer is c
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Which of the following is a compound?<br><br> (A) Brass<br> (B) Rust<br> (C) Iron<br> (D) Steel
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Rust is a compound of iron and oxygen.

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Pls help me with it..fast as possible...pls .....​
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Because the light only spears to part of the water so it would appear less deep

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A wind turbine is rotating at 15 rpm under steady winds flowing through the turbine at a rate of 42,000 kg/s. The tip velocity o
zzz [600]

Answer:

a) 5.22 m/s

b) 31.4 %

Explanation:

f = rotating speed = 15 rpm = 15/60 =0.25 rps

m = Mass flow rate of air = 42000 kg/s

v = Tip velocity = 250 km/h = 250/3.6 = 69.44 m/s

W = Work output = 180 kW

A = Swept area of wind turbine

r = Radius of wind turbine

η = Efficiency

r=\frac{v}{2\pi f}\\\Rightarrow r=\frac{250\div 3.6}{2\pi 0.25}=\frac{250}{1.8\pi}

A=\pi r^2\\\Rightarrow A=\pi\left(\frac{250}{1.8\pi}\right)^2\\\Rightarrow A=\left(\frac{250}{1.8}\right)^2\frac{1}{\pi}

m=\rho V\\\Rightarrow m=\rho vA\\\Rightarrow v=\frac{m}{\rho A}\\\Rightarrow v=\frac{42000}{1.31 \left(\frac{250}{1.8}\right)^2\frac{1}{\pi}}\\\Rightarrow v=5.22\ m/s

∴ The average velocity of the air is 5.22 m/s

E=m\frac{v^2}{2}=42000\frac{5.22^2}{2}\\\Rightarrow E=572538.92

\eta=\frac{W}{E}=\frac{180000}{572538.92}\\\Rightarrow \eta =0.314

∴ Conversion efficiency of the turbine is 0.314 or 31.4 %

7 0
3 years ago
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a tennis ball is thrown straight up with an initial velocity of 22.5 m/s how much total time is the ball in the air
lutik1710 [3]

Answer:

4.6s

Explanation:

v=u+at

0=22.5+(-9.8)t

-22.5=-9.8t

t=-22.5/-9.8

t=2.295 s

The total time will double

2.295×2=4.59s

=4.6s

6 0
3 years ago
A V = 108-V source is connected in series with an R = 1.1-kΩ resistor and an L = 34-H inductor and the current is allowed to rea
soldi70 [24.7K]

Answer:

Explanation:

Given an RL circuit

A voltage source of.

V = 108V

A resistor of resistance

R = 1.1-kΩ = 1100 Ω

And inductor of inductance

L = 34 H

After he inductance has been fully charged, the switch is open and it connected to the resistor in their own circuit, so as to discharge the inductor

A. Time the inductor current will reduce to 12% of it's initial current

Let the initial charge current be Io

Then, final current is

I = 12% of Io

I = 0.12Io

I / Io = 0.12

The current in an inductor RL circuit is given as

I = Io ( 1—exp(-t/τ)

Where τ is time constant and it is given as

τ = L/R = 34/1100 = 0.03091A

So,

I = Io ( 1—exp(-t/τ))

I / Io = ( 1—exp(-t/τ))

Where I/Io = 0.12

0.12 = 1—exp(-t/τ)

0.12 — 1 = —exp(-t/τ)

-0.88 = -exp(-t/0.03091)

0.88 = exp(-t/0.03091)

Take In of both sides

In(0.88) = In(exp(-t/0.03091)

-0.12783 = -t/0.030901

t = -0.12783 × 0.030901

t = 3.95 × 10^-3 seconds

t = 3.95 ms

B. Energy stored in inductor is given as

U = ½Li²

So, the current at this time t = 3.95ms

I = Io ( 1—exp(-t/τ))

Where Io = V/R

Io = 108/1100 = 0.0982 A

Now,

I = Io ( 1—exp(-t/τ))

I = 0.0982(1 — exp(-3.95 × 10^-3 / 0.030901))

I = 0.0982(1—exp(-0.12783)

I = 0.0982 × 0.12

I = 0.01178

I = 11.78mA

Therefore,

U = ½Li²

U = ½ × 34 × 0.01178²

U = 2.36 × 10^-3 J

U = 2.36 mJ

8 0
3 years ago
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