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3 years ago

5x - 1 1 - 2x3511.152<8+2

Physics

1 answer:

zheka24 [161]3 years ago

6 0

Answer:

x≥0

Explanation:

Let's find the critical points of the inequality.

−2x3511.152+5x−11=10

−2x3511.152+5x−11+−5x=10+−5x(Add -5x to both sides)

−2x3511.152−11=−5x+10

−2x3511.152−11+11=−5x+10+11(Add 11 to both sides)

−2x3511.152=−5x+21

−2x3511.152

−2=−5x+21/−2

(Divide both sides by -2)

x3511.152=5/2x+−21/2

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3 years ago

Read 2 more answers

A resonant circuit using a 286-nFnF capacitor is to resonate at 18.0 kHzkHz. The air-core inductor is to be a solenoid with clos

lukranit [14]

Answer:

The inductor contains $N = 523962.32$ loops

Explanation:

From the question we are told that

The capacitance of the capacitor is $C = 286nF = 286 * 10^{-9} \ F$

The resonance frequency is $f = 18.0 kHz = 18*10^{3} Hz$

The diameter is $d = 1.1 mm = \frac{1.1 }{1000} = 0.00011 \ m$

The of the air-core inductor is $l = 12 \ m$

The permeability of free space is $\mu_o = 4 \pi *10^{-7} \ T \cdot m/A$

Generally the inductance of this air-core inductor is mathematically represented as

$L = \frac{\mu_o * N^2 \pi d^2}{4 l}$

This inductance can also be mathematically represented as

$L = \frac{1}{w^2}$

Where $w$ is the angular speed mathematically given as

$w = 2 \pi f$

$L = \frac{1}{4 \pi ^2 f^2}$

Now equating the both formulas for inductance

$\frac{\mu_o * N^2 \pi d^2}{4 l} = \frac{1}{4 \pi ^2 f^2}$

making N the subject of the formula

$N = \sqrt{\frac{1}{(2 \pi f)^2} * \frac{4 * l }{\mu_o * \pi d^2 C} }$

$N = \frac{1}{2 \pi f} * \frac{2}{d} * \sqrt{\frac{l}{\pi * \mu_o * C} }$

Substituting value

$N = \frac{1}{ 3.142 * 18*10^{3} * 0.00011 } \sqrt{\frac{12}{ 3.142 * 4 \pi *10^{-7}* 286 *10^{-9}} }$

$N = 523962.32$ loops

4 0

3 years ago

A balloon filled with helium gas at 20°C occupies 4.91 L at 1.00 atm. The balloon is immersed in liquid nitrogen at -196°C, whil

mrs_skeptik [129]

Answer:

0.25 L

Explanation:

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$T_2$ = Final Temperature = -196 °C

$V_2$ = Final volume

From ideal gas law we have

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7 0

4 years ago

A model engine accelerated forward from rest along a straight track at 10.0 m/s2 for 3.0 seconds. It then accelerates coward at

Mrac [35]

Answer:

17.2 seconds

Explanation:

Given:

v₀ = 0 m/s

a₁ = 10.0 m/s²

t₁ = 3.0 s

a₂ = 16 m/s²

t₂ = 5.0 s

a₃ = -12 m/s²

v₃ = 0 m/s

Find: t

First, find v₁:

v₁ = a₁t₁ + v₀

v₁ = (10.0 m/s²) (3.0 s) + (0 m/s)

v₁ = 30 m/s

Next, find v₂:

v₂ = a₂t₂ + v₁

v₂ = (16 m/s²) (5.0 s) + (30 m/s)

v₂ = 110 m/s

Finally, find t₃:

v₃ = a₃t₃ + v₂

(0 m/s) = (-12 m/s²) t₃ + (110 m/s)

t₃ = 9.2 s

The total time is:

t = t₁ + t₂ + t₃

t = 3.0 s + 5.0 s + 9.2 s

t = 17.2 s

Round as needed.

4 0

3 years ago

5x - 1 1 - 2x3511.152<8+2​

5x - 1 1 - 2x3511.152<8+2