Need help with these 3 questions pls help

If you touch a downed power line, covered or bare, what's the likely outcome?

olya-2409 [2.1K]

Answer:

you get electrocuted...........

5 0

3 years ago

Read 2 more answers

Can i join three 12 volts batteriesto give me 24 volts output

bulgar [2K]

Answer:

YES

Explanation:

If we connect batteries in series then the output voltage is the sum of the individual voltage of each battery i.e if you connect three 12 volts batteries in series then their output voltage will be 12+12+12=36 volts, but the current rating of the batteries in series will be same of the individual battery rating in 'mah'.

But when we connect the batteries in parallel their voltage is not added but their current rating in mah is addition of their individual rating.

So, If you want 24 volts from three 12 volts battery then you can connect two of them in series and the other one in parallel with them this will give 24 volts and the current will be addition of the two series batteries and the third which is in parallel with them. You can use this configuration if current value is not a big factor.

8 0

3 years ago

Please answer fast. With full step by step solution.

lina2011 [118]

Let f(z) = (4z ² + 2z) / (2z ² - 3z + 1).

First, carry out the division:

f(z) = 2 + (8z - 2) / (2z ² - 3z + 1)

Observe that

2z ² - 3z + 1 = (2z - 1) (z - 1)

so you can separate the rational part of f(z) into partial fractions. We have

(8z - 2) / (2z ² - 3z + 1) = a / (2z - 1) + b / (z - 1)

8z - 2 = a (z - 1) + b (2z - 1)

8z - 2 = (a + 2b) z - (a + b)

so that a + 2b = 8 and a + b = 2, yielding a = -4 and b = 6.

So we have

f(z) = 2 - 4 / (2z - 1) + 6 / (z - 1)

or

f(z) = 2 - (2/z) (1 / (1 - 1/(2z))) + (6/z) (1 / (1 - 1/z))

Recall that for |z| < 1, we have

$\displaystyle\frac1{1-z}=\sum_{n=0}^\infty z^n$

Replace z with 1/z to get

$\displaystyle\frac1{1-\frac1z}=\sum_{n=0}^\infty z^{-n}$

so that by substitution, we can write

$\displaystyle f(z) = 2 - \frac2z \sum_{n=0}^\infty (2z)^{-n} + \frac6z \sum_{n=0}^\infty z^{-n}$

Now condense f(z) into one series:

$\displaystyle f(z) = 2 - \sum_{n=0}^\infty 2^{-n+1} z^{-(n+1)} + 6 \sum_{n=0}^\infty z^{-n-1}$

$\displaystyle f(z) = 2 - \sum_{n=0}^\infty \left(6+2^{-n+1}\right) z^{-(n+1)}$

$\displaystyle f(z) = 2 - \sum_{n=1}^\infty \left(6+2^{-(n-1)+1}\right) z^{-n}$

$\displaystyle f(z) = 2 - \sum_{n=1}^\infty \left(6+2^{2-n}\right) z^{-n}$

So, the inverse Z transform of f(z) is $\boxed{6+2^{2-n}}$ .

4 0

3 years ago

An inductor (L = 400 mH), a capacitor (C = 4.43 µF), and a resistor (R = 500 Ω) are connected in series. A 44.0-Hz AC generator

MakcuM [25]

Answer:

(A) Maximum voltage will be equal to 333.194 volt

(B) Current will be leading by an angle 54.70

Explanation:

We have given maximum current in the circuit $i_m=385mA=385\times 10^{-3}A=0.385A$

Inductance of the inductor $L=400mH=400\times 10^{-3}h=0.4H$

Capacitance $C=4.43\mu F=4.43\times 10^{-3}F$

Frequency is given f = 44 Hz

Resistance R = 500 ohm

Inductive reactance will be $x_l=\omega L=2\times 3.14\times 44\times 0.4=110.528ohm$

Capacitive reactance will be equal to $X_C=\frac{1}{\omega C}=\frac{1}{2\times 3.14\times 44\times 4.43\times 10^{-6}}=816.82ohm$

Impedance of the circuit will be $Z=\sqrt{R^2+(X_C-X_L)^2}=\sqrt{500^2+(816.92-110.52)^2}=865.44ohm$

So maximum voltage will be $\Delta V_{max}=0.385\times 865.44=333.194volt$

(B) Phase difference will be given as $\Phi =tan^{-1}\frac{X_C-X_L}{R}=\frac{816.92-110.52}{500}=54.70$

So current will be leading by an angle 54.70

5 0

3 years ago

Explain the conductivity results observed for ionic compounds in the solid state and in aqueous solution.i.Solid ionic compounds

Svetach [21]

Answer:

Aqueous solution of ionic compounds conduct electricity while solid ionic compounds don't.

Explanation:

Ionic compound conduct electricity when liquid or in aqueous solution that is resolved in water because the ionic bonds of the compound become weak and the ions are free to move from place to place.

Ionic compounds don't conduct electricity while in solid state because the ionic bonds are to strong and ions cannot move around with lack of space for movement which makes the electric conductivity zero.

8 0

2 years ago