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3 years ago

Mercury is in the 80th position in the periodic table. How many protons does it have?

Physics

2 answers:

Verdich [7]3 years ago

7 0

Mercury has 80 protons. Ironic?

Amanda [17]3 years ago

6 0

Mercury has 80 protons between 120-121 neutron and 80 electron

Hope I help u

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A feeling of worry or uneasiness, often triggered by an event with an uncertain outcome is defined as:

juin [17]

The answer is Anxiety. Hope this helps.

4 0

4 years ago

An AC source operating at 59 Hz with a maximum voltage of 170 V is connected in series with a resistor (R = 1.2 kΩ) and an induc

Alexxandr [17]

I = V/Z

V = voltage, I = current, Z = impedance

First let's find the total impedance of the circuit.

The impedance of the resistor is:

$Z_{R}$ = R

R = resistance

Given values:

R = 1200Ω

Plug in:

$Z_{R}$ = 1200Ω

The impedance of the inductor is:

$Z_{L}$ = j2πfL

f = source frequency, L = inductance

Given values:

f = 59Hz, L = 2.4H

Plug in:

$Z_{L}$ = j2π(59)(2.4) = j889.7Ω

Add up the individual impedances to get the Z, and convert Z to polar form:

Z = $Z_{R}$ + $Z_{L}$

Z = 1200 + j889.7

Z = 1494∠36.55°Ω

I = V/Z

Given values:

V = 170∠0°V (assume 0 initial phase)

Z = 1494∠36.55°Ω

I = 170∠0°/1494∠36.55°Ω

I = 0.1138∠-36.55°A

Round the magnitude of I to 2 significant figures and now you have your maximum current:

I = 0.11A

5 0

3 years ago

How does`ezzzzz321erfq23c2f

Nataliya [291]

Lol what???? i don’t understand

7 0

3 years ago

An 800-kHz radio signal is detected at a point 2.7 km distant from a transmitter tower. The electric field amplitude of the sign

dimaraw [331]

Answer:

Option D is correct: 170 µW/m²

Explanation:

Given that,

Frequency f = 800kHz

Distance d = 2.7km = 2700m

Electric field Eo = 0.36V/m

Intensity of radio signal

The intensity of radial signal is given as

I = c•εo•Eo²/2

Where c is speed of light

c = 3×10^8m/s

εo = 8.85 × 10^-12 C²/Nm²

I = 3×10^8 × 8.85×10^-12 × 0.36²/2

I = 1.72 × 10^-4W/m²

I = 172 × 10^-6 W/m²

I = 172 µW/m²

Then, the intensity of the radio wave at that point is approximately 170 µW/m²

7 0

3 years ago

A certain radioactive nuclide decays with a disintegration constant of 0.0178 h-1.

umka21 [38]

Explanation:

Given that,

The disintegration constant of the nuclide, $\lambda=0.0178\ h^{-1}$

(a) The half life of this nuclide is given by :

$t_{1/2}=\dfrac{ln(2)}{\lambda}$

$t_{1/2}=\dfrac{ln(2)}{0.0178}$

$t_{1/2}=38.94\ h$

(b) The decay equation of any radioactive nuclide is given by :

$N=N_oe^{-\lambda t}$

$\dfrac{N}{N_o}=e^{-\lambda t}$

Number of remaining sample in 4.44 half lives is :

$t_{1/2}=4.44\times 38.94$

$t_{1/2}=172.89\ h^{-1}$

So, $\dfrac{N}{N_o}=e^{-0.0178\times 172.89}$

$\dfrac{N}{N_o}=0.046$

$t_{1/2}=14.6\times 24$

$t_{1/2}=350.4\ h^{-1}$

So, $\dfrac{N}{N_o}=e^{-0.0178\times 350.4}$

$\dfrac{N}{N_o}=0.0019$

Hence, this is the required solution.

4 0

3 years ago