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Mazyrski [523]
2 years ago
13

If you touch the two terminals of a power supply with your two fingertips on opposite hands, the potential difference will produ

ce a current through your torso. The maximum safe current is approximately 5 {\rm mA}.If your hands are completely dry, the resistance of your body from fingertip to fingertip is approximately 500 {\rm k\Omega}. If you accidentally touch both terminals of your 120 {\rm V} household electricity supply with dry fingers, will you receive a dangerous shock?If your hands are moist, your resistance drops to approximately 1 {\rm k\Omega}. If you accidentally touch both terminals of your 120 {\rm V} household supply with moist fingers, will you receive a dangerous shock?
Physics
1 answer:
LiRa [457]2 years ago
7 0

Answer:

Yes the body will receive a dangerous shock in both cases.

Explanation:

Different parts of the body has different resistance. skin has the high resistance as compared to other organs of the body.

Dry skin has high resistance than wet skin this is because water is relatively good conductor of electricity, it adds parallel path to the current flow and hence reduces skin resistance.

Dry hands body has approximately 500 kΩ resistance and if 120 V electricity supply current received will be:

I = V/R= 120/ 500*10^3

I= 0.24 mA

Even the current seems is much lower than the safe zone but this is the case in case of DC voltage in case of AC voltage the body will receive a shock this is because the skin pass more current when the voltage is changing i.e. AC.

Similarly for wet hands body resistance is 1 kΩ. so the current through the body seems to be:

I = 120 / 1000

I = 12 mA

The current is higher than safe zone so the body will receive a dangerous shock.

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Solution: The correct order is: C, A, B

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How can we prove Earth is round and calculate its circumference?

Hypotheis:

If the sun casts shadows at different angles at the same time of day in different places, we can determine how much Earth curves.

If the Earth was flat, the angle measured at different places at the same time of the day would be same.

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In Syene, the sun's rays are vertical at noon. At the same time in Alexandria, the rays are 7.2 degrees from the vertical.

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2 years ago
Explain why the motto "Do whatever it takes to win!" may not be an ethical guideline to follow.
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If you did this then it could lead to cheating or someone else getting hurt.
4 0
2 years ago
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A particle in the first excited state of a one-dimensional infinite potential energy well (with U = 0 inside the well) has an en
nataly862011 [7]

Answer:

The energy of this particle in the ground state is E₁=1.5 eV.

Explanation:

The energy E_{n} of a particle of mass <em>m</em> in the <em>n</em>th energy state of an infinite square well potential with width <em>L </em>is:

                                                    E_{n}=\frac{n^{2}h^{2}}{8mL^{2}}

In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

                                                    E_{1}=\frac{h^{2}}{8mL^{2}}            

                                                    E_{2}=\frac{h^{2}}{2mL^{2}}

So we can rewrite the energy in the ground state as:

                                                   E_{1}=\frac{1}{4}(\frac{h^{2}}{2mL^{2}})

                                                      E_{1}=\frac{1}{4} E_{2}

                                                   E_{1}=\frac{1}{4} ( 6.0\ eV)

Finally

                                                    E_{1}=1.5\ eV

                                                   

                                                   

6 0
2 years ago
In a circus act, an acrobat rebounds upward from the surface of a trampoline at the exact moment that another acrobat, perched 9
slega [8]

Answer:

1.6 secs

Explanation:

In a circus act, an acrobat upwards from the surface of a trampoline

At that same moment another acrobat perched 9.0m above him

A ball is released from rest

While still in motion the acrobat catches the ball

He left the ball with a trampoline of 5.6m/s

Since the ball is falling downwards from a distance then acceleration will be negative

a= -9.8

s= d

s= 1/2at^2

= 1/2 × (-9.8)t^2

= 0.5× (-9.8)t^2

d = -4.9t^2

Therefore the time the acrobat stays in the air before catching the ball can be calculated as follows

9 - 4.9t^2= 5.6t + 1/2(-9.8)t^2

9 - 4.9t^2= 5.6t + (-4.9)t^2

9 - 4.9t^2= 5.6t - 4.9t^2

9= 5.6t

t= 9/5.6

t= 1.6 secs

6 0
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Answer:

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8 0
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