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AveGali [126]
3 years ago
9

Electric current passing through a human body can be dangerous, even fatal, depending on the amount of current, the duration of

the current, and the region of the body through which the current passes. However, currents less than about 0.5 mA are typically imperceptible. A current caused by a low potential difference typically travels through the outer layer of the skin. The resistance of the skin therefore determines how much current flows for a given potential difference. The resistance can be influenced by a number of factors, including whether the skin is wet or dry. For example, suppose an electric current of 89.0 µA follows a path through the thumb and index finger. When the skin is dry, the resistance along this path is 4.70 ✕ 105 Ω. What voltage (in V) is required for this current, in the case of dry skin? V When the skin is wet, the resistance is lowered to 2,200 Ω. What voltage (in V) is required for the same current, in the case of wet skin? V
Physics
1 answer:
Sati [7]3 years ago
7 0

Answer:

V for dry skin = 41.83 volts

V for wet skin = 0.196 volts

Explanation:

The relationship between current and voltage is given as follows by Ohm's Law:

V = IR

where,

V = voltage

I = Current

R = Resistance

FOR DRY SKIN:

V = Vd = ?

I = 89 μA = 89 x 10⁻⁶ A

R = 4.7 x 10⁵ Ω

Therefore,

Vd = (89 x 10⁻⁶ A)(4.7 x 10⁵ Ω)

<u>Vd = 41.83 volts</u>

FOR WET SKIN:

V = Vw = ?

I = 89 μA = 89 x 10⁻⁶ A

R = 2200 Ω

Therefore,

Vd = (89 x 10⁻⁶ A)(2200 Ω)

<u>Vd = 0.196 volts</u>

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A river flows due east at 1.70 m/s. A boat crosses the river from the south shore to the north shore by maintaining a constant v
ad-work [718]

Answer:

a)

v = 14.1028 m/s  

∅ = 83.0765° north of east

b)

the required distance is 40.98 m

Explanation:

Given that;

velocity of the river u = 1.70 m/s

velocity of boat v = 14.0 m/s

Now to get the velocity of the boat relative to shore;

( north of east), we say

a² + b² = c²

(1.70)² + (14.0)² = c²

2.89 + 196 = c²

198.89 = c²

c = √198.89

c = 14.1028 m/s  

tan∅ = v/u = 14 / 1.7 =  8.23529

∅ = tan⁻¹ ( 8.23529 ) = 83.0765° north of east

Therefore, the velocity of the boat relative to shore is;

v = 14.1028 m/s  

∅ = 83.0765° north of east

b)  

width of river = 340 m,

ow far downstream has the boat moved by the time it reaches the north shore in meters = ?

we say;

340sin( 90° - 83.0765°)

⇒ 340sin( 6.9235°)

= 40.98 m

Therefore, the required distance is 40.98 m

5 0
3 years ago
Lab: Thermal Energy Transfer What is the purpose of the lab, the importance of the topic, and the question you are trying to ans
VladimirAG [237]

Answer:

it's important because it shows how thermal energy transforms or continues to be all around us in everything

6 0
3 years ago
A flatbed truck is carrying a 20.0 kg crate along a level road. the coefficient of static friction between the crate and the bed
Dahasolnce [82]
<span>3.92 m/s^2 Assuming that the local gravitational acceleration is 9.8 m/s^2, then the maximum acceleration that the truck can have is the coefficient of static friction multiplied by the local gravitational acceleration, so 0.4 * 9.8 m/s^2 = 3.92 m/s^2 If you want the more complicated answer, the normal force that the crate exerts is it's mass times the local gravitational acceleration, so 20.0 kg * 9.8 m/s^2 = 196 kg*m/s^2 = 196 N Multiply by the coefficient of static friction, giving 196 N * 0.4 = 78.4 N So we need to apply 78.4 N of force to start the crate moving. Let's divide by the crate's mass 78.4 N / 20.0 kg = 78.4 kg*m/s^2 / 20.0 kg = 3.92 m/s^2 And you get the same result.</span>
6 0
3 years ago
Why do the passengers on a high-flying airplane not appear weightless, similar to the astronauts on the space station?
sergey [27]

<span>Even in space, there is still presence of gravity. The cause of weightlessness is not how far above the earth the space shuttle is but rather how fast it is travelling. The shuttle is in free fall causing weightlessness, but it is travelling fast enough to miss the earth as it falls. Similarly, the airplane could also provide weightlessness if it went free fall as well. However, that ends as the plane hits the ground. </span>

4 0
3 years ago
A race car accelerates from 0 m/s to 30.0 m/s with a displacement of
worty [1.4K]

Answer:

4. 10.0 m/s²

Explanation:

I) if initial velocity is 'v₀', the final velocity is 'v', the accelaration is 'a', the distance is 'L' and elapsed time if 't', then:

1. \ a=\frac{v-v_0}{t};

2. \ L=\frac{at^2}{2}.

II) using these two equations after substitution v₀=0; v=30 and L=45:

\left \{{{45 =\frac{at^2}{2}} \atop {a=\frac{30-0}{t} }} \right.

\left \{ {{at^2=90} \atop {at=30}} \right. \  \ \left \{ {{a=10} \atop {t=3}} \right. => \ a=10\frac{m}{s^2}

6 0
3 years ago
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