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Ad libitum [116K]
2 years ago
6

1. A current of 0.001 A can be felt by the human body. 0.005 A can produce a pain response. 0.015 A can cause a loss of muscle c

ontrol. In the procedures of this lesson, over 0.030 A of current traveled in the three-battery circuit. Why was this circuit safe to handle with dry hands?
Physics
1 answer:
katrin2010 [14]2 years ago
5 0

Answer:

It was safe to handle the circuit with dry hands because dry skin body resistance is very high, measuring up to 500,000 ohms.

Explanation:

Given;

Current of 0.001 A to be felt

Current of 0.005 A can produce a pain response

Current of 0.015 A can cause a loss of muscle control

Total current that traveled in the three-battery circuit = 0.03 A

Thus, we can conclude that, it was safe to handle the above mentioned circuit with dry hands because dry skin body resistance is very high, measuring up to 500,000 ohms.

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You’ve been given the challenge of balancing a uniform, rigid meter-stick with mass M = 95 g on a pivot. Stacked on the 0-cm end
Mariulka [41]

Answer: d = 4750n/3.1+95n

Explanation:

Using the principle of moment to solve the question.

Sum of clockwise moments = sum of anti clockwise moments

Since there are n identical coins with mass 3.1g placed at point 0cm, 1 coin will have mass of 3.1/n grams

Taking moment about the pivot,

Mass 3.1/n grams will move anti-clockwisely while the mass 95g will move in the clockwise direction.

Since its a meter rule (100cm) the distance from the center mass(95g) to the pivot will be 50-d (check attachment for diagram).

To get 'd'

We have 3.1/n × d = 95 × (50-d)

3.1d/n = 4750-95d

3.1d = 4750n-95dn

3.1d+95dn=4750n

d(3.1+95n) = 4750n

d = 4750n/3.1+95n

6 0
2 years ago
A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turn
Cerrena [4.2K]

Answer:

e. Not enough information to determine

Explanation:

This question is incomplete. Here is the complete question with my solution afterwards;

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turntable (initially at rest) begins to rotate with its rate of rotation constantly increasing.

What is the first event that will occur?(Assume non-zero frictional force and the same coefficients of friction for both bugs.)

a. The ladybug begins to slide

b. The gentleman bug begins to slide

c. Both bugs begin to slide at the same time

d. Nothing ever happens

e. Not enough information to determine

The centripetal force acting on a rotating body or bugs can be written as,

F=mrw^2

m= mass of the corresponding bugs

r= corresponding radial distance of each bug

w= angular speed of the turntable

The centripetal force tries to slide the bugs in an outward direction and it is directly proportional to the products of its mass and radial distance from the axis of rotation of the turntable

F ∝ mr

Since the radial distance from the axis of rotation of the turntable for each bug is given, but the mass is not given, the given information is therefore not enough to determine which bugs will slide first.

Option "e" is correct.

7 0
2 years ago
Read 2 more answers
You have a resistor and a capacitor of unknown values. First, you charge the capacitor and discharge it through the resistor. By
Fittoniya [83]

Answer:

The frequency is    f  = 0.221 \ Hz

Explanation:

From the question we are told that  

     The  time taken for it to decay to half its original size is t  =  3.40 \ ms  =  3.40 *10^{-3} \ s

Let the voltage of the capacitor when it is fully charged be  V_o

Then the voltage of the capacitor at time t is  said to be  V  =  \frac{V_o}{2}

   Now  this voltage can be  mathematical represented as

      V  =  V_o  * e ^{-\frac{t}{RC} }

Where  RC  is the time constant

   substituting values  

    \frac{V_o}{2}  =  V_o  *  e ^{-\frac{3.40 *10^{-3}}{RC} }

    0.5  =  e^{-\frac{3.40 *10^{-3}}{RC} }

    - \frac{0.5}{RC}  =  ln (0.5)

     -\frac{0.5}{RC} =  -0.6931

     RC  =  0.721

Generally the cross-over frequency for a low pass filter is mathematically represented as

          f  = \frac{1}{2 \pi  * RC  }

substituting values  

           f  = \frac{1}{2*  3.142  * 0.72  }

           f  = 0.221 \ Hz

7 0
2 years ago
A child is sliding a toy block (with mass = m) down a ramp. The coefficient of static friction between the block and the ramp is
tiny-mole [99]

Answer:

F=mg(sin(\theta )-0.25 cos(\theta ))

Explanation:

The free body diagram of the block on the slide is shown in the below figure

Since the block is in equilibrium we apply equations of statics to compute the necessary unknown forces

N is the reaction force between the block and the slide

For equilibrium along x-axis we have

\sum F_{x}=0\\\\mgsin(\theta )-\mu N-F=0\\\therefore F=mgsin(\theta)-\mu N......(\alpha )\\Similarly\\\sum F_{y}=0\\\\N-mgcos(\theta )=0\\\therefore N=mgcos(\theta ).......(\beta )\\\\

Using value of N from equation β in α we get value of force as

F=mg(sin(\theta )-\mu cos(\theta ))

Applying values we get

F=mg(sin(\theta )-0.25 cos(\theta ))

8 0
2 years ago
Read 2 more answers
Find the net electric force that the two charges would exert on an electron placed at point on the xx-axis at xx = 0.200 mm. Exp
UkoKoshka [18]

Answer:

The question has some details missing, here is the complete question ; A -3.0 nC point charge is at the origin, and a second -5.0nC point charge is on the x-axis at x = 0.800 m. Find the net electric force that the two charges would exert on an electron placed at point on the x-axis at x = 0.200 m.

Explanation:

The application of coulonb's law is used to approach the question as shown in the attached file.

6 0
2 years ago
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