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steposvetlana [31]
3 years ago
8

1)why do we get electric shock while holding a live wire barefooted and not when wearing rubber shoes?

Physics
1 answer:
katrin2010 [14]3 years ago
5 0
Electricity is always going to take the path of least resistance to ground. The rubber in your shoes is not a conductor of electricity, therefore you are not completing the circuit and you don't get shocked. Your bare feet, on the other hand ARE conductors of electricity, so when you hold the wire, you complete the circuit and become the path of least resistance to ground... ZAP!
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A street musician sounds the A string of his violin, producing a tone of 440 Hz, What frequency does a bicyclist hear as he a) a
DanielleElmas [232]

Answer:

a) f_o=454.11Hz

b)f_o=425.89Hz

Explanation:

Let´s use Doppler effect, in order to calculate the observed frequency by the byciclist. The Doppler effect equation for a general case is given by:

f_o=\frac{v\pm v_o}{v\pm v_s} *f_s

where:

f_o=Observed\hspace{3}frequency

f_s=Actual\hspace{3}frequency

v=Speed\hspace{3}of\hspace{3}the\hspace{3}sound\hspace{3}waves

v_s=Velocity\hspace{3}of\hspace{3}the\hspace{3}source

v_o=Velocity\hspace{3}of\hspace{3}the\hspace{3}observer

Now let's consider the next cases:

+v_o\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}observer\hspace{3}moves\hspace{3}towards\hspace{3}the\hspace{3}source

-v_o\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}observer\hspace{3}moves\hspace{3}away\hspace{3}from\hspace{3}the\hspace{3}source

-v_s\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}source\hspace{3}moves\hspace{3}towards\hspace{3}the\hspace{3}observer

+v_s\hspace{3}is\hspace{3}used\hspace{3}when\hspace{3}the\hspace{3}source\hspace{3}moves\hspace{3}away\hspace{3}from\hspace{3}the\hspace{3}observer

The data provided by the problem is:

f_s=440Hz\\v_o=11m/s

The problem don't give us aditional information about the medium, so let's assume the medium is the air, so the speed of sound in air is:

v=343m/s

Now, in the first case the observer alone is in motion towards to the source, hence:

f_o=\frac{v+v_o}{v}*f_s=\frac{343+11}{343} *440=454.1107872Hz

Finally, in the second case the observer alone is in motion away from the source, so:

f_o=\frac{v-v_o}{v}*f_s=\frac{343-11}{343} *425.8892128Hz

6 0
4 years ago
Why does an ice cube melt when you hold it in your hand? Question 1 options: Heat from the ice cube is transferred to your hand.
Sunny_sXe [5.5K]
Your hand transferred temperature to he ice cube
8 0
3 years ago
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A 30.4-newton force is used to slide a 40.0-newton crate a distance of 6.00 meters at constant speed along an incline to a verti
Angelina_Jolie [31]

Hi there!

Since the crate is being slid at a constant speed, the forces sum to 0 N. In this instance, the following forces occur in the axis of interest:

Wsinθ = downward acceleration along incline due to gravity (N)

Fκ = kinetic friction force along incline (N)

A = applied force (N)

The acceleration due to gravity and friction force act in the same direction, so:

Wsinθ + Fκ = A

Solve for sinθ using right triangle trigonometry:

sinθ = O/H = 3/6 = 0.5

Rearrange the equation for the force of kinetic friction and solve:

Fκ = A - 0.5W

Fκ = 30.4 - 20 = 10.4 N

Now, recall that:

Work = Force × displacement (W = F × d)

Since the box's displacement is in the same axis as the force but OPPOSITE direction, we must use:

W = Fdcosθ

Angle between displacement and friction force is 180°.

cos(180) = -1

Work done by friction = -Fd = -10.4(6) = -62.4 J

8 0
3 years ago
The pupil of a cat's eye narrows to a slit width of 0.5 mm in daylight. What is the angular resolution of the cat's eye in dayli
Elenna [48]

Answer:

C. 10⁻³ rads

Explanation:

Here, we shall use Rayleigh's Criterion to find out the angular resolution of Cat's eye during day light. Rayleigh's Criterion is written as follows:

θ = λ/a

where,

θ = angular resolution of Cat's eye = ?

λ = wavelength = 500 nm = 5 x 10⁻⁷ m

a = slit width of eye = 0.5 mm = 5 x 10⁻⁴ m

Therefore,

θ = (5 x 10⁻⁷ m/5 x 10⁻⁴ m)

Therefore,

θ = 0.001

θ = Sin⁻¹(0.001)

θ = 0.001 rad = 1 x 10⁻³ rad

Hence, the correct answer is:

<u>C. 10⁻³ rads</u>

4 0
3 years ago
An object on a straight line travels from point A to point B at a constant speed of 60 kmh-1 and returns to point A at a constan
Kamila [148]

Answer:

55kmh

Explanation:

first of all according to formula add both of the velocity and divide it by 2

5 0
3 years ago
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