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3 years ago

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Suppose that after walking across a carpeted floor you reach for a doorknob and just before you touch it a spark jumps 0.95cm fr

om your finger to the knob. Find the minimum voltage needed between your finger and the doorknob to generate this spark. The minimum field that can produce a spark in air is 3.0x10^6 V/m. V = _______ V

1 answer:

alexandr402 [8]3 years ago

8 0

To solve this problem we will consider the definition of Minimum Energy, as the change between the minimum voltage and its distance. Mathematically it can be written as

$E_{min} = \frac{V_{min}}{r}$

Rearranging to find the Voltage,

$V_{min} = E_{min} \times r$

Replacing,

$V_{min} = (3.0) \times (0.95*10^4)$

$V_{min} = 28500V$

Therefore the minimum voltage needed between your finger and the doorknob to generate this spark is 28.5kV

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A transverse wave on a string is described by the wave functiony(x, t) = 0.350 sin (1.25x + 99.6t)where x and y are in meters an

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<h3>How to solve for the time interval</h3>

We have y = 0.175

y(x, t) = 0.350 sin (1.25x + 99.6t) = 0.175

sin (1.25x + 99.6t) = 0.175

sin (1.25x + 99.6t) = 0.5

99.62 = pi/6

t1 = 5.257 x 10⁻³

99.6t = pi/6 + 2pi

= 0.0683

The time interval that is between the first two instants when the element has a position of 0.175 is 0.0683.

b. we have k = 1.25, w = 99.6t

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s = vt

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brainly.com/question/25699025

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complete question

A transverse wave on a string is described by the wave function y(x, t) = 0.350 sin (1.25x + 99.6t) where x and y are in meters and t is in seconds. Consider the element of the string at x=0. (a) What is the time interval between the first two instants when this element has a position of y= 0.175 m? (b) What distance does the wave travel during the time interval found in part (a)?

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2 years ago

The position of a particle moving along an x axis is given by x = 14.0t2 - 2.00t3, where x is in meters and t is in seconds. det

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a) x = $14t^{2} -2t^{3}$

at t = 5s

$x = 14*5^{2} -2*5^{3} = 14*25 - 2*125 = 350 - 250 = 100 m$

b) v = $\frac{dx}{dt}$

= $28t - 6t^2$

at t = 5 s

v = $28*5-6*25 = 140 - 150 = -10 m/s$

c) a = $\frac{dv}{dt}$

= 28 - 12t

at t = 5 s

a = 28 -12*5= 28-60= -32 m/ $s^2$

d) At maximum positive coordinate velocity = 0

So, $0 = 28t - 6t^2$

$t = \frac{28}{6} = \frac{14}{3} = 4.66 s$

At t = 4.66 s

$X = 14 * 4.66^2 - 2* 4.66^3 = 202.39 m$

e) At t = 4.66 s

f) At maximum positive velocity a = 0

$0=28-12t$

$t = \frac{28}{12} = \frac{7}{3} = 2.33 s$

At t = 2.33 s

V = $28*2.33- 6*2.33^2= 32.67 m/s$

g) t = 2.33 s

h) When particle is not moving v = 0

So $0= 28t - 6 t^2$

$t = \frac{28}{6} = 4.66 seconds$

At t = 4.66 s

a = $28 - 12 * 4. 66 = -27.93m/s^2$

i) At t = 0s, X =0m

t = 5s, X = 100m

So, Displacement = 100m

Velocity = $\frac{Displacement}{Time} = \frac{100}{5} = 20m/s$

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3 years ago