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2 years ago

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Most workers in nanotechnology are actively monitored for excess static charge buildup. the human body acts like an insulator as

one walks across a carpet, collecting –50 nc per step. what charge buildup will a worker in a manufacturing plant accumulate if she walks 31 steps?

1 answer:

irga5000 [103]2 years ago

7 0

Sometimes arithmetic problems can be solved much more easily using the dimensional analysis approach. You focus on the units of the given information. Then, you manipulate them applying the laws of algebra where like units cancel, in order to end up with the unit of the unknown.

Given:
-50 nc/step
31 steps
Unknown: charge

Thus,
Charge = -50 nc/step * 31 steps =<em> -1550 nc</em>

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An object is moving in the plane according to these parametric equations:

aniked [119]

A. The horizontal velocity is
vx = dx/dt = π - 4πsin (4πt + π/2)
vx = π - 4π sin (0 + π/2)
vx = π - 4π (1)
vx = -3π

b. vy = 4π cos (4πt + π/2)
vy = 0

c. m = sin(4πt + π/2) / [<span>πt + cos(4πt + π/2)]

d. m = </span>sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]

e. t = -1.0

f. t = -0.35

g. Solve for t
vx = π - 4πsin (4πt + π/2) = 0
Then substitute back to solve for vxmax

h. Solve for t
vy = 4π cos (4πt + π/2) = 0
The substitute back to solve for vymax

i. s(t) = [<span>x(t)^2 + y</span>(t)^2]^(1/2)

h. s'(t) = d [x(t)^2 + y(t)^2]^(1/2) / dt

k and l. Solve for the values of t
d [x(t)^2 + y(t)^2]^(1/2) / dt = 0
And substitute to determine the maximum and minimum speeds.

5 0

2 years ago

Read 2 more answers

The surface charge density on an infinite charged plane is - 2.10 ×10−6C/m2. A proton is shot straight away from the plane at 2.

inn [45]

Explanation:

Formula to calculate electric field because of the plate is as follows.

E = $\frac{\sigma}{2 \times \epsilon_{o}}$

= $\frac{2.10 \times 10^{-6}}{2 \times 8.85 \times 10^{-12}}$

= $1.18 \times 10^{5} N/C$

Now, we will consider that equilibrium of forces are present there. So,

ma = qE

a = $\frac{1.6 \times 10^{-19} \times 1.18 \times 10^{5}}{1.67 \times 10^{-27}}$

= $1.13 \times 10^{13} m/s^2$

According to the third equation of motion,

$v^{2} = 2 \times a \times d$

or, d = $\frac{v^{2}}{2d}$

= $\frac{(2.4 \times 10^{6})^{2}}{2 \times 1.13 \times 10^{13}}$

= 0.254 m

Thus, we can conclude that the proton will travel 0.254 m before reaching its turning point.

7 0

2 years ago

Two resistors of resistances R1 and R2, with R2>R1, are connected to a voltage source with voltage V0. When the resistors are

ehidna [41]

Answer

The Value of r = 0.127

Explanation:

The mathematical representation of the two resistors connected in series is

$R_T = R_1 +R_2$

And from Ohm law

$I_s =\frac{ V}{R_T}$

$I_s = \frac{V_0}{R_1 +R_2} ---(1)$

The mathematical representation of the two resistors connected in parallel is

$R_T = \frac{1}{R_1} +\frac{1}{R_2}$

$= \frac{R_1 R_2}{R_1 +R_2}$

From the question $I_p =10I_s$

=> $I_p =10I_s = \frac{V_0 }{\frac{R_1R_2}{R_1 +R_2} } = \frac{V_0 (R_1 +R_2)}{R_1 R_2}---(2)$

Dividing equation 2 with equation 1

=> $\frac{10I_s}{I_s} =\frac{\frac{V_0 (R_1 +R_2)}{R_1 R_2}}{\frac{V_0}{R_1 +R_2}}$

$10 = \frac{(R_1+R_2)^2}{R_1 R_2}----(3)$

We are told that $r = \frac{R_1}{R_2} \ \ \ \ \ = > R_1 = rR_2$

From equation 3

$10 = \frac{(1-r)^2}{r}$

$=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1+r^2 + 2r = 10r$

$=> \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ r^2 -8r+1 = 0$

Using the quadratic formula

$r =\frac{-b\pm \sqrt{(b^2 - 4ac)} }{2a}$

a = 1 b = -8 c =1

$= \frac{8 \pm\sqrt{((-8)^2- (4*1*1))} }{2*1}$

$r= \frac{8+ \sqrt{60} }{2} \ or \ r = \frac{8 - \sqrt{60} }{2}$

$r = \ 7.87\ or \ r \ = \ 0.127$

Now r = 0.127 because it is the least value among the obtained values

4 0

2 years ago

A thin, horizontal, 18-cm-diameter copper plate is charged to -3.8 nC. Assume that the electrons are uniformly distributed on th

son4ous [18]

Answer:

Part a)

$E = 8436.7 N/C$

Part b)

$E = 8436.7 N/C$

Explanation:

Part a)

Electric field due to large sheet is given as

$E = \frac{\sigma}{2\epsilon_0}$

$\sigma = \frac{Q}{A}$

$Q = -3.8 nC$

$A = \pi(0.09)^2$

$A = 0.025 m^2$

$\sigma = \frac{-3.8\times 10^{-9}}{0.025}$

$\sigma = -1.5 \times 10^{-7} C/m^2$

now the electric field is given as

$E = \frac{-1.5 \times 10^{-7}}{2(8.85 \times 10^{-12})}$

$E = 8436.7 N/C$

Part b)

Now since the electric field is required at same distance on other side

so the field will remain same on other side of the plate

$E = 8436.7 N/C$

5 0

2 years ago

what is the resistance of a car light bulb that conducts 0.025A current when connected to a 12V car accumulator? is the current

Sophie [7]

One form of Ohm's Law says . . . . . Resistance = Voltage / Current .

R = V / I

R = (12 v) / (0.025 A)

R = (12 / 0.025) (V/I)

<em>R = 480 Ohms</em>

I don't know if the current in the bulb is steady, because I don't know what a car's "accumulator" is. (Floogle isn't sure either.)

If you're referring to the car's battery, then the current is quite steady, because the battery is a purely DC storage container.

If you're referring to the car's "alternator" ... the thing that generates electrical energy in a car to keep the battery charged ... then the current is pulsating DC, because that's the form of the alternator's output.

7 0

2 years ago