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Butoxors [25]
3 years ago
14

2. How can we make the percent error value as minimal as possible?

Physics
2 answers:
andriy [413]3 years ago
8 0

Answer:

Percent error can be reduced by improving both your accuracy and precision.

Molodets [167]3 years ago
3 0

Answer:

Percent error can be reduced by improving both your accuracy and precision.

Explanation:

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The small currents in axons corresponding to nerve impulses produce measurable magnetic fields. a typical axon carries a peak cu
Gemiola [76]

Answer:

6.66\cdot 10^{-12}T

Explanation:

The magnetic field produced by a current-carrying wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0 is the vacuum permeability

I is the current

r is the distance from the wire

In this problem we have

I=0.040 \mu A=4\cdot 10^{-8}A

r = 1.2 mm = 0.0012 m

So the magnetic field strength is

B=\frac{(4\pi \cdot 10^{-7} H/m)(4\cdot 10^{-8}A)}{2\pi (0.0012 m)}=6.66\cdot 10^{-12}T

5 0
3 years ago
A 2.02 nF capacitor with an initial charge of 4.55 µC is discharged through a 1.22 kΩ resistor. (a) Calculate the current in the
Mandarinka [93]

Answer:

a) 0.048A

b) 0.18µC

c) 1.85A

Explanation:

The discharged current of the capacitor as a function of time is given by:

i=\frac{q_o}{RC}*e^{-\frac{t}{\tau}}\\where:\\\tau=RC\\

\tau=1.22*10^3*2.02*10^{-9}\\\tau=2.46*10^{-6}s

a)

i=\frac{4.55\µC}{2.46\µs}*e^{-\frac{9\µs}{2.46\µs}}\\\\i=0.048A

b)

q=q_o*e^{-\frac{t}{\tau}}

q=4.55\µC*e^{-\frac{8\µs}{2.46\µs}}\\q=0.18\µC

c) the maximum current occurs when t=0

i=\frac{4.55\µC}{2.46\µs}*e^{-\frac{0\µs}{2.46\µs}}\\\\i=1.85A

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Two ways that radio waves are used for transmitting information. Help me please.
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