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1 year ago

In the sum of 4.34 + 5.66, the number of significant figures is

Physics

1 answer:

kifflom [539]1 year ago

4 0

Answer:

Explanation:

4.34 + 5.66 = 10

number of significant figures in 10: 1

(the significant figure here is 1)

here's a clear explanation from Northern Kentucky University (NKU)

[To determine the number of significant figures in a number use the following 3 rules:

Non-zero digits are always significant

Any zeros between two significant digits are significant

A final zero or trailing zeros in the decimal portion ONLY are significant

Example: .500 or .632000 the zeros are significant

.006 or .000968 the zeros are NOT significant ]

[For addition and subtraction use the following rules:

Count the number of significant figures in the decimal portion ONLY of each number in the problem

Add or subtract in the normal fashion

Your final answer may have no more significant figures to the right of the decimal than the LEAST number of significant figures in any number in the problem.]

(https://www.nku.edu/~intsci/sci110/worksheets/rules_for_significant_figures.html#:~:text=To%20determine%20the%20number%20of,decimal%20portion%20ONLY%20are%20significant)

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2 years ago

A 217 Ω resistor, a 0.875 H inductor, and a 6.75 μF capacitor are connected in series across a voltage source that has voltage a

Nataly [62]

For an AC circuit:

I = V/Z

V = AC source voltage, I = total AC current, Z = total impedance

Note: We will be dealing with impedances which take on complex values where j is the square root of -1. All phasor angles are given in radians.

For a resistor R, inductor L, and capacitor C, their impedances are given by:

$Z_{R}$ = R

R = resistance

$Z_{L}$ = jωL

ω = voltage source angular frequency, L = inductance

$Z_{C}$ = -j/(ωC)

ω = voltage source angular frequency, C = capacitance

Given values:

R = 217Ω, L = 0.875H, C = 6.75×10⁻⁶F, ω = 220rad/s

Plug in and calculate the impedances:

$Z_{R}$ = 217Ω

$Z_{L}$ = j(220)(0.875) = j192.5Ω

$Z_{C}$ = -j/(220×6.75×10⁻⁶) = -j673.4Ω

Add up the impedances to get the total impedance Z, then convert Z to polar form:

Z = $Z_{R}$ + $Z_{L}$ + $Z_{C}$

Z = 217 + j192.5 - j673.4

Z = (217-j480.9)Ω

Z = (527.6∠-1.147)Ω

Back to I = V/Z

Given values:

V = (30.0∠0+220t)V (assume 0 initial phase, and t = time)

Z = (527.6∠-1.147)Ω (from previous computation)

Plug in and solve for I:

I = (30.0∠0+220t)/(527.6∠-1.147)

I = (0.0569∠1.147+220t)A

To get the voltages of each individual component, we'll just multiply I and each of their impedances:

$v_{R}$ = I× $Z_{R}$

$v_{L}$ = I× $Z_{L}$

$v_{C}$ = I× $Z_{C}$

Given values:

I = (0.0569∠1.147+220t)A

$Z_{R}$ = 217Ω = (217∠0)Ω

$Z_{L}$ = j192.5Ω = (192.5∠π/2)Ω

$Z_{C}$ = -j673.4Ω = (673.4∠-π/2)Ω

Plug in and calculate each component's voltage:

$v_{R}$ = (0.0569∠1.147+220t)(217∠0) = (12.35∠1.147+220t)V

$v_{L}$ = (0.0569∠1.147+220t)(192.5∠π/2) = (10.95∠2.718+220t)V

$v_{C}$ = (0.0569∠1.147+220t)(673.4∠-π/2) = (38.32∠-0.4238+220t)V

Now we have the total and individual voltages as functions of time:

V = (30.0∠0+220t)V

$v_{R}$ = (12.35∠1.147+220t)V

$v_{L}$ = (10.95∠2.718+220t)V

$v_{C}$ = (38.32∠-0.4238+220t)V

Plug in t = 22.0×10⁻³s into these values and take the real component (amplitude multiplied by the cosine of the phase) to determine the real voltage values at this point in time:

V = 30.0cos(0+220(22.0×10⁻³)) = 3.82V

$v_{R}$ = 12.35cos(1.147+220(22.0×10⁻³)) = 11.8V

$v_{L}$ = 10.95cos(2.718+220(22.0×10⁻³)) = 3.19V

$v_{C}$ = 38.32cos(-0.4238+220(22.0×10⁻³)) = -11.2V

4 0

2 years ago

A person travels from New York to Washington at the rate of 45 km per hour and

mr_godi [17]

Answer:

The average speed for the whole journey is 49.5 miles per hour.

Explanation:

Step 1 :

Here, both the ways, he covers the same distance. Then, the formula to find average speed is

= 2xy / (x+y)

Step 2 :

x ----> Rate at which he travels from New York to Washington

x = 45

y ----> Rate at which he travels from New York to Washington

y = 55

Step 3 :

So, the average speed is

= (2 ⋅ 45 ⋅ 55) / (45 + 55)

= 4950 / 100

= 49.5

8 0

2 years ago