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2 years ago

How many significant figures are in the following measurements ?

2 answers:

5 0

The number of significant figures are counted by,
non-zeros are significant
Zeros in between non-zeros are significant.
Zeros after the decimal and non zeros are significant.

a. 300000000m/s = 1 significant figure
b. 3.00x10(8th power) = 3 significant figures
c.25.030*c = 4 significant figures
d. 0.006070 *c = 5 significant figures
e. 1.004 J = 4 significant figures
f. 1.30520 MHz =5 significant figures

kifflom [539]2 years ago

4 0

The number of significant figures are based on the rules:non-zeros are significant Zeros in between non-zeros are also significant. Trailing zeros are considered not significant. Zeros after the decimal and non zeros are significant.

a. 1
b. 3
c. 5
d. 4
e. 4
f. 6

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3 years ago

What is the frequency, in units of kiloHertz, of an AC waveform that has a period of 12 microseconds?

Alik [6]

Answer:

83.3 kHz

Explanation:

The frequency of a waveform is equal to the reciprocal of its period:

$f=\frac{1}{T}$

where

f is the frequency

T is the period

In this problem, we have

$T=12 \mu s=12\cdot 10^{-6} s$

so, the frequency of the waveform is

$f=\frac{1}{12 \cdot 10^{-6} s}=8.33\cdot 10^4 Hz$

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2 years ago

Beginning 145 miles directly south of the city of Hartville, a car travels due west. If the car is travelling at a speed of 42 m

ziro4ka [17]

Answer:

The rate of change of the distance is 14.89.

Explanation:

Given that,

Distance = 145 miles

Speed of car = 42 miles/hr

Distance covered by car = 55 miles

We need to calculate the the rate of change of the distance

According to figure,

Let OA is x, and AB is y.

Now, using Pythagorean theorem

$x^2=y^2+145^2$

On differentiating

$2x\dfrac{dx}{dt}=2y\dfrac{dy}{dt}$

$\dfrac{dx}{dt}=\dfrac{y}{x}\dfrac{dy}{dt}$

$\dfrac{dx}{dt}=\dfrac{55\times42}{\sqrt{55^2+145^2}}$

$\dfrac{dx}{dt}=14.89\ miles/hr$

Hence, The rate of change of the distance is 14.89.

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3 years ago

Why do scientist use models to study atoms

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3 years ago

If 35 J of work was performed in 70 seconds, how much power was used to do this task?

Irina18 [472]

$P = w \times delta \: T$
P = Power
W = Work
Delta T = Change of Time

So:
P = 35 Joules
Delta T = 70 secs
W = ?

$P = 35 \div 70 \\ P = 0.5 \: watts$

* Joule/Sec = Watt

Answer: The power used to do this task was 0.5 Watts.

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2 years ago