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Anuta_ua [19.1K]
3 years ago
8

If you were to measure the width of your lab manual with a meter stick, and then the length (roughly 50% longer), by what fracti

on would the absolute uncertainty differ in the two cases?
b) If you were to measure the width of your lab manual with a meterstick, and then the length (roughly 50% longer), by what fraction would the percent uncertainty differ in the two cases?

c) If you knew the radius of a circle with 1% accuracy, how accurate would your calculation of the area be?
Physics
1 answer:
Leona [35]3 years ago
7 0

Answer:

Explanation:

a )  Let breadth be L , then length will be 1.5 L

Since it is a meter scale , the least count is 1 m.m. So

maximum possible error in a measurement is 1 mm.

In both the measurement of length and breadth , absolute

uncertainty will be 1mm.

Hence there is no difference in  absolute uncertainty of their measurement.

b ) percent uncertainty in the measurement of length

\frac{1 mm}{1.5L\times1000} \times100

= 1 / 15L%

percent uncertainty in the measurement of breadth

\frac{1 mm}{L\times1000} \times100

1 / 10L%

In breadth it will be higher by

( 1/10L - 1/15L)/ 1/10L

= 1/3

c ) Uncertainty  in the measurement of R= 1 %

Area A = π R²

ΔA/A X100 = 2 X ΔR/R X 100

Percentage uncertainty in A = 2 X percentage uncertainty in R

= 2 X 1

=2 %

Therefore percentage  accuracy in the calculation of area

= 2 %.

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Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The d
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Answer:

a)Car E = Car D  > (Car F = Car B = Car C) > Car A

b)Car E = Car D  > (Car F = Car B = Car C) > Car A

Explanation:

Car A: mass = 500 kg; speed = 10 m/s

Car B: mass = 2000 kg;speed = 5 m/s

Car C:mass = 500 kg; speed = 20 m/s

Car D: mass = 1000 kg; speed = 20 m/s

Car E:mass = 4000 kg; speed = 5 m/s

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Part a) Now we know that momentum of each car is product of mass and velocity , so we will have

<em>CarA:</em>

P_1 = m \times v\\P_1 = (500)(10)\\P_1 = 5 \times 10^3 kg m/s

<em>Car B:</em>

P_2 = m v\\P_2 = (2000)(5)\\P_2 = 10^4 kg m/s

Car C:

P_3 = m v\\P_3 = (500)(20)\\P_3 = 10^4 kg m/s

Car D:

P_4 = m v\\P_4 = (1000)(20)\\P_4 = 2\times 10^4 kg m/s

Car E:

P_5 = m v\\P_5 = (4000)(5)\\P_5 = 2\times 10^4 kg m/s

Car F:

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Car E = Car D  > (Car F = Car B = Car C) > Car A

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3 years ago
What is the acceleration of an object that takes 20 sec to change from a speed of 200 m/s to 300 m/s ?
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<u>Given</u><u>:</u>

  • initial velocity, u = 200 m/s

  • Final velocity, v = 300 m/s

  • Time taken, t = 20 sec

<u>To</u><u> </u><u>be</u><u> </u><u>calculated</u><u>:</u>

Calculate the acceleration of given object ?

<u>Formula</u><u> </u><u>used</u><u>:</u>

Acceleration = v - u / t

<u>Solution</u><u>:</u>

We know that,

Acceleration = v - u / t

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4 years ago
Help please, It's for science :&gt;
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Answer:

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This is amazing.  When you read the quest ion, you wouldn't think there's enough information there to find an answer.  But there is !

-- When the block is sliding along the flat surface, its kinetic energy is (1/2)(Mass·v²).

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(1/2)(Mass·v²)  =  (2.5m)·(Mass·g)

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