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tiny-mole [99]
3 years ago
11

Calculate the maximum absolute uncertainty for R if:

Physics
1 answer:
Radda [10]3 years ago
3 0

Answer:

ΔR = 9 s

Explanation:

To calculate the propagation of the uncertainty or absolute error, the variation with each parameter must be calculated and the but of the cases must be found, which is done by taking the absolute value

           

The given expression is      R = 2A / B

the uncertainty is                 ΔR = | \frac{dR}{dA} | ΔA + | \frac{ dR}{dB} | ΔB

we look for the derivatives

     \frac{dR}{dA} = 9 / B

     \frac{dR}{dB} = 9A ( - \frac{1}{B^2 } )

we substitute

     ΔR = \frac{9}{B}  ΔA + \frac{9A}{B^2}  ΔB

the values ​​are

     ΔA = 2 s

     ΔB = 3 s

 

     ΔR = \frac{9}{11}   2 + \frac{9 \ 32}{11^2 }  3

     ΔR = 1.636 + 7.14

     ΔR = 8,776 s

the absolute error must be given with a significant figure

     ΔR = 9 s

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Please show work : A particle with mass 2.00 μg and a charge of – 200 nC has a velocity of 3000 m/s in the x-direction. There is
irga5000 [103]

Answer:

 x =4.5 10⁴ m

Explanation:

To find the distance that the particle moves we must use the equations of motion in one dimension and to find the acceleration of the particle we will use Newton's second law

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     q = -200 nc (1C / 10 9 nC) = -200 10-9 C

Let's calculate the acceleration

     F = ma

     F = q E

     a = qE / m

     a = -200 10⁻⁹ 1000 / 2.00 10⁻⁶

     a = 1 10² m / s²

Let's use kinematics to find the distance traveled before stopping, where it has zero speed (Vf = 0)

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     0 = Vo² - 2 a x

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This is the distance the particule stop, after this distance in the field accelerates in the opposite direction of the initial

Second part

In this case Newton's second law is applied on the y axis

      F -W = 0

      F = w = mg

      E q = mg

      E = mg / q

      E = 2.00 10⁻⁶ 9.8 / 200 10⁻⁹

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Answer:

7.45

Explanation:

At 3.0 m PE = m*g*3.0, KE = m*g*(4.0 - 3.0)

At 2.0 m PE = m*g*2.0, KE = m*g*(4.0 - 2.0)

At 3.0 m PE = m*g*1.0, KE = m*g*(4.0 - 1.0)

At 0 m PE = 0, KE = m*g*h = total energy = 7.45 J

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