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2 years ago

If a measurement is 4.2 cm and its possible error is 0.04 cm, the relative error is?

Physics

1 answer:

galben [10]2 years ago

3 0

Given the value of possible error, the relative error of the measurement is 0.0095.

Given the data in the question;

Measurement taken; $m = 4.2cm$

Absolute or possible erorr; $e = 0.04m$

Relative error; $RE = \ ?$

<h3>Relative Error</h3>

Relative error is simply the ratio of the absolute or possible error of a measurement to the measurement its self.

Relative error is expressed as;

$RE = \frac{e}{m}$

Where e is the absolute or possible error and m is the measurement being taken.

To determine the relative error, we substitute our given values into the expression above.

$RE = \frac{0.04cm}{4.2cm} \\\\RE = 0.0095$ %

Therefore, given the possible error, the relative error of the measurement is 0.0095.

Learn more about relative error: brainly.com/question/17016021

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Answer:

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Explanation:

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3 years ago

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3 years ago

North Dakota Electric Company estimates its demand trend line (in millions of kilowatt hours) to be: D = 75.0 + 0.45Q, whe

Alborosie

Answer:

The demand forecast for winter is 96.36 millions KWH

The demand forecast for spring is 145.08 millions KWH

The demand forecast for summer is 169.89 millions KWH

The demand forecast for fall is 73.08 millions KWH

Explanation:

Given that,

The demand trend line is

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

We need to calculate the demand forecast for winter

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times101)\times0.80$

$D=96.36\ millions\ KWH$

We need to calculate the demand forecast for spring

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times102)\times1.20$

$D=145.08\ millions\ KWH$

We need to calculate the demand forecast for summer

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times103)\times1.40$

$D=169.89\ millions KWH$

We need to calculate the demand forecast for fall

Using given formula

$D=(75.0+0.45Q)\times multiplicative\ seasonal\ factors$

Put the value into the formula

$D=(75.0+0.45\times104)\times0.60$

$D=73.08\ millions KWH$

Hence, The demand forecast for winter is 96.36 millions KWH

The demand forecast for spring is 145.08 millions KWH

The demand forecast for summer is 169.89 millions KWH

The demand forecast for fall is 73.08 millions KWH

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3 years ago

What would a series circuit be used for?

igor_vitrenko [27]

Answer:

Explanation:

a series circuit would be an odd choice to power a battery or light a lamp when a direct would be much more efficient, and it's not converting types of energy, so C is the best possible answer

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3 years ago

Un ladrillo se le imparte una velocidad inicial de 6m/s en su trayectoria hacia abajo. ¿cual sera su velocidad final despues de

marshall27 [118]

Saludos!

Respuesta:

28,64 m/s.

Explicación:

Datos:

Altura o distancia recorrida: 40 m
Vo: 6 m/s
Aceleración de la gravedad: 9,81 m/s²

El ejercicio puede ser resuelto facilmente utilizando la siguiente formula, sin embargo es posible realizarlo utilizando formulas diferentes.

Entonces tenemos que:

$Vf ^{2} -Vo ^{2} =2 x g x h$

Es importante saber que al estar lanzando el ladrillo hacia abajo, el sentido del movimiento sigue el sentido de la gravedad, es decir es necesario que tomes el valor de la gravedad como positivo (+) y no negativo (-) como normalmente se usa.

Sustituyendo tenemos que:

$Vf x^{2} =(6 m/s) ^{2} +(2)x(9,81 m/s ^{2})x(40m) \\ Vf ^{2} =820,8 m^{2} /s ^{2} \\ \sqrt{Vf ^{2} } = \sqrt{820,8m^{2}/s ^{2} } \\ Vf=28,64 m/s$

Que tengas un buen día!

6 0

3 years ago

If a measurement is 4.2 cm and its possible error is 0.04 cm, the relative error is?​

If a measurement is 4.2 cm and its possible error is 0.04 cm, the relative error is?