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2 years ago

10

The field used in the Canadian football League (CFL) has the midfield marker at the 55 yard line.how long is the fiend from goal

line to goal line?

1 answer:

kogti [31]2 years ago

3 0

Answer:

110 yds

Explanation:

Well if 55 yards is 1/2 of the field then 2 x 55 = 110 yards is total field length

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1- ¿Cuál sería la energía de un objeto de 50 newton de peso que se encuentra sobre una estantería de 3 metros de altura? ¿Qué ti

Liula [17]

Responder:

<h3>150 Nm </h3><h3>Energía potencial </h3>

Explicación:

El tipo de energía que posee el objeto se conoce como energía potencial. <u>La energía potencial es la energía que posee un objeto, mi virtud de su posición. </u>

Energía potencial = masa * aceleración debido a la gravedad * altura

Dado que Force = masa * aceleración debido a la gravedad

Energía potencial = Fuerza * altura

Fuerza dada = 50N y altura = 3 m

Energía potencial = 50 * 3

Energía potencial = 150 Nm

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3 years ago

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s

yawa3891 [41]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

Data:

$E_{pe}\:(elastic\:potential\:energy) = 5184\:J$

$K\:(constant) = 16200\:N/m$

$x\:(displacement) =\:?$

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

$E_{pe} = \dfrac{k*x^2}{2}$

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:

$E_{pe} = \dfrac{k*x^2}{2}$

$5184 = \dfrac{16200*x^2}{2}$

$5184*2 = 16200*x^2$

$10368 = 16200\:x^2$

$16200\:x^2 = 10368$

$x^{2} = \dfrac{10368}{16200}$

$x^{2} = 0.64$

$x = \sqrt{0.64}$

$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark$

Answer:

The displacement of the spring = 0.8 m

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

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4 years ago

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Answer:

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Although planets orbit the Sun in ellipses, all the planetary orbits are fairly close to circular and not very eccentric.

Nutka1998 [239]

Answer:

False

Explanation:

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