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3 years ago

85POINTS ASAP!!!!!!!!!!!!!!

Physics

2 answers:

Charra [1.4K]3 years ago

8 0

The value of x is 3 to balance the equation

alexgriva [62]3 years ago

6 0

The andwer of tye question is 3O2

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What is a major goal of the federal government when managing land resources?

oksian1 [2.3K]

Answer:

I just took the quiz and got 100% when choosing A.Conservation. Hope this helps:)

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3 years ago

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Cuando una persona sube y baja una escalera, Cuanto vale su desplazamiento y cual es la medida de su trayectoria.

adoni [48]

Answer:

Primero, definimos el desplazamiento como la distancia entre la posición final y la posición inicial.

Así, si comenzamos abajo, luego subimos la escalera, y luego bajamos, la posición final y la posición inicial serán la misma

por lo que el desplazamiento es igual a cero.

La medida recorrida es el espacio total recorrido.

Es decir, si entre el principio y el final de la escalera hay una distancia D.

La persona que sube y baja, recorre esta distancia dos veces.

Entonces cuando una persona sube y baja la escalera, la medida de su trayectoria será 2*D.

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3 years ago

What acceleration will a force of 20 newtons cause if applied to a go-kart with a mass of 20 kilograms?

iren2701 [21]

Answer:

1m/s is the acceleration used. C

Explanation:

please mark brainliest

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3 years ago

Out of aluminum,copper,steel,and glass.Which material do you think will be the best thermal conductor?

Nastasia [14]

I believe it is copper

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3 years ago

A bungee cord is 30m long and when stretched a distance x itexerts a restoring force of magnitude kx. Your father-in-law (mass95

madam [21]

Answer:

The distance the bungee cord that would be stretched 0.602 m, should be selected when pulled by a force of 380 N.

Explanation:

As from the given data

the length of the rope is given as l=30 m

the stretched length is given as l'=41m

the stretched length required is give as y=l'-l=41-30=11m

the mass is m=95 kg

the force is F=380 N

the gravitational acceleration is g=9.8 m/s2

The equation of k is given by equating the energy at the equilibrium point which is given as

$U_{potential}=U_{elastic}\\mgh=\dfrac{1}{2} k y^2\\k=\dfrac{2mgh}{y^2}$

Here

m=95 kg, g=9.8 m/s2, h=41 m, y=11 m so

$k=\dfrac{2mgh}{y^2}\\k=\dfrac{2\times 95\times 9.8\times 41}{11^2}\\k=630.92 N/m$

Now the force is

$F=kx\\$ or

$x=\dfrac{F}{k}\\$

So here F=380 N, k=630.92 N/m

$x=\dfrac{F}{k}\\x=\dfrac{380}{630.92}\\x=0.602 m$

So the distance is 0.602 m

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3 years ago