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3 years ago

Determine the magnitude of the effective value of g⃗ at a latitude of 60 ∘ on the earth. assume the earth is a rotating sphere.

1 answer:

5 0

In addition to acceleration of gravity we experience centrifugal acceleration away from the axis of rotation of the earth. this additional acceleration has value ac = r w^2 where w = angular velocity and r is distance from your spot on earth to the earth's axis of rotation so r = R cos(l) where l = 60 deg is the lattitude and R the earth's radius and w = 1 / (24hr x 3600sec/hr)
<span>now you look up R and calculate ac then you combine the centrifugal acc. vector ac with the gravitational acceleration vector ag = G Me/R^2 to get effective ag' = ag -</span>

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Alex17521 [72]

Answer:

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Explanation:

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Force constant, k=1300N/m

Restoring force, Fx=6500 N

Average friction force, f=50 N

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y=2.5 m

Initial velocity, u=0

$F_x=kx$

Substitute the values

$6500=1300x$

$x=\frac{6500}{1300}=5$ m

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$W_f=fscos\theta$

We have $\theta=180^{\circ}$

Substitute the values

$W_f=50\times 5cos180^{\circ}$

$W_f=-250J$

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Initial gravitational energy, $U_{grav,1}=0$ \

Initial elastic potential energy

$U_{el,1}=\frac{1}{2}kx^2=\frac{1}{2}(1300)(5^2)$

$U_{el,1}=16250J$

Final elastic energy, $U_{el,2}=0$

Final kinetic energy, $K_f=\frac{1}{2}(60)v^2=30v^2$

Final gravitational energy, $U_{grav,2}=mgh=60\times 9.8\times 2.5$

Final gravitational energy, $U_{grav,2}=1470J$

Using work-energy theorem

$K_i+U_{grav,1}+U_{el,1}+W_f=K_f+U_{grav,2}+U_{el,2}$

Substitute the values

$0+0+16250-250=30v^2+1470+0$

$16000-1470=30v^2$

$14530=30v^2$

$v^2=\frac{14530}{30}$

$v=\sqrt{\frac{14530}{30}}$

$v=22m/s$

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Answer:

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