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xz_007 [3.2K]
3 years ago
5

When a pendulum with a period of 2.00000 s is moved to a new location from one where the acceleration due to gravity was 9.80 m/

s2, its new period becomes 1.99824 s. By how much does the acceleration due to gravity differ at its new location? How is the period of the pendulum related to the length of the pendulum and the acceleration due to gravity at the location of the pendulum? What happens to the length of the pendulum as it is moved to the new location?
Physics
1 answer:
Fynjy0 [20]3 years ago
8 0

1. By 0.02 m/s^2

The period of a pendulum is given by:

T=2 \pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the gravitational acceleration

Initially, we know:

T = 2.00000 s is the period of the pendulum

g = 9.80 m/s^2 is the acceleration due to gravity at the original location

We can solve the equation for L in order to find the length of the pendulum:

L=\frac{T^2}{(2 \pi)^2}g=\frac{(2.0000 s)^2}{(2 \pi)^2}(9.80 m/s^2)=0.99396 m

The length of the pendulum does not change when it is moved to the new location, so we can use the same equation with T=1.99824 s (the new period) and solving it for g to find the acceleration due to gravity at the new location:

g=L\frac{(2 \pi)^2}{T^2}=(0.99396 m)\frac{(2 \pi)^2}{(1.99824 s)^2})=9.82 m/s^2

So, the change in gravitational acceleration is

\Delta g = g_2 - g_1 = 9.82 m/s^2-9.80 m/s^2 = 0.02 m/s^2

2) the period of the pendulum is directly proportional to the square root of the length, L, and inversely proportional to the square root of the gravitational acceleration, g.

The period of a pendulum is given by:

T=2 \pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the gravitational acceleration

So, we see that the period of the pendulum is directly proportional to the square root of the length, L, and inversely proportional to the square root of the gravitational acceleration, g.

3) The length of the pendulum does not change

The length of the pendulum does not depend on the location: in fact, only the value of the gravitational acceleration, g, depends on the location, therefore the length of the pendulum, L, does not change.

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I = 2.7A    V=120V , t=?

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Hi there!

(a)

Recall that:
W = F \cdot d = Fdcos\theta

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

W =248(56)cos(30) = 12027.36 J

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The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
\boxed{W_g = 0 J}

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Similarly, the normal force is perpendicular to the displacement, so:
\boxed{W_N = 0 J}

(d)

Recall that the force of kinetic friction is given by:
F_{f} =\mu_k mg

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
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Nearest multiple of ten:
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Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
F_{net} = F_{Ax} - F_f

W = F_{net} \cdot d = (F_{Ax} - F_f)

W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}

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