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The problem involves the motion of a simple pendulum, where the number of oscillations \( N \) made in time \( t \) varies directly with \( t \) and inversely as the square root of the length \( l \). This can be described by the relationship:

\[
N \propto \frac{t}{\sqrt{l}}
\]

Given the information that:

- A pendulum with length \( l = 248 \, \text{mm} = 0.248 \, \text{m} \) makes 10.0 oscillations in 10 seconds.

We can use this information to solve for the constant of proportionality \( k \) in the general equation:

\[
N = k \cdot \frac{t}{\sqrt{l}}
\]

Now, let’s solve each part.

### Part (a)
**Find the number of oscillations made in 14 seconds by a pendulum of length 3.98 m.**

We can use the equation:

\[
N = k \cdot \frac{t}{\sqrt{l}}
\]

We first need to determine \( k \) using the provided data.

### Part (b)
**Find the length of a pendulum that makes 50.0 oscillations in 1.00 minute (60 seconds).**

Rearranging the equation for \( l \):

\[
l = \left(\frac{k \cdot t}{N}\right)^2
\]

### Part (c)
**Find how long it would take a pendulum of length 95.0 cm (0.95 m) to make 15.0 oscillations.**

Rearranging the equation for \( t \):

\[
t = N \cdot \frac{\sqrt{l}}{k}
\]

I will calculate each part step by step.Here are the results for each part:

### Part (a):
The number of oscillations made in 14 seconds by a pendulum of length 3.98 m is approximately **3.49 oscillations**.

### Part (b):
The length of a pendulum that makes 50 oscillations in 1.00 minute (60 seconds) is approximately **0.357 meters** (or 35.7 cm).

### Part (c):
The time it would take a pendulum of length 95.0 cm to make 15.0 oscillations is approximately **29.36 seconds**.

Would you like more details or further clarification on any part?

Here are some related questions for further exploration:
1. How does changing the length of the pendulum affect the number of oscillations in a given time?
2. What would happen to the oscillations if the time period were doubled for the same pendulum length?
3. How does gravitational acceleration influence the oscillation frequency of a pendulum?
4. Can the same formula be used for pendulums on other planets with different gravitational forces?
5. What assumptions are made in deriving the formula for simple pendulum oscillations?

**Tip**: In a pendulum, the period of oscillation is largely influenced by the square root of its length, which makes longer pendulums oscillate more slowly than shorter ones.

The number of oscillations N made in a given time period t by a simple pendulum varies directly as t and inversely as the square root of the length of the pendulum. Given that a pendulum of length l = 248 mm makes 10.0 oscillations in 10 seconds, determine: (a) the number of oscillations made in 14 seconds by a pendulum of length 3.98 m, (b) the length of a pendulum that makes 50.0 oscillations in 1.00 minute, (c) how long it would take a pendulum of length 95.0 cm to make 15.0 oscillations.

Solve a pendulum oscillation problem involving direct proportionality with time and inverse square root of length. Learn how to find the number of oscillations, pendulum length, and time needed for a pendulum of various lengths. Suitable for high school physics students.

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