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22 hours ago

What is the sources of error and suggestion on how to overcome it in the hooke's law experiment?

Physics

1 answer:

Yuliya22 [1.1K]22 hours ago

6 0

In the study of physics, Hooke's law can be expressed as:

F = kx

This law indicates that the spring force F is proportional to the extension x, with k being the spring constant.

In experiments, this is often examined using the setup illustrated in the included figure. The spring is tested, and a known weight is applied underneath it. This weight exerts a gravitational pull, essentially its weight, on the spring. While the spring elongates, the displacement can be measured using a ruler.

Several potential errors can arise during this experiment. Firstly, the person's measurement reading may be faulty. Digital scales offer greater accuracy as they reduce human error, while ruler readings can be subjective, especially if not viewed at eye level. Additionally, the object's weight may be inaccurately measured if the scale is untrustworthy. Lastly, the measuring equipment may not be correctly calibrated.

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A 10-turn conducting loop with a radius of 3.0 cm spins at 60 revolutions per second in a magnetic field of 0.50T. The maximum e

kicyunya [1025]

Answer:

Maximum emf = 5.32 V

Explanation:

Provided data includes:

Number of turns, N = 10

Radius of loop, r = 3 cm = 0.03 m

It made 60 revolutions each second

Magnetic field, B = 0.5 T

We are tasked to determine the maximum emf produced in the loop, which is founded on Faraday's law. The induced emf can be calculated by:

$\epsilon=\dfrac{d(NBA\cos\theta)}{dt}\\\\\epsilon=NBA\dfrac{d(\cos\theta)}{dt}\\\\\epsilon=NBA\omega \sin\omega t\\\\\epsilon=NB\pi r^2\omega \sin\omega t$

For the maximum emf, $\sin\omega t=1$

Therefore,

$\epsilon=NB\pi r^2\omega \\\\\epsilon=NB\pi r^2\times 2\pi f\\\\\epsilon=10\times 0.5\times \pi (0.03)^2\times 2\pi \times 60\\\\\epsilon=5.32\ V$

Hence, the maximum emf generated in the loop is 5.32 V.

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8 days ago

For a projectile, which of the following quantities are constant during the flight: x, y, vx, vy, v, ax, ay? Check all that appl

Yuliya22 [1153]

Response:

C. vx

F. ax

G. ay

Clarification:

The projectile follows a curved trajectory toward the ground, causing changes in x and y positions.

Since there is no external force acting in the x-direction, the acceleration in x remains at zero. Consequently, ax and vx remain unchanged.

The projectile is subject to the force of gravity, directed downwards, leading to an increase in its velocity due to the rise in its y-component.

Meanwhile, the y-component of acceleration remains constant due to gravitational acceleration.

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7 days ago

Astronomers determine that a certain square region in interstellar space has an area of approximately 2.4 \times 10^72.4×10 7

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Answer:

1.5 × 10³⁶ light-years

Explanation:

A particular square area in interstellar space measures roughly 2.4 × 10⁷² (light-years)². To find the area of a square, the following formula is utilized:

A = l²

where,

A represents the area of the square

l denotes the length of one side of the square

Thus, l = √A = √2.4 × 10⁷² (light-years)² = 1.5 × 10³⁶ light-years

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6 days ago

Consider a perfectly insulated cup (no

ValentinkaMS [1149]

Answer:

When ice is subjected to heat, it melts; however, the temperature remains constant at 0◦ C.

Explanation:

Solution

The heat supplied by the heater is solely utilized for the melting of the ice, thus maintaining the temperature at 0◦ C.

Once all the ice has liquefied, the temperature of the resulting water will start to rise over time.

Note: please see the attached document with solutions featuring diagrams related to this explanation

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15 days ago

Question 1

Keith_Richards [1034]

Answer:

1) g = 4π² / m, 3) on the x-axis we have the pendulum lengths, while the y-axis shows the squared periods.

Explanation:

a) learners can model this system as a simple pendulum, where the angular velocity is given by

w = √ g / l

Here, angular velocity, frequency, and period are interconnected:

w = 2π f = 2π / T

Substituting yields:

T = 2π√ l / g

Using this formula, students can calculate the gravitational acceleration by measuring the period for several pendulum lengths and plotting:

T² = 4π² l / g

We plot T² against l.

This represents a linear equation where T² is on the y-axis and l is on the x-axis:

y = (4π² / g) l

The slope is given by:

m = 4π² / g

Solving for g gives:

g = 4π² / m

The slope is determined from the line's values rather than experimental data.

2) To perform the experiment, the string is secured to the sphere, then the pendulum length from the pivot to the sphere's center is measured using a tape measure. A slight angle (less than 10 degrees) is released, allowing the first swing to occur. Generally, the time for several oscillations, usually 10 or 20, is tracked to find the period:

T = t / n

Next, a table is created comparing T² to the length, plotted with length on the x-axis to find the slope, from which the gravitational acceleration is derived.

3) The independent variable, which is the length of the pendulums, is plotted on the x-axis, while the dependent variable, the squared period, is on the y-axis.

4) Referring to the line equation:

m = 4π² / g

resulting in:

g = 4π² / m

5) Once the spring is cut, the sphere continues to be influenced by gravitational acceleration. The harmonic motion ceases, and the sphere moves vertically.

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