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Scrat
5 days ago
10

Which statements about inertia and centripetal force are correct? Check all that apply. Inertia is always present. Inertia cause

s objects to continue moving in circles. Centripetal force is always present. Centripetal force pulls objects toward the center of a circle. Inertia and centripetal force cause circular motion.
Physics
2 answers:
Sav [1.1K]5 days ago
7 0

Inertia is always there.

Centripetal force draws objects toward the center of a circle.

Both inertia and centripetal force contribute to circular movement.

kicyunya [1K]5 days ago
6 0
Inertia is universally present. It's important to note that inertia doesn't serve as the force keeping objects in circular paths; that role belongs to centripetal force, which is not always present. Centripetal force actively pulls objects towards the center of a circle. Both inertia and centripetal force contribute to the phenomenon of circular motion. Thank you, and enjoy your day;)
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Answer:

Explanation:

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Regarding carbon and nitrogen, the air inhaled by cows contains nitrogen, oxygen, carbon dioxide, and other gases. These molecules enter the cow through respiration.

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Explanation:

Here’s a revised version of the requirements;

Fill in the blanks with the appropriate terms. Picture a force gauge fixed between the rope and the saddle of the chain carousel. If you keep your feet off the ground while the vehicle is not in motion, the dynamometer shows A / B. When the carousel is spinning, you’ll see C / D displayed on the dynamometer.

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15 days ago
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Keith_Richards [1034]

Response:

Physical well-being

Clarification:

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17 days ago
The iron ball shown is being swung in a vertical circle at the end of a 0.7-m string. how slowly can the ball go through its top
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6 0
6 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.

5 0
15 days ago
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