Answer:
1) Hence, the period is 0.33 s.
2) The amplitude is 10 cm.
Explanation:
1) The period is given by:

Where:
f: is the frequency = 3 bob up and down each second = 3 s⁻¹ = 3 Hz
Hence, the period is 0.33 s.
2) The amplitude is the distance between the equilibrium position and the maximum position traveled by the spring. Since the spring is moving up and down over a distance of 20 cm, then the amplitude is:
Therefore, the amplitude is 10 cm.
I hope it helps you!
As the water plunges, its velocity increases. Its potential energy<span> becomes kinetic</span>energy<span>. The law of conservation of </span>energy<span> states that when one form of </span>energy<span> is</span>transformed<span> to another, no </span>energy<span> is destroyed in the process. ... So the total amount of </span>energy<span> is the same before and after any </span>transformation<span>.
hope it helps
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The magnitude of the electric field for 60 cm is 6.49 × 10^5 N/C
R(radius of the solid sphere)=(60cm)( 1m /100cm)=0.6m

Since the Gaussian sphere of radius r>R encloses all the charge of the sphere similar to the situation in part (c), we can use Equation (6) to find the magnitude of the electric field:

Substitute numerical values:

The spherical Gaussian surface is chosen so that it is concentric with the charge distribution.
As an example, consider a charged spherical shell S of negligible thickness, with a uniformly distributed charge Q and radius R. We can use Gauss's law to find the magnitude of the resultant electric field E at a distance r from the center of the charged shell. It is immediately apparent that for a spherical Gaussian surface of radius r < R the enclosed charge is zero: hence the net flux is zero and the magnitude of the electric field on the Gaussian surface is also 0 (by letting QA = 0 in Gauss's law, where QA is the charge enclosed by the Gaussian surface).
Learn more about Gaussian sphere here:
brainly.com/question/2004529
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Answer:
The work done on the wagon is 37 joules.
Explanation:
Given that,
The force applied by Charlie to the right, F = 37.2 N
The force applied by Sara to the left, F' = 22.4 N
We need to find the work done on the wagon after it has moved 2.50 meters to the right. The net force acting on the wagon is :



Work done on the wagon is given by the product of net force and displacement. It is given by :


W = 37 Joules
So, the work done on the wagon is 37 joules. Hence, this is the required solution.
Answer:
Explanation:
<u>1. Name of the variables:</u>

<u>2. Formulae:</u>




<u>3. Solution (calculations)</u>



