Simple cells have liquid chemicals, making it harder for it to carry. While as dry cells have no liquid chemicals, making it easier to carry.
The period will be the same if the amplitude of the motion is increased to 2a
What is an Amplitude?
Amplitude refers to the maximum extent of a vibration or oscillation, measured from the position of equilibrium.
Here,
mass m is attached to the spring.
mass attached = m
time period = t
We know that,
The time period for the spring is calculated with the equation:

Where k is the spring constant
Now if the amplitude is doubled, it means that the distance from the equilibrium position to the displacement is doubled.
From the equation, we can say,
Time period of the spring is independent of the amplitude.
Hence,
Increasing the amplitude does not affect the period of the mass and spring system.
Learn more about time period here:
<u>brainly.com/question/13834772</u>
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Turn off lights when leaving rooms.
Unplug unused appliances. Even when powered off these appliances use electricity.
Replace regular light bulbs with energy saving bulbs.
Answer:
x = 0.775m
Explanation:
Conceptual analysis
In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.
We apply Coulomb's law to calculate the electrical forces on q₃:
(Electric force of q₂ over q₃)
(Electric force of q₁ over q₃)
Known data
q₁ = 15 μC = 15*10⁻⁶ C
q₂ = 6 μC = 6*10⁻⁶ C
Problem development
F₂₃ = F₁₃
(We cancel k and q₃)

q₂(2-x)² = q₁x²
6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)
6(2-x)² = 15(x)²
6(4-4x+x²) = 15x²
24 - 24x + 6x² = 15x²
9x² + 24x - 24 = 0
The solution of the quadratic equation is:
x₁ = 0.775m
x₂ = -3.44m
x₁ meets the conditions for the forces to cancel in q₃
x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel
The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.
Answer:
The gravitational force changing velocity is

Explanation:
The expression for the gravitational force is

Differentiate the above equation

The velocity is the distance in at time so


