The distance between two particles that are <em>in phase</em>
None of the choices is a force. 'a' and 'b' are speeds. 'C' and 'd' are accelerations. ... The steady force of gravity is 9.8 newtons PER KILOGRAM of mass. ... The question is written by someonewho very much wants to discourage anyone interested in Physics.
Time period remains the same in both the experiment as change in amplitude does not affect time period.
What are the factors on which time period depends in SHM?
Time period is given by:
where,
T = time period
m = mass
k = spring constant
In a straightforward harmonic motion, we see from the preceding formula that the time period depends only on the object's mass and spring constant (SHM). The time period will adjust to any variations in the object's mass or the spring constant.
What is Spring Constant?
A spring's "spring constant" is a property that quantifies the relationship between the force acting on the spring and the displacement it produces. In other words, it characterises a spring's stiffness and the extent of its range of motion.
Learn more about SHM here:
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14 because the horses name is friday and sprite cranberry is coming out with a new jersey
Answer:
ee that the lens with the shortest focal length has a smaller object
Explanation:
For this exercise we use the constructor equation or Gaussian equation
where f is the focal length, p and q are the distance to the object and the image respectively.
Magnification a lens system is
m = = -
h ’= -\frac{h q}{p}
In the exercise give the value of the height of the object h = 0.50cm and the position of the object p =∞
Let's calculate the distance to the image for each lens
f = 6.0 cm
as they indicate that the light fills the entire lens, this indicates that the object is at infinity, remember that the light of the laser rays is almost parallel, therefore p = inf
q = f = 6.0 cm
for the lens of f = 12.0 cm q = 12.0 cn
to find the size of the image we use
h ’= h q / p
where p has a high value and is the same for all systems
h ’= h / p q
Thus
f = 6 cm h ’= fo 6 cm
f = 12 cm h ’= fo 12 cm
therefore we see that the lens with the shortest focal length has a smaller object