The two ladybugs have same rotational (angular) speed
Explanation:
The rotational (angular) speed of an object in circular motion is defined as:
where
is the angular displacement
t is the time interval considered
Here we have two ladybugs, which are located at two different distances from the axis. In particular, ladybug 1 is halfway between ladybug 2 and the axis of rotation. However, since they rotate together with the disk, and the disk is a rigid body, every point of the disk cover the same angle in the same time : this means that every point along the disk has the same angular speed, and therefore the two ladybugs also have the same angular speed.
On the other hand, the linear speed of the two ladybugs is different, because it follows the equation:
where r is the distance from the axis: and since the two ladybugs are located at different , they have different linear speed.
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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.
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Answer:
We know that for a pendulum of length L, the period (time for a complete swing) is defined as:
T = 2*pi*√(L/g)
where:
pi = 3.14
L = length of the pendulum
g = gravitational acceleration = 9.8 m/s^2
Now, we can think on the swing as a pendulum, where the child is the mass of the pendulum.
Then the period is independent of:
The mass of the child
The initial angle
Where the restriction of not swing to high is because this model works for small angles, and when the swing is to high the problem becomes more complex.
Answer:
C
Explanation:
If the theory were to be proved you you need to repeat the experiment over and over again so that way you can prove that it is true wuth the same results.
Answer:
Density of liquid = 4730 kg/m³
Atmospheric pressure on planet X = 8401.7 N/m²
Explanation:
Pressure, P = ρgh where ρ = density of liquid, g =9.8 m/s² and h = height of column at earth's surface = 2185 mm. Since P = atmospheric pressure, for mercury, P = ρ₁gh₁ where ρ₁ = 13.6 g/cm³ and h₁ = 760 mm
So, ρgh = ρ₁gh₁
ρ = ρ₁h₁/h = 13.6 g/cm³ × 760/2185 = 4.73 g/cm³ = 4730 kg/m³
The atmospheric pressure on planet X
P = ρg₁h₃ g₁ = g/4 and h₃ = 725 mm = 0.725 m
on planet X
P = ρg₁h₃ = (4730 kg/m³ × 9.8 m/s² × 0.725 m)/4 = 8401.7 N/m²