If you are building a clock by using a pendulum's oscillations to keep time and you find the period is a little too small (the c
1 answer:
Time period of pendulum is given by
Here,
L is length
g is acceleration due to gravity
Therefore we can determine that the period of the pendulum is directly proportional to the square root of the length of the rope.
Therefore, it is necessary to increase the length of the rope to increase the period so that it counts the seconds more slowly. The correct answer is A.
You might be interested in
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
Answer:
work done is -2.8 × 10⁻⁶ J
Explanation:
Given the data in the question;
mass of the pendulum m = 6 kg
Length of core = 1.7 m
Now, case1, mass is pulled aside a small distance of 7.6 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.
so, θ₁ = ( 7.6 × 10⁻² m / 1.7 m ) = 0.045 rad
In case2, mass is pulled aside a small distance of 8 cm and released from rest. so let θ₁ be the angle made by mass with vertical axis.
so, θ₂ = ( 8 × 10⁻² m / 1.7 m ) = 0.047 rad.
Now, the required work done will be;
W = -cosθ
W = 6 × 9.8 × 1.7 × [ cos( 0.047 ) - cos( 0.045 ) ]
W = 6 × 9.8 × 1.7 × [ -2.8 × 10⁻⁸ ]
W = -2.8 × 10⁻⁶ J
Therefore, work done is -2.8 × 10⁻⁶ J