Answer:
12.17 m/s²
Explanation:
The formula of period of a simple pendulum is given as,
T = 2π√(L/g)........................ Equation 1
Where T = period of the simple pendulum, L = length of the simple pendulum, g = acceleration due to gravity of the planet. π = pie
making g the subject of the equation,
g = 4π²L/T²................... Equation 2
Given: T = 1.8 s, l = 1.00 m
Constant: π = 3.14
Substitute into equation 2
g = (4×3.14²×1)/1.8²
g = 12.17 m/s²
Hence the acceleration due to gravity of the planet = 12.17 m/s²
Common symbol of the volume (L)
Answer:
A) 0.50 mV
Explanation:
In this problem, we can think the wings of the bird as a metal rod moving across a magnetic field. So, and emf will be induced into the wings of the bird, according to the formula:

where
is the strength of the magnetic field
v = 13 m/s is the speed of the bird
L = 1.2 m is the wingspan of the bird
is the angle between the direction of motion and the direction of the magnetic field
Substituting numbers into the formula, we find

To solve this problem we will derive the expression of the precession period from the moment of inertia of the given object. We will convert the units that are not in SI, and finally we will find the precession period with the variables found. Let's start defining the moment of inertia.

Here,
M = Mass
R = Radius of the hoop
The precession frequency is given as

Here,
M = Mass
g= Acceleration due to gravity
d = Distance of center of mass from pivot
I = Moment of inertia
= Angular velocity
Replacing the value for moment of inertia


The value for our angular velocity is not in SI, then


Replacing our values we have that


The precession frequency is




Therefore the precession period is 5.4s