Answer:
The moment of inertia is 
Explanation:
From the question we are told that
The frequency is 
The mass of the pendulum is 
The location of the pivot from the center is 
Generally the period of the simple harmonic motion is mathematically represented as
Where I is the moment of inertia about the pivot point , so making I the subject of the formula it
=> ![I = [ \frac{T}{2 \pi } ]^2 * m* g * d](https://tex.z-dn.net/?f=I%20%3D%20%20%5B%20%5Cfrac%7BT%7D%7B2%20%5Cpi%20%7D%20%5D%5E2%20%2A%20%20m%2A%20%20g%20%2A%20d)
But the period of this simple harmonic motion can also be represented mathematically as
substituting values
So
![I = [ \frac{2.174}{2 * 3.142 } ]^2 * 2.40* 9.8 * 0.380](https://tex.z-dn.net/?f=I%20%3D%20%20%5B%20%5Cfrac%7B2.174%7D%7B2%20%2A%203.142%20%7D%20%5D%5E2%20%2A%20%20%202.40%2A%20%209.8%20%2A%200.380)
Answer:
Total distance = 700 m
Displacement = 500 m
Explanation:
Notice that Jed travelled a total of 3 x 100 m = 300 m in the North direction, and 300 m + 100 m = 400 m in the East direction. Therefore the total distance he travelled is: 300 + 400 = 700 m.
But the actual displacement is given by the Pythagorean theorem as the hypotenuse of a right angle triangle of legs 300 m and 400 m:
displacement = 
Answer:
Explanation:
If an object has a higher density than the fluid it is in (fluid can mean liquid or gas), it will sink. If it has a lower density, it will float. Density is determined by an object's mass and volume. If two objects take up the same volume, but have one has more mass, then it also has a higher density.
Answer:
The launching point is at a distance D = 962.2m and H = 39.2m
Explanation:
It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.
X axis
x = Vox t
t = x / vox
t = 7.1 / 340
t = 2.09 10-2 s
In this same time the height of the window fell
Y = Voy t - ½ g t²
Let's calculate the initial vertical speed, this speed is in the window
Voy = (Y + ½ g t²) / t
Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209
Voy = 27.7 m / s
We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s
Vy = Voy - gt₂
Vy = 0 -g t₂
t₂ = Vy / g
t₂ = 27.7 / 9.8
t₂ = 2.83 s
This is the time it also takes to travel the horizontal and vertical distance
X = Vox t₂
D = 340 2.83
D = 962.2 m
Y = Voy₂– ½ g t₂²
Y = 0 - ½ g t2
H = Y = - ½ 9.8 2.83 2
H = 39.2 m
The launching point is at a distance D = 962.2m and H = 39.2m
