Answer:
Change in potential energy = 7350 Joules
Explanation:
It is given that,
Side of cube, a = 0.5 m
Density of cube, 
The cube is lifted vertically by a crane to a height of 3 m
We know that, density 
So, m = d × V (V = volume of cube = a³)
m = 250 kg
We have to find the change in potential energy of the cube. At ground level, the potential energy is equal to 0.
Potential energy at height h is given by :
PE = 250 kg × 9.8 m/s² ×3 m
PE = 7350 Joules
So, change in potential energy of the cube is 7350 Joules.
Answer:
x ’= 1,735 m, measured from the far left
Explanation:
For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.
Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive
They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,
the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar
x_{cm} = 1.2 -1
x_ {cm} = 0.2 m
Σ τ = 0
w₁ 1.2 + mg 0.2 - W₂ x = 0
x = 
x = 
let's calculate
x = 
2.9 1.2 + 4 0.2 / 8
x = 0.535 m
measured from the pivot point
measured from the far left is
x’= 1,2 + x
x'= 1.2 + 0.535
x ’= 1,735 m
Answer:
The smallest part of a millimeter that can be read with a digital caliper with a four digit display is 0.02mm. Thus, it has to be converted to centimetre. So, divide by 10, we then have 0.02/10= *0.002cm* not mm.
I think by using data collected by Tycho Brahe
