Answer:
a) see attached, a = g sin θ
b)
c) v = √(2gL (1-cos θ))
Explanation:
In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by
Wₓ = m a
W sin θ = m a
a = g sin θ
b) The diagram is the same, the only thing that changes is the angle that is less
θ' = 9/2 θ
c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.
The easiest way to find linear speed is to use conservation of energy
Highest point
Em₀ = mg h = mg L (1-cos tea)
Lowest point
Emf = K = ½ m v²
Em₀ = Emf
g L (1-cos θ) = v² / 2
v = √(2gL (1-cos θ))
Answer:
Uncertainty in position of the bullet is 
Explanation:
It is given that,
Mass of the bullet, m = 35 g = 0.035 kg
Velocity of bullet, v = 709 m/s
The uncertainty in momentum is 0.20%. The momentum of the bullet is given by :
Uncertainty in momentum is,
We need to find the uncertainty in position. It can be calculated using Heisenberg uncertainty principal as :
Hence, this is the required solution.
Answer:
Density is defined as:
Density = Mass/Volume
Now, density is an intensive property, this means that if you have 10 grams of a given material or 1000 grams of the same material, in both cases you will find the same density.
Then a roll of 50 pennies has the same density that a single penny.
The measures of a single penny are:
Mass = 2.5 g
Thickness = 1.52 mm
Radius = 9.525 mm
The coin is a cylinder, and the volume of a cylinder is:
V = pi*r^2*h
where:
pi = 3.14
r = radius = 9.525mm
h = thikness = 1.52mm
The volume is:
V = 3.14*(9.525mm)^2*1.52mm = 433.015 mm^3
The density will be:
D = 2.5g/433.015mm^3 = 0.00577 g/mm^3
The answer is C. elastic potential energy
Answer:
A. 1.172 metres
B. 6.82 Ns
C. 4.796 m/s
Explanation:
The total initial momentum is gotten by multiplying the mass and initial velocity of the both bodies.
The 1.40 kg block is at rest so velocity is zero and has no momentum.
The bullet of mass 22 g = 0.022 kg with velocity of 310 m/s
Momentum = 310*0.022
Momentum = 6.82 Ns.
If the bullet gets embedded they will both have common velocity v
6.82 = (0.022+1.40)v
6.82 = 1.422v
V = 6.82/1.422
V = 4.796 m/s
How high the block will rise after the bullet is embedded is given by
H = (U²Sin²tita)/2g
Where tita is 90°
H = (4.796² * sin²(90))/(2*9.81)
H =( 23.001616*1)/19.62
H = 1.172 metres
