Answer:
v₀ = 280.6 m / s
Explanation:
we have the shock between the bullet and the block that we can work with at the moment and another part where the assembly (bullet + block) compresses a spring, which we can work with mechanical energy,
We write the mechanical energy when the shock has passed the bodies
Em₀ = K = ½ (m + M) v²
We write the mechanical energy when the spring is in maximum compression
½ (m + M) v² = ½ k x²
Let's calculate the system speed
v = √ [k x² / (m + M)]
v = √[152 ×0.78² / (0.012 +0.109) ]
v = 27.65 m / s
This is the speed of the bullet + Block system
Now let's use the moment to solve the shock
Before the crash
p₀ = m v₀
After the crash
The system is formed by the bullet and block assembly, so the forces during the crash are internal and the moment is preserved
m v₀ = (m + M) v
v₀ = v (m + M) / m
let's calculate
v₀ = 27.83 (0.012 +0.109) /0.012
v₀ = 280.6 m / s
N2(g)<span> + 3H</span>2(g)<span> → 2NH</span><span>3(g) Is the answer. </span>
Given parameters:
Mass of object = 6.7kg
Velocity = 8m/s
Unknown parameter:
Kinetic energy = ?
Energy is defined as the ability to do work. There are two forms of energy;
Kinetic and potential energy.
Kinetic energy is the energy due to the motion of a body. Whereas, potential energy is the energy due to the position of a body usually at rest.
Kinetic energy is mathematically expressed as;
Kinetic energy = 
where m is the mass of the body
v is the velocity of the body
Since we have been given both mass and velocity, input the parameter to solve for the unknown;
Kinetic energy = 
x 6.7 x 8² = 214.4J
So the kinetic energy of the body is 214.4J
True
In fact, the weight of an object on the surface of the Earth is given by:
where m is the mass of the object and
is the gravitational acceleration on Earth's surface. If we use the mass of the object, m=3.0 kg, we find
Answer:
P = 1 (14,045 ± 0.03 ) k gm/s
Explanation:
In this exercise we are asked about the uncertainty of the momentum of the two carriages
Δ (Pₓ / Py) =?
Let's start by finding the momentum of each vehicle
car X
Pₓ = m vₓ
Pₓ = 2.34 2.5
Pₓ = 5.85 kg m
car Y
Py = 2,561 3.2
Py = 8,195 kgm
How do we calculate the absolute uncertainty at the two moments?
ΔPₓ = m Δv + v Δm
ΔPₓ = 2.34 0.01 + 2.561 0.01
ΔPₓ = 0.05 kg m
Δ
= m Δv + v Δm
ΔP_{y} = 2,561 0.01+ 3.2 0.001
ΔP_{y} = 0.03 kg m
now we have the uncertainty of each moment
P = Pₓ /
ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²
ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²
ΔP = 0.006 + 0.0026
ΔP = 0.009 kg m
The result is
P = 14,045 ± 0.039 = (14,045 ± 0.03 ) k gm/s
