Answer:
H / R = 2/3
Explanation:
Let's work this problem with the concepts of energy conservation. Let's start with point P, which we work as a particle.
Initial. Lowest point
Em₀ = K = 1/2 m v²
Final. In the sought height
= U = mg h
Energy is conserved
Em₀ =
½ m v² = m g h
v² = 2 gh
Now let's work with the tire that is a cylinder with the axis of rotation in its center of mass
Initial. Lower
Em₀ = K = ½ I w²
Final. Heights sought
Emf = U = m g R
Em₀ =
½ I w² = m g R
The moment of inertial of a cylinder is
I = 
+ ½ m R²
I= ½ + ½ m R²
Linear and rotational speed are related
v = w / R
w = v / R
We replace
½ w² + ½ m R² w² = m g R
moment of inertia of the center of mass
= ½ m R²
½ ½ m R² (v²/R²) + ½ m v² = m gR
m v² ( ¼ + ½ ) = m g R
v² = 4/3 g R
As they indicate that the linear velocity of the two points is equal, we equate the two equations
2 g H = 4/3 g R
H / R = 2/3
Answer:
Frequency
Explanation:
Doesn't say how often he exercises
Answer:
is the compression in the spring
Explanation:
Given:
- mass of the bullet,
- mass of block,
- stiffness constant of the spring,
- initial velocity of the spring just before it hits the block,
<u>Now since the bullet-mass gets embed into the block, we apply the conservation of momentum as:</u>
Now this kinetic energy of the combined mass gets converted into potential energy of the spring.
is the compression in the spring
Answer:
a) F₁₂₀ = 1.34 pa A , b) F₂₀ = 0.746 pa A
Explanation:
Part. A
. The definition of pressure is
P = F / A
As the air can approach an ideal gas we can use the ideal gas equation
P V = n R T
Let's write this equation for two temperatures
P₁ V = n R T₁
P₂2 V = n R T₂
P₁ / P₂ = T₁ / T₂
point 1 has a pressure of P₁ = pa and a temperature of (20 + 273) K, point 2 is at (120 + 273) K, we calculate the pressure P₂
P₂ = P₁ T₂ / T₁
P₂ = pa 393/293
P₂ = 1.34 pa
We calculate the strength
P₂ = F₁₂₀ / A
F₁₂₀ = 1.34 pa A
Part B
In this case the data is
Point 1 has a temperature of 393K and an atmospheric pressure (P₁ = pa), point 2 has a temperature of 293K, let's calculate its pressure
P₁ / P₂ = T₁ / T₂
P₂ = P₁ T₂ / T₁
P₂ = pa 293/393
P₂ = 0.746 pa
Let's calculate the force (F20), from this point
F₂₀ / A = 0.746 pa
F₂₀ = 0.746 pa A
