Answer:
0.61 second.
Explanation:
So, from the question we are given the following useful parameters or data or information that is going to assist us in solving this question and they are;
(1). Inclination of plane, θ = 50°.
(2). ''The horizontal and a cylindrical metal is allowed to roll between two fixed height 92cm apart on the plane" that is height = 0.92 m(92cm).
(3). The "acceleration of the cylinder to be 2/3 × g Sin (theta) and g = Acceleration due to gravity = 9.8m/s^2
Soz let us delve right into the solution to the question above. We will be making use of the mathematical representation or Equation below;
s = (1/2) a × t².
Where s = height, a = acceleration due to gravity and t = time.
0.92 = (9.8 × sin 50°/ 3) × t².
t² = 0.37
t = √ 0.37.
t =0.61 seconds.
Explanation:
Pressure = force / area
P = (68 kg × 9.8 m/s²) / (2 × (0.04 m)²)
P = 208,250 Pa
Converting to psi:
P = 208,250 N/m² × (0.225 lbf/N) × (0.0254 m/in)²
P = 30.2 psi
It could be stress or strain
Answer:
375 and 450
Explanation:
The computation of the initial and the final temperature is shown below:
In condition 1:
The efficiency of a Carnot cycle is 
So, the equation is

For condition 2:
Now if the temperature is reduced by 75 degrees So, the efficiency is 
Therefore the next equation is

Now solve both the equations
solve equations (1) and (2)

T_2 + 450 = 75
T_2 = 375
Now put the T_2 value in any of the above equation
i.e
T_1 = T_2 + 75
T_1 = 375 + 75
= 450
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