Answer:
Time, t = 8 seconds
Explanation:
An object is thrown upward from the top of a 128-foot building with an initial velocity of 112 feet per second. The height h as a function of time t is given by :

We need to find the time when the object will hit the ground. When it will hit the ground, h = 0
So,

On solving the above quadratic equation, we get the value of t = 8 seconds. So, after 8 seconds the object will hit the ground. Hence, this is the required solution.
Answer:
The distribution is as depicted in the attached figure.
Explanation:
From the given data
- The plane wall is initially with constant properties is initially at a uniform temperature, To.
- Suddenly the surface x=L is exposed to convection process such that T∞>To.
- The other surface x=0 is maintained at To
- Uniform volumetric heating q' such that the steady state temperature exceeds T∞.
Assumptions which are valid are
- There is only conduction in 1-D.
- The system bears constant properties.
- The volumetric heat generation is uniform
From the given data, the condition are as follows
<u>Initial Condition</u>
At t≤0

This indicates that initially the temperature distribution was independent of x and is indicated as a straight line.
<u>Boundary Conditions</u>
<u>At x=0</u>
<u />
<u />
This indicates that the temperature on the x=0 plane will be equal to To which will rise further due to the volumetric heat generation.
<u>At x=L</u>
<u />
<u />
This indicates that at the time t, the rate of conduction and the rate of convection will be equal at x=L.
The temperature distribution along with the schematics are given in the attached figure.
Further the heat flux is inferred from the temperature distribution using the Fourier law and is also as in the attached figure.
It is important to note that as T(x,∞)>T∞ and T∞>To thus the heat on both the boundaries will flow away from the wall.
Answer:
0.832 m/s
Explanation:
The work done by the spring W equals the kinetic energy of the object K
The work done by the spring W = 1/2k(x₀² - x₁²) where k = spring constant, x₀ = initial compression = 0.065 m and x₁ = final compression = 0.032 m
The kinetic energy of the object, K = 1/2mv² where m = mass of object and v = speed of object
Since W = K,
1/2k(x₀² - x₁²) = 1/2mv²
k(x₀² - x₁²) = mv²
mv² = k(x₀² - x₁²)
v² = [(k/m)(x₀² - x₁²)]
taking square root of both sides, we have
v = √[(k/m)(x₀² - x₁²)] since ω = angular frequency = √(k/m),
v = √[(k/m)√(x₀² - x₁²)]
v = ω√(x₀² - x₁²)]
Since ω = 14.7 rad/s, we substitute the other variables into the equation, so we have
v = 14.7 rad/s × √((0.065 m)² - (0.032 m)²)]
v = 14.7 rad/s × √(0.004225 m² - 0.001024 m²)]
v = 14.7 rad/s × √(0.003201 m²)
v = 14.7 rad/s × 0.056577
v = 0.832 m/s
Answer:
I = 3.6 kg•m²
Explanation:
Conservation of angular momentum
Let's assume CW is the positive direction
3.4(-6.1) + I(9.3) = 3.4(1.8) + I(1.8)
I(9.3 - 1.8) = 3.4(1.8 + 6.1)
I(7.5) = 3.4(7.9)
I = 3.4(7.9)/(7.5) = 3.5813333333...