Answer:
does this even make sense
Explanation:
Answer:
v = 5.34[m/s]
Explanation:
In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the sum of the mechanical energy in the initial state plus the work on or performed by a body must be equal to the mechanical energy in the final state.
Mechanical energy is defined as the sum of energies, kinetic, potential, and elastic.
E₁ = mechanical energy at initial state [J]

In the initial state, we only have kinetic energy, potential energy is not had since the reference point is taken below 1.5[m], and the reference point is taken as potential energy equal to zero.
In the final state, you have kinetic energy and potential since the car has climbed 1.5[m] of the hill. Elastic energy is not available since there are no springs.
E₂ = mechanical energy at final state [J]

Now we can use the first statement to get the first equation:

where:
W₁₋₂ = work from the state 1 to 2.


where:
h = elevation = 1.5 [m]
g = gravity acceleration = 9.81 [m/s²]

![58 = v^{2} +29.43\\v^{2} =28.57\\v=\sqrt{28.57}\\v=5.34[m/s]](https://tex.z-dn.net/?f=58%20%3D%20v%5E%7B2%7D%20%2B29.43%5C%5Cv%5E%7B2%7D%20%3D28.57%5C%5Cv%3D%5Csqrt%7B28.57%7D%5C%5Cv%3D5.34%5Bm%2Fs%5D)
<em>meter per</em><em> </em><em>second</em><em> </em><em>is </em><em>the </em><em>main </em><em>answer </em><em>of</em><em> </em><em>both</em>
Answer:

Explanation:
Here we can use energy conservation
As per energy conservation conditions we know that work done by external source is converted into kinetic energy of the disc
Now we have

now we know that work done is product of force and displacement
so here we have


now for moment of inertia of the disc we will have



now from above equation we will have


Answer:
The rate of heat conduction through the layer of still air is 517.4 W
Explanation:
Given:
Thickness of the still air layer (L) = 1 mm
Area of the still air = 1 m
Temperature of the still air ( T) = 20°C
Thermal conductivity of still air (K) at 20°C = 25.87mW/mK
Rate of heat conduction (Q) = ?
To determine the rate of heat conduction through the still air, we apply the formula below.


Q = 517.4 W
Therefore, the rate of heat conduction through the layer of still air is 517.4 W