A box is sliding up an inclined plane with friction, and it is speeding up.What free-body diagrams matches this description?
1 answer:
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Refer to the diagram shown below.
We want to find y in terms of d, φ and θ.
By definition,
Therefore
y = x tan(θ) (1)
y = (x - d) tan(φ) (2)
Equate (1) and (2).
![(x - d) \, tan(\phi) = x \, tan(\theta) \\ x[tan(\phi) - tan(\theta)] = d \, tan(\phi) \\ x= \frac{d tan(\phi)}{tan(\phi)-tan(\theta)}](https://tex.z-dn.net/?f=%28x%20-%20d%29%20%5C%2C%20tan%28%5Cphi%29%20%3D%20x%20%5C%2C%20tan%28%5Ctheta%29%20%5C%5C%20x%5Btan%28%5Cphi%29%20-%20tan%28%5Ctheta%29%5D%20%3D%20d%20%5C%2C%20tan%28%5Cphi%29%20%5C%5C%20x%3D%20%5Cfrac%7Bd%20tan%28%5Cphi%29%7D%7Btan%28%5Cphi%29-tan%28%5Ctheta%29%7D%20)
From (1), obtain the required expression for y.
Answer:
We want to find y in terms of d, φ and θ.
By definition,
Therefore
y = x tan(θ) (1)
y = (x - d) tan(φ) (2)
Equate (1) and (2).
From (1), obtain the required expression for y.
Answer:
<u>Answer:</u> The energy released in the given nuclear reaction is 3.526 MeV.
<u>Explanation:</u>
For the given nuclear reaction:
We are given:
Mass of = 41.962403 u
Mass of = 41.958618 u
To calculate the mass defect, we use the equation:
Putting values in above equation, we get:
To calculate the energy released, we use the equation:
(Conversion factor:
)
Hence, the energy released in the given nuclear reaction is 3.526 MeV.