If an air parcel is given a small push upward and it falls back to its original position, the atmosphere is said to be Stable.
A Stable atmospheric condition is a state where a body thrown like an parcel tends to return to its original position where it was present earlier even after being disturbed externally.
Stable conditions tends to oppose the forces that lead to the displacement of the object from original position.
When a parcel is given a small push upward and it falls back to its original position in air. This shows Stability in the atmosphere which bring back the sir parcel where it was present initially.
A body under the action of gravity in case of stable equilibrium conditions shows returning back of the body after sometime on reaching a certain height.
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Answer:
The answer is C because there is no friction there will be no friction force only applied and since its on ice you have to account for gravity
Explanation:
Answer:
4 m/s
Explanation:
Momentum is conserved.
m₁ v₁ + m₂ v₂ = (m₁ + m₂) v
(50)(5) + (20)(1.5) = (50 + 20) v
v = 4
The final velocity is 4 m/s.
Answer:
Explanation:
Use that:
![F=-\vec{\nabla}U(x,y)=-\left(\hat{i} \frac{\partial U}{\partial x}+\hat{j}\frac{\partial U}{\partial y}\right)=-11.6x\hat{i}+10.8y^{2}\hat{j}](https://tex.z-dn.net/?f=F%3D-%5Cvec%7B%5Cnabla%7DU%28x%2Cy%29%3D-%5Cleft%28%5Chat%7Bi%7D%20%5Cfrac%7B%5Cpartial%20U%7D%7B%5Cpartial%20x%7D%2B%5Chat%7Bj%7D%5Cfrac%7B%5Cpartial%20U%7D%7B%5Cpartial%20y%7D%5Cright%29%3D-11.6x%5Chat%7Bi%7D%2B10.8y%5E%7B2%7D%5Chat%7Bj%7D)
Then use the 2nd Newton's Law of Motion:
![\vec{a}=\frac{\vec{F}}{m}=\frac{-11.6x\hat{i}+10.8y^{2}\hat{j}}{0.04}=-290x\hat{i}+270y^{2}\hat{j}](https://tex.z-dn.net/?f=%5Cvec%7Ba%7D%3D%5Cfrac%7B%5Cvec%7BF%7D%7D%7Bm%7D%3D%5Cfrac%7B-11.6x%5Chat%7Bi%7D%2B10.8y%5E%7B2%7D%5Chat%7Bj%7D%7D%7B0.04%7D%3D-290x%5Chat%7Bi%7D%2B270y%5E%7B2%7D%5Chat%7Bj%7D)
At x = 0.3 and y = 0.6, we can find the acceleration as:
(in SI unit)
Then the magnitude of the acceleration on that point is:
(SI Unit)
Answer:
P = (2 + 3) * V where V is their initial speed (total momentum)
P = 2 * 10 + 3 * Vx where Vx here would be V3
If the initial momentum is not known how can one determine the final velocity of the 3 kg obj.
Also work depends on the sum of the velocities
W (initial) = 1/2 (2 + 3) V^2 the initial kinetic energy
W (final) = 1/2 * 2 * V2^2 + 1/2 * 3 * V3^2
It appears that more information is required for this problem