The equation to be used is for the rectilinear motion at constant acceleration:
x = v₀t + 0.5at²
a = (v-v₀)/t
where
x is distance
v and v₀ is the final and initial velocity
t is time
a is acceleration
Because the acceleration is decelerating, that would be reported as -7.5 m/s². Substituting,
-7.5 = (0 - v₀)/t
v₀ = 7.5 t --> eqn 1
x = v₀t + 0.5at²
60 = (7.5t)(t) + 0.5(-7.5)(t²)
Solving for t,
t = 4s
Thus,
v₀ = 7.5 m/s² * 4s
v₀ = 30 m/s
When plane is flying along the wind then we can say



Now when its going against the wind the speed is given by



Now by the above two equations we will have


Answer:
W=76.55 miles.metric tons
Explanation:
Given that
Weight on the earth = 12 tons
So weight on the moon =12/6 = 2 tons
( because at moon g will become g/6)
As we know that

Here x= 1100 miles
F 2 tons

So

We know that
Work = F. dx


![W=-2.4\times 10^6\left[\dfrac{1}{x}\right]_{1100}^{1140}](https://tex.z-dn.net/?f=W%3D-2.4%5Ctimes%2010%5E6%5Cleft%5B%5Cdfrac%7B1%7D%7Bx%7D%5Cright%5D_%7B1100%7D%5E%7B1140%7D)
![W=-2.4\times 10^6\left[\dfrac{1}{1140}-\dfrac{1}{1100}\right]](https://tex.z-dn.net/?f=W%3D-2.4%5Ctimes%2010%5E6%5Cleft%5B%5Cdfrac%7B1%7D%7B1140%7D-%5Cdfrac%7B1%7D%7B1100%7D%5Cright%5D)
W=76.55 miles.metric tons
Because one pole of the Earth's axis of rotation (the North one) points
almost exactly toward Polaris.
If Polaris had a pimple or a bump somewhere on its edge, you'd see
the bump rotate around the whole edge, like a clock, once a day. But
the whole star appears to stay in one place, because our axis points to it.
Answer:
A) The particle will accelerate in the direction of point C.
Explanation:
As we know that
potential at points A, B,C and D as V_A, V_B, V_C, V_D and it is clear from the question that
V_A>V_B>V_C
And we know that flow is always from higher to lower potential (for positive charge due to positive potential energy).
So the charge will accelerate from B toward C.
Hence, the correct option is A.