Eddie the eagle, british olympic ski jumper, is attempting his most mediocre jump yet. after leaving the end of the ski ramp, he
1 answer:
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From laws of motion:
Where S is the distance/displacement (as you would call it) which is unknown
v = final velocity which is 0m/s (this is because the car stops)
u = initial velocity which is 36m/s (from the data given)
t = time taken for the distance to be covered and it is 6s
Substitute the values, hence:
But this is merely the distance he travelled in the 6 seconds he was trying to stop the car.
Therefore, the distance between the car and the cows = 160-108
Distance = 52m
Where S is the distance/displacement (as you would call it) which is unknown
v = final velocity which is 0m/s (this is because the car stops)
u = initial velocity which is 36m/s (from the data given)
t = time taken for the distance to be covered and it is 6s
Substitute the values, hence:
But this is merely the distance he travelled in the 6 seconds he was trying to stop the car.
Therefore, the distance between the car and the cows = 160-108
Distance = 52m
Answer:
same value in R and 2R E = E₀ = σ / 2ε₀
Explanation:
For this exercise we use Gauss's law
Ф = E. dA = /ε₀
We define a Gaussian surface with a cylinder with the base being parallel to the load sheet, so the electic field line and the normal line to the base are parallel and the scalar product is reduced to the algebraic product, in the parts the angle is 90º and the dot product is zero
As the sheet has two faces
2E A = q_{int} /ε₀
The charge inside the cylinder is
σ = q_{int} / A
q_{int} = σ A
We substitute
E = σ / 2ε₀
We see that this expression is independent of the distance, so it has the same value in R and 2R
E = E₀ = σ / 2ε₀