Answer:
d = 4 d₀o
Explanation:
We can solve this exercise using the relationship between work and the variation of kinetic energy
W = ΔK
In that case as the car stops v_f = 0
the work is
W = -fr d
we substitute
- fr d₀ = 0 - ½ m v₀²
d₀ = ½ m v₀² / fr
now they indicate that the vehicle is coming at twice the speed
v = 2 v₀
using the same expressions we find
d = ½ m (2v₀)² / fr
d = 4 (½ m v₀² / fr)
d = 4 d₀o
The electric field strength of a point charge is inversely proportional to the square of the distance from the charge ... a lot like gravity.
If the magnitude of the field is (2E) at the distance 'd', then at the distance '2d', it'll be (2E)/(2²). That's (2E)/4 = 0.5E .
Answer:
radius comes out to be 3 m
height of the cylinder comes out to be 3m
Explanation:
given
volume of cylinder = 27π m³
π r² h = 27π
r² h = 27.............(1)
surface area of cylinder open at the top
S = 2πrh + π r²
for least amount of material requirement.
hence radius comes out to be 3 m
for height put the value in the equation 1
so, height of the cylinder comes out to be 3m
Strange as it may seem, the object would keep moving, in a straight line and at the same speed, until it came near another object. Its momentum and kinetic energy would never change. It might continue like that for a billion years or more.
Have a look at Newton's first law of motion.
