The answer is 0.981 J
E = m · g · h<span>
E - energy
m - mass
g - gravitational acceleration
h - height
We know:
E = ?
m = 0.10 kg
g = 9.81 m/s</span>²
h = 1 m
E = 0.10 kg * 9.81 m/s² * 1 m = 0.981 J
Explanation:
a) d = ½.a.t²
200 = ½(4)t²
200 = 2t²
t² = 200/2
t² = 100
t =√100 = 10 s
b) Vt = a. t
= 4(10)
= 40 m/s
c) V av. = d/t = 200/10 = 20m/s
Answer:
Explanation:
F = ma
<u>Assuming</u> the 20° is angle θ measured to the horizontal
mgsinθ - μmgcosθ = ma
g(sinθ - μcosθ) = a
at constant velocity, a = 0
g(sinθ - μcosθ) = 0
sinθ - μcosθ = 0
sinθ = μcosθ
μ = sinθ/cosθ
μ = tanθ
μ = tan20
μ = 0.3639702342...
μ = 0.36
Answer:3W
If it takes an amount of work W to move two q point charges from infinity to a distance d apart from each other, then how much work should it take to move three q point charges from infinity to a distance d apart from each other?
A) 2W
B) 3W
C) 4W
D) 6W
Explanation: calculating work done,W, in moving two positive q point charges from infinity to a valued distance d from each other is
W = k(+q)(+q)/ d
k is couloumb's constant
work done in moving 3 equal positive charges from infinity to a finite distance is given by
W₂=W₄=W₆=k(+q)(+q)/ d
Total work done, W' =k(+q)(+q)/ d + k(+q)(+q)/ d + k(+q)(+q)/ d
= W + W + W = 3W
<span>EP (potential energy) = mgy -> (59)(9.8)(-5) = -2,891
EP + EK (kinetic energy) = 0; but rearranging it for EK makes it EK = -EP, such that EK = 2891 when plugged in.
EK = 0.5mv^2, but can also be v = sqrt(2EK/m).
Plugging that in for sqrt((2 * 2891)/59), we get 9.9 m/s^2 with respect to significant figures.</span>
