Answer:
So A we cant sadly do because we cant draw. B is going to be kinetic. Thats because static friction means it stays in one place, for kinetic it means moving. So it will be 0.05 as the coefficient of the friction. Sadly, I cannot calculate C. You will have to use trigonemetry but I cannot fit that big an explanation.
Answer to A: the free body diagram would be the ski things inclined with gravity, friction, and air resistance. I except you know which directions
Answer to B: Kinetic friction is the answer.
Answer to C: Find on own, I cannot write super big explanations - use trigonometry.
Answer:
y = 33.93 10⁵ m
Explanation:
This is an interference exercise, for the contributory interference is described by the expression
d sin θ = m λ
let's use trigonometry for the angle
tan θ = y / L
how the angles are small
tan θ = sin θ / cos tea = sin θ
we substitute
sin θ = y / L
d y / L = m λ
y = m λ L / d
the light fulfills the relation of the waves
c = λ f
λ = c / f
λ = 3 10⁸ /375
λ = 8 10⁵ m
first order m = 1
let's calculate
y = 1 8 10⁵ 4030 10-9 / 950 10-9
y = 33.93 10⁵ m
12N because you are just adding those two up on the same side
Answer:
176.58 m
Explanation:
t = Time taken = 6 seconds
u = Initial velocity = 0
v = Final velocity
s = Displacement
g = Acceleration due to gravity = 9.81 m/s² = a
Equation of motion
The object travels 176.58 m from the cliff in 6 seconds.
How much work in J does the string do on the boy if the boy stands still?
<span>answer: None. The equation for work is W = force x distance. Since the boy isn't moving, the distance is zero. Anything times zero is zero </span>
<span>--------------------------------------... </span>
<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m away from the kite? </span>
<span>answer: might be a trick question since his direction away from the kite and his velocity weren't noted. Perhaps he just set the string down and walked away 11m from the kite. If he did this, it is the same as the first one...no work was done by the sting on the boy. </span>
<span>If he did walk backwards with no velocity indicated, and held the string and it stayed at 30 deg the answer would be: </span>
<span>4.5N + (boys negative acceleration * mass) = total force1 </span>
<span>work = total force1 x 11 meters </span>
<span>--------------------------------------... </span>
<span>How much work does the string do on the boy if the boy walks a horizontal distance of 11m toward the kite? </span>
<span>answer: same as above only reversed: </span>
<span>4.5N - (boys negative acceleration * mass) = total force2 </span>
<span>work = total force2 x 11 meters</span>
