Answer:
3.82 Ns
Explanation:
Time varying horizontal Force is given as
F(t) = A t⁴ + B t²
F(t) = 4.50 t⁴ + 8.75 t²
Impulse imparted is given as
I'd have to say that the list of choices doesn't go far enough.
Advances in Astronomy have been occurring for at least the past two millennia (2000 years). Maybe longer.
Answer:
μ = 0.37
Explanation:
For this exercise we must use the translational and rotational equilibrium equations.
We set our reference system at the highest point of the ladder where it touches the vertical wall. We assume that counterclockwise rotation is positive
let's write the rotational equilibrium
W₁ x/2 + W₂ x₂ - fr y = 0
where W₁ is the weight of the mass ladder m₁ = 30kg, W₂ is the weight of the man 700 N, let's use trigonometry to find the distances
cos 60 = x / L
where L is the length of the ladder
x = L cos 60
sin 60 = y / L
y = L sin60
the horizontal distance of man is
cos 60 = x2 / 7.0
x2 = 7 cos 60
we substitute
m₁ g L cos 60/2 + W₂ 7 cos 60 - fr L sin60 = 0
fr = (m1 g L cos 60/2 + W2 7 cos 60) / L sin 60
let's calculate
fr = (30 9.8 10 cos 60 2 + 700 7 cos 60) / (10 sin 60)
fr = (735 + 2450) / 8.66
fr = 367.78 N
the friction force has the expression
fr = μ N
write the translational equilibrium equation
N - W₁ -W₂ = 0
N = m₁ g + W₂
N = 30 9.8 + 700
N = 994 N
we clear the friction force from the eucacion
μ = fr / N
μ = 367.78 / 994
μ = 0.37
In a series circuit, a common current flows through all the components of the circuit. While in a parallel circuit, a different amount of current flows through each parallel branch of the circuit. Whereas in the parallel circuit, the same voltage exists across the multiple components in the circuit.
Hope It Helps!
The solution would be like
this for this specific problem:
V^2 = 2AS = 2FS/M
V = sqrt(2FS/M) =
sqrt(2*105*.75/.087) = 44.52817783 = 42.5 mps
So the speed of the arrow as it leaves the bow
is 42.5 mps.
I am hoping that this answer has
satisfied your query and it will be able to help you in your endeavor, and if
you would like, feel free to ask another question.
