Supposing that the spring is un stretched when θ = 0, and has a toughness of k = 60 N/m.It seems that the spring has a roller support on the left end. This would make the spring force direction always to the left
Sum moments about the pivot to zero.
10.0(9.81)[(2sinθ)/2] + 50 - 60(2sinθ)[2cosθ] = 0 98.1sinθ + 50 - (120)2sinθcosθ = 0 98.1sinθ + 50 - (120)sin(2θ) = 0
by iterative answer we discover that
θ ≈ 0.465 radians
θ ≈ 26.6º
Answer:
x ≤ 3.6913 m
Explanation:
Given
Mrod = 44.0 kg
L = 4.90 m
Tmax = 1450 N
Mman = 69 kg
A: left end of the rod
B: right end of the rod
x = distance from the left end to the man
If we take torques around the left end as follows
∑τ = 0 ⇒ - Wrod*(L/2) - Wman*x + T*Sin 30º*L = 0
⇒ - (Mrod*g)*(L/2) - (Mman*g)*x + Tmax*Sin 30º*L = 0
⇒ - (44*9.8)*(4.9/2) - (69*9.8)*x + (1450)*(0.5)*(4.9) = 0
⇒ x ≤ 3.6913 m
It would be D (hoping the picture you posted is what you needed help with
Answer:
7 hours
Explanation:
The total charge contained in the battery is
And this is a charge, since it is a current multiplied by a time:
The current drawn by each headlight (which is the charge consumed per second) is
Since we have two headlight, the total current drawn is
Therefore, the total charge consumed by the two headlights each hour is
So, the time before the battery is dead is
Answer:
Im pretty sure its B im very sorry if its wrong.
