Answer:
This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap
Explanation:
We must work on this problem using the rotational equilibrium equations and then they compared the tension values that the cable supports.
Let's start with fixing a reference system on the hinge of the flag, we take as positive the anti-clockwise turn
They indicate the weight of the pole W₁ = 120 lb and a length of L = 9 ft, the weight of the man W₂ = 150, we assume that the cable is at the tip of the pole
- 
L + W₂ L + W₁ L / 2 = 0
T_{y} = W₂ + W₁ / 2
T_{y} = 120 + 150/2
T_{y} = 195 lb
we use trigonometry to find the cable tension
sin 30 = T_{y} / T
T = T_{y} / sin 30
T = 195 / sin 30
T = 390 lb
This value is less than the maximum tension of 500 lbs, making it safe for man to go to the tip flap
T < 500 lb
Answer:
V
I and II
III and IV
Explanation:
The impulse is equal to the change in momentum of the object involved, so we can calculate the change in momentum in each situation and compare them all.
Taking always east as positive direction, and labelling
u the initial velocity
v the final velocity
m = 1000 kg the mass (which is always equal)
We find:
(i)
u = 25 m/s
v = 0
(II)
u = 25 m/s
v = 0
(III)
In this case,
F = 2000 N is the force
is the time
So the magnitude of the impulse is
(IV)
F = 2000 N is the force
is the time
So the magnitude of the impulse is
(V)
u = 25 m/s
v = -25 m/s
So the ranking from largest to smallest is:
V
I and II
III and IV
Answer:
0 kg m/s before and after collision
Explanation:
Let m, v be the mass and speed of the 2 balls, respectively, before the collision. Since they have the same mass and same speed but in opposite direction, the total momentum of the system would be:
P = mv - mv = 0 kg m/s
As the collision is elastic. The total momentum after the collision is the same as the total momentum before the collision, which is 0.
he speed of the student relative to shore is
v_ up = v- vs
v _down = v+ vs
The time required to travel distance d upstream
is
t_up = d/ v_up = d/ v- vs
(2)
The time required to swim the same distance d downstream is
t_down = d/ v_down = d/ v+ vs
Answer: 10 and 35 degrees
Explanation: Localizers width below 10 degree and 35 degree signal arc is unreliable and considered unusable for navigation and as a result, aircrafts may loose alignment
